ch-17-mpa
Content
Ch 17 HW
Due: 11:59pm on Tuesday, September 8, 2015
To understand how points are awarded, read the Grading Policy for this assignment.
Heat Radiated by a Person
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.
Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a
human body may be considered to be that of the sides of a cylinder of length L = 2.0 m and circumference
C = 0.8 m.
For the StefanBoltzmann constant use σ
= 5.67 × 10
−8
W/m
2
/K
4
.
Part A
If the surface temperature of the skin is taken to be T body
described in the introduction radiate?
Take the emissivity to be e
∘
, how much thermal power P rb does the body
= 30 C
.
= 0.6
Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.
Hint 1. Area of person
Find the area A of the person, ignoring the ends of the cylinder that represents the person's area.
Express the area in terms of L, C , and any constants.
ANSWER:
A
=
LC
Hint 2. Symbolic expression
Find P rb , the total power radiated into the room by the body of the person.
Express the total power radiated in terms of the StefanBoltzmann constant σ, the body area A , the
body temperature T body , and the emissivity e .
ANSWER:
P rb
=
eσAT body
ANSWER:
P rb
= 460 W
Correct
Typesetting math: 38%
4
Part B
The basal metabolism of a human adult is the total rate of energy production when a person is not performing
significant physical activity. It has a value around 125 W, most of which is lost by heat conduction to the surrounding
air and especially to the exhaled air that was warmed while inside the lungs. Given this energy production rate, it
would seem impossible for a human body to radiate 460 W as you calculated in the previous part.
Which of the following alternatives seems to best explain this conundrum?
ANSWER:
The human body is quite reflective in the infrared part of the spectrum (where it radiates) so e is in fact
less than 0.1.
The surrounding room is near the temperature of the body and radiates nearly the same power into the
body.
Correct
The human body contains significant amounts of water and organic compounds. These typically have many
absorption bands in the infrared part of the spectrum where room temperature objects radiate thermal energy.
Consequently, human skin has a high emissivity in this part of the spectrum regardless of the emissivity of the
skin in the visible part of the spectrum.
Part C
Now calculate P br , the thermal power absorbed by the person from the thermal radiation field in the room, which is
assumed to be at T room = 20∘ C. If you do not understand the role played by the emissivities of room and person,
be sure to open the hint on that topic.
Express the thermal power numerically, giving your answer to the nearest 10 W.
Hint 1. Role of emissivities
The emissivity e is the complement of the reflectivity. The power incident on the person is partly absorbed
and partly reflected. The fraction absorbed is proportional to the emissivity of the person and thus is less
than or equal to the thermal power striking the person by the amount of this power that is reflected.
Because the room is closed, no allowance has to be made for its emissivity. If the room has an emissivity of
0.7, then when you look at the wall (with an infrared viewer) 0.7 of what you see is the power radiated by that
spot on the wall; the remaining 0.3 of what you see is what is reflected by the wall. But in a closed container
like a room, this reflection is simply another part of the wall (since the room is closed). Being at the same
temperature, this other part of the wall radiates power at the same rate as the spot you are looking at. When
all the emissions and reflections are added together, the net effect is that the radiation from any part of the
wall has the same total thermal power as if it were perfectly black. This accounts for the remarkable fact that
the thermal radiation field in any closed cavity is independent of the material of the cavity and depends only
on the temperature.
Such radiation is called cavity radiation or blackbody radiation. The fact that the intensity at each color
depended only on the temperature was so striking to physicists that the explanation of this phenomenom
was assiduously studied until it unlocked the key to quantum mechanics and revealed a fundamental
quantum of nature: Planck's constant h .
A good example of this is a kiln. Just after a part is placed in a kiln, the heaters are much hotter than the cool
part, and the part is visible through the peephole. As thermal equilibrium is reached everything glows a
uniform red or orange, and the part disappears into the background color, becomming invisible.
Hint 2. Area of person
Find the area A of the person.
Express the area in terms of L, C , and any constants.
ANSWER:
A
=
LC
Hint 3. Symbolic expression
Find P br , the total power absorbed by the person.
Express your answer in terms of the StefanBoltzmann constant σ, the body area A , the room
temperature T room , and the emissivity e .
ANSWER:
P br
=
eσAT room
4
ANSWER:
P br
= 400 W
Correct
Part D
Find P net , the net power radiated by the person when in a room with temperature T room
Express the net radiated power numerically, to the nearest 10 W.
Hint 1. How to set up the problem
The net power radiated is simply the power radiated minus the power absorbed.
ANSWER:
P net
= 60 W
∘
.
= 20 C
Correct
This net thermal power radiated is less than half of the basal metabolic rate. Under these circumstances a
person would feel cool, but not cold (if we neglect loss of heat by conduction). However, if the surrounding
temperature is much cooler than this, the net radiative loss will become a significant loss mechanism for body
heat. This explains why lightweight survival blankets have a shiny side: The low emissivity dramatically
reduces radiative heat loss (and the air trapped between blanket and body reduces heat loss by conduction).
Exercise 17.6
Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales.
Part A
The midday temperature at the surface of the moon (400 K ).
ANSWER:
T
= 127 ∘ C
Correct
Part B
The midday temperature at the surface of the moon (400 K ).
ANSWER:
T
= 260 ∘ F
Correct
Part C
The temperature at the tops of the clouds in the atmosphere of Saturn (95 K ).
ANSWER:
T
= 178 ∘ C
Correct
Part D
K
The temperature at the tops of the clouds in the atmosphere of Saturn (95 K ).
ANSWER:
T
= 289 ∘ F
Correct
Part E
The temperature at the center of the sun (1.55 × 107 K) .
ANSWER:
T
= 1.55×107 ∘ C
Correct
Part F
The temperature at the center of the sun (1.55 × 107 K) .
ANSWER:
T
= 2.79×107 ∘ F
Correct
Rankine Temperature Scale
Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0
∘
∘
R). However, the units of this scale are the same size as those on the Fahrenheit scale ( F ) rather than the Celsius
∘
scale ( C).
Part A
Given that water at standard pressure freezes at 0∘ C, which corresponds to 32∘ F , and that it boils at 100∘ C,
which corresponds to 212∘ F , calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a
temperature difference of 1 K on the Kelvin scale.
Give your answer to two significant figures.
Hint 1. Relation of Celsius and Kelvin temperature scales
A temperature increase of one kelvin corresponds to a temperature increase of one degree also on the
Celsius scale. The Kelvin temperature scale and the Celsius temperature scale differ only in their zero point.
ANSWER:
= 1.8 ∘ F
ΔT = 1 K
Correct
Part B
What is the numerical value of the triplepoint temperature T triple of water on the Rankine scale?
Give your answer to three significant figures.
Hint 1. Triplepoint temperature
On the Kelvin temperature scale, water freezes and coexists in three phases (solid, liquid, and vapor) at
273.16 K at standard pressure. This temperature is known as the triple point.
ANSWER:
T triple
= 492 ∘ R
Correct
Exercise 17.18
A steel tank is completely filled with 2.50 m 3 of ethanol when both the tank and the ethanol are at a temperature of 35.0
∘
C .
Part A
When the tank and its contents have cooled to 16.0 ∘ C , what additional volume of ethanol can be put into the
tank?
Express your answer using two significant figures.
ANSWER:
ΔV
= 3.4×10−2 m 3
Correct
Hot Rods
Two circular rods, both of length L and having the same diameter, are placed end to end between rigid supports with no
initial stress in the rods.
The coefficient of linear expansion and Young's modulus for rod A are α A and Y A respectively; those for rod B are α B
and Y B respectively. Both rods are "normal" materials with α > 0.
The temperature of the rods is now raised by ΔT .
Part A
After the rods have been heated, which of the following statements is true?
Choose the best answer.
ANSWER:
The length of each rod is still L.
The length of each rod changes but the combined length of the rods is still 2L.
Correct
The length of the combined rod remains the same, but because the rods have different expansion coefficients,
the lengths of the individual rods change. In other words, ΔLA + ΔLB = 0 even though ΔLA ≠ 0 and
ΔLB ≠ 0 .
Part B
After the rods have been heated, which of the following statements is true?
Choose the best answer.
ANSWER:
The stress in each rod remains zero.
A compressive stress arises that is the same for both rods.
A compressive stress arises that is different for the two rods.
A tensile stress arises that is the same for both rods.
A tensile stress arises that is different for the two rods.
A compressive stress arises in one rod and a tensile stress arises in the other rod.
Correct
Stress is a force per unit area. By Newton's 3rd law, the force on rod A due to rod B is the same as that on rod
B due to rod A. Since the rods have the same diameter, their crosssectional area is the same. Therefore, the
stress on each rod must be the same.
Part C
What is the stress F /A in the rods after heating?
Y
Y
ΔT
Express the stress in terms of α A , α B , Y A , Y B , and ΔT .
Hint 1. How to approach the problem
You know that the length of the two rods remains equal to 2L during the heating process. Find the change in
length of each rod as a function of the stress on the rods, then set the total change in length to zero and
solve for the stress.
Hint 2. Change in length for each rod
The change in length for each rod is due to both thermal as well as compressive stresses. The
corresponding formulas are
ΔLthermal = αLΔT
and
ΔL compressive
stress
The sum gives the net change in length for each rod.
=L
F
AY
.
ANSWER:
F
A
=
−(α A +α B )ΔT
1
Y
A
+
1
Y
B
Correct
Another way of thinking about this is that the combination of rods has a net thermal expansion coefficient
and a net Young's modulus given by
α = α1 + α2
Y =
Y1Y2
Y 1 +Y 2
.
Video Tutor: Heating Water and Aluminum
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an
experiment. Then, close the video window and answer the question on the right. You can watch the video again at any
point.
Part A
Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the
video, the mass of the metal is the same as that of the water. Room temperature is about 20∘ C before the start of
the experiment. The water heats up to 40∘ C, and the mystery metal heats up to 80∘ C. Compared to that of water,
the heat capacity of our mystery metal is
Hint 1. How to approach the problem
Recall that the heat Q delivered to a substance can be written Q = mc ΔT , where m is the mass of the
substance and c is its heat capacity.
ANSWER:
half as great.
three times greater.
onethird as great.
two times greater.
the same.
Correct
Given the same input of energy, the temperature of the metal increased three times as much as the
temperature of the water. Therefore, the metal has onethird the heat capacity of water. (Recall that the heat Q
delivered to a substance can be written Q = mc ΔT , where m is the mass of the substance and c is its heat
capacity.)
Exercise 17.27
An aluminum tea kettle with mass 1.55 kg and containing 1.70 kg of water is placed on a stove.
Part A
If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 21.0 ∘ C to 80.0
∘
C
∘
C
?
ANSWER:
Q
= 5.03×105 J
Answer Requested
Thermal Energy from Friction on a Rope
A capstan is a rotating drum or cylinder over which a rope or cord slides to provide a great amplification of the rope's
tension while keeping both ends free . Since the added
tension in the rope is due to friction, the capstan generates
thermal energy.
Part A
If the difference in tension (T 0 − T ′ ) between the two ends of the rope is 520 N and the capstan has a diameter
of 10.0 cm and turns once in 0.90 \;{\rm s}, find the rate \texttip{P_{\rm thermal}}{P_thermal} at which thermal
energy is being generated.
Give a numerical answer, in watts, rounded to the nearest 10 W.
Hint 1. Torque applied to the rope
In this case, the contact force between the capstan and the rope is not localized at a single point but is
distributed along the rope. (This is more realistic than assuming that forces act at a single point.) However,
we can still say that the net torque is the product of the difference between the tensions (T_0T') and the
radius. (Finding a numerical value for this torque can certainly be done as an intermediate calculation, but it
doesn't help much in reaching the answer.)
Hint 2. Power due to the torque
If you've found the torque, either symbolically or numerically, the power is given by P=\tau\omega, where
\omega=2\pi/t_0, with \texttip{t_{\rm 0}}{t_0} as the time for one revolution of the capstan.
Hint 3. Power from force
We should be able to find the power as the product of a force and a relative speed, and of course we can.
The net force on the rope is T_0T' and the relative speed is the circumference of the capstan divided by the
period of the capstan's rotation.
Hint 4. How both methods work equally well
If you've done this problem symbolically using torques, you've found that
P_{\rm thermal}=\tau\omega=(Fr)\left({2\pi\over t_0}\right)={2\pi r\left(T_0T'\right)\over t_0}\;.
If you've done this problem symbolically using forces, you've found that
P_{\rm thermal}= Fv=\left(T_0T'\right){2\pi r\over t_0}.
The fact that these expressions are the same should not be surprising.
ANSWER:
\texttip{P_{\rm thermal}}{P_thermal} = 180 W
Correct
The answer does not depend on the number of turns, only on the difference in tensions! A larger number of
turns might well create a larger tension difference, but once this difference has been achieved, the work done
per rotation of the capstan will not depend on the number of turns.
Part B
If the capstan is made of iron (with a specific heat capacity C_{\rm iron}=470\;{\rm J/(kg}\cdot ^\circ{\rm C}) and has
a mass of 6.00 \;{\rm kg}, at what rate d\Theta/ dt does its temperature rise? Assume that the temperature in the
capstan is uniform and that all the thermal energy generated flows into it.
Note that \texttip{\Theta }{Theta} is a temperature.
Give a numerical answer, in degrees Celsius per second, rounded to two significant figures.
Hint 1. Use the chain rule
In Part A, you found the rate at which heat is added to the capstan. This added heat will of course cause a
temperature rise, and this rate of temperature rise will be proportional to \texttip{P_{\rm thermal}}{P_thermal},
the proportionality being the reciprocal of the heat capacity. Symbolically,
P_{\rm thermal}={dQ\over dt}={dQ\over d\Theta}{d\Theta\over dt}=(mC){d\Theta\over dt},
where \texttip{C}{C} is the specific heat capacity, so that
{d\Theta\over dt}={P_{\rm thermal}\over mC}.
ANSWER:
{d\Theta\over dt} = 0.064 ^\circ {\rm C/s}
Correct
Video Tutor: Water Balloon Held over Candle Flame
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an
experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.
Part A
Which of the following changes would make the water balloon more likely to pop? (Ignore effects of convection
within the fluid.)
Select all that apply.
Hint 1. How to approach the problem.
Recall that in the video, the water balloon did not pop because most of the candle's thermal energy went into
heating the water. Let's think about the process in more detail.
The high heat capacity of water means that the water in the balloon warmed only slowly. (Recall that the
higher a substance's heat capacity, the less the substance's temperature will rise in response to a given
input of heat.) Contact with this relatively cool water kept the balloon's thin rubber skin from getting hot
enough to melt.
Rubber conducts heat more slowly than water. If you make the rubber skin of a water balloon thicker, will the
skin be more likely or less likely to melt? Think about the conduction of heat across the skin between the hot
side facing the candle and the cooler side in contact with water.
ANSWER:
Use a liquid that has a lower heat capacity than water.
Use a thinner balloon.
Use a thicker balloon.
Use a liquid that has a higher heat capacity than water.
Correct
Using a liquid with a lower heat capacity than water means that the candle flame will raise the liquid's
temperature more quickly than for water. A thicker balloon will conduct heat through to the liquid more slowly,
so that the outer part of the balloon wall will heat up more quickly to its melting point.
Exercise 17.41
A copper pot with a mass of 0.500 {\rm {\rm kg}} contains 0.170 {\rm {\rm kg}} of water, and both are at a temperature of
20.0 {\rm ^\circ C} . A 0.250 {\rm {\rm kg}} block of iron at 85.0 {\rm ^\circ C} is dropped into the pot.
Part A
Find the final temperature of the system, assuming no heat loss to the surroundings.
ANSWER:
T = 27.5 {\rm ^\circ C}
Correct
Steam vs. HotWater Burns
Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat
input to your skin from steam as opposed to hot water at the same temperature.
Assume that water and steam, initially at 100^\circ{\rm C}, are cooled down to skin temperature, 34^\circ{\rm C}, when
they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a
constant specific heat capacity c = 4190\; {\rm J/(kg \cdot K)} for both liquid water and steam.
Part A
Under these conditions, which of the following statements is true?
ANSWER:
Steam burns the skin worse than hot water because the thermal conductivity of steam is much higher than
that of liquid water.
Steam burns the skin worse than hot water because the latent heat of vaporization is released as well.
Hot water burns the skin worse than steam because the thermal conductivity of hot water is much higher
than that of steam.
Hot water and steam both burn skin about equally badly.
Correct
The key point is that the latent heat of vaporization has to be taken into account for the steam.
Part B
How much heat \texttip{H_{\rm 1}}{H_1} is transferred to the skin by 25.0 {\rm g} of steam onto the skin? The latent
heat of vaporization for steam is L = 2.256 \times 10^6 \; {\rm J/kg}.
Express the heat transferred, in joules, to three significant figures.
Hint 1. Determine the heat transferred from steam to skin
Find an expression for the heat \texttip{H_{\rm 1}}{H_1} transferred from the vapor to the skin.
Express your answer in terms of \texttip{c}{c}, the mass of steam \texttip{m}{m}, the latent heat of
vaporization \texttip{L}{L}, and the temperature difference \texttip{\Delta T}{DeltaT} between the
initial temperature of the steam and the skin temperature.
ANSWER:
\texttip{H_{\rm 1}}{H_1} = m \left(c {\Delta{T}}+L\right)
ANSWER:
\texttip{H_{\rm 1}}{H_1} = 6.33×104 {\rm J}
Correct
Here we assumed that the skin continues to remain at 34^{\circ}{\rm C}. Actually the local temperature in the
area where the steam condenses can be raised quite significantly.
Part C
How much heat \texttip{H_{\rm 2}}{H_2} is transferred by 25.0 {\rm g} of water onto the skin? To compare this to the
result in the previous part, continue to assume that the skin temperature does not change.
Express the heat transferred, in joules, to three significant figures.
Hint 1. Determine the heat transferred from water to skin
Find an expression for the heat \texttip{H_{\rm 2}}{H_2} transferred from the hot water to the skin.
Express your answer in terms of \texttip{c}{c}, the mass of water \texttip{m}{m}, and the temperature
difference \texttip{\Delta T}{DeltaT} between the initial temperature of the water and the skin
temperature.
ANSWER:
\texttip{H_{\rm 2}}{H_2} = m c {\Delta{T}}
ANSWER:
\texttip{H_{\rm 2}}{H_2} = 6910 {\rm J}
Correct
The amount of heat transferred to your skin is almost 10 times greater when you are burned by steam versus
hot water. The temperature of steam can also potentially be much greater than 100^\circ{\rm C}. For these
reasons, steam burns are often far more severe than hotwater burns.
An Electric Water Heater
An engineer is developing an electric water heater to provide a continuous ("on demand") supply of hot water. One trial
design is shown in the figure. Water is flowing at the rate
\texttip{F}{F}, the inlet thermometer registers \texttip{T_{\rm
1}}{T_1}, the voltmeter reads \texttip{V}{V}, and the ammeter
reads current \texttip{I}{I}. Then the power (i.e., the heat
generated per unit time by the heating element) is VI.
Assume that the heat capacity of water is \texttip{C}{C} and
that the heat capacity of the heater apparatus is
\texttip{C_{\rm h}}{C_h}.
Part A
When a steady state is finally reached, what is the temperature reading \texttip{T_{\rm 2}}{T_2} of the outlet
thermometer?
Express the outlet temperature in terms of \texttip{T_{\rm 1}}{T_1}, \texttip{F}{F}, \texttip{C}{C}, and any
other given quantities.
Hint 1. How to approach the problem
Determine the amount of heat input per unit mass of water that flows through the heater. Then use the
definition of heat capacity to compute how much the temperature of a unit mass of water will increase for a
given quantity of heat. Combine these two calculations to determine the increase in temperature of the water
as it flows through the heater.
Hint 2. Find heat input per unit mass of water
How much heat \texttip{Q}{Q} is delivered by the resistor per unit mass \texttip{m}{m} of water flowing
through the heater?
Express the heat per unit mass in terms of quanitities given in the problem introduction.
Hint 1. What relationships are involved
The power \texttip{P}{P} dissipated in the resistor is equal to the energy dissipated per unit time. The
flow rate \texttip{F}{F} is equal to the mass of the water that flows through the heater per unit time.
Thus, if all of the energy dissipated by the resistor goes into heating the water, then
\large{\frac{P}{F}=\frac{\frac{\mbox{heat}}{\mbox{time}}}{\frac{\mbox{mass}}
{\mbox{time}}}=\frac{\mbox{heat}}{\mbox{mass}}}.
Hint 2. Electrical power
The electrical power disspated in a device is given by the product of the voltage and the current:
P = V I.
ANSWER:
Q/m = \large{\frac{V I}{F}}
Hint 3. Find the change in water temperature
If an amount of heat \texttip{Q}{Q} is added uniformly to a substance of mass \texttip{m}{m} and heat
capacity \texttip{C}{C}, by how much will the temperature of that substance increase?
Express the temperature increase \texttip{\Delta T}{DeltaT} in terms of \texttip{Q}{Q}, \texttip{m}{m},
and \texttip{C}{C}.
ANSWER:
\texttip{\Delta T}{DeltaT} = \large{\frac{Q}{m} \frac{1}{C}}
ANSWER:
\texttip{T_{\rm 2}}{T_2} = \large{\frac{VI}{FC}+T_{1}}
Correct
Part B
Why is it unnecessary to take into account the heat capacity of the apparatus itself?
Choose the best explanation.
ANSWER:
In steady state the temperature of the apparatus doesn't change. Hence its heat capacity is irrelevant.
The heat capacity of any of the materials used in a water heater is much smaller than that of water.
Since the actual heating unit is immersed in the flow all the heat goes directly into the water.
Correct
Imagine that the input temperature of the water is T_1 = 18^\circ {\rm C}, the ammeter reads I = 15.0 \;{\rm A}, the
voltmeter reads V = 120\; {\rm V}, and the flow rate is F= 0.500\; {\rm kg/min}. The heat capacity of water C = 4200\; {\rm
J/(kg \cdot K)}.
Part C
What is the power \texttip{P}{P} at which the heater operates?
Express your answer numerically, in watts.
Hint 1. Electrical power
The electrical power disspated in a device is given by the product of the voltage and the current:
P = V I.
ANSWER:
\texttip{P}{P} = 1800 W
Correct
This is a large, but reasonable, power requirement for a household appliance.
Part D
Calculate the temperature \texttip{T_{\rm 2}}{T_2} of the water leaving the heater. Note that \texttip{F}{F} is defined
as the number of kilograms of water flowing through the heater per minute, whereas the power is measured in watts
(joules/second).
Express your answer numerically, in degrees Celsius, to the nearest integer.
ANSWER:
\texttip{T_{\rm 2}}{T_2} = 69 ^\circC
Correct
Part E
Consider using this heater to generate "on demand" hot water for a home shower. The input water temperature
(during the winter) can be as low as T_1=5.00^\circ {\rm C} and the output temperature should be at least
T_2=40.0^\circ {\rm C} for a moderately warm shower. Assume a conservative flow rate of F=8.00\; {\rm kg/min}
(corresponding to about 2.5 gallons per minute).
If the heater is to operate on a US standard V = 120 \;{\rm V} wall plug, how much current \texttip{I}{I} would it need
to draw to meet the design requirements?
Express the current numerically, in amperes, to the nearest integer.
ANSWER:
\texttip{I}{I} = 163 A
Correct
163 A is a tremendous amount of current for a home appliance. Most homes in the United States are wired for
a maximum of 200 A for the entire house, and only 20 A on any particular outlet. For the electric heater to meet
the modest design requirements given, it would consume almost 20 kW of power!
For this reason, most "on demand" water heaters generate heat by burning a fuel such as propane, rather than
using electricity.
Exercise 17.71
A picture window has dimensions of 1.40 {\rm m} \times2.50 {\rm m} and is made of glass 5.60 {\rm {\rm mm}} thick. On
a winter day, the outside temperature is 17.0 {\rm {\rm ^\circ C}} , while the inside temperature is a comfortable 20.5
{\rm {\rm ^\circ C}} .
Part A
At what rate is heat being lost through the window by conduction?
Express your answer using three significant figures.
ANSWER:
H = 1.88×104 {\rm W}
Correct
Part B
At what rate would heat be lost through the window if you covered it with a 0.750 {\rm mm}thick layer of paper
(thermal conductivity 0.0500 {\rm W/m \cdot K})?
Express your answer using three significant figures.
ANSWER:
H = 5970 {\rm W}
Correct
Heat Flow through Three Rods
Three identical rods are welded together to form a Yshaped figure. The crosssectional area of each rod is \texttip{A}
{A}, and they have length \texttip{L}{L} and thermal conductivity \texttip{k}{k}.
The free end of rod 1 is maintained at \texttip{T_{\rm 1}}{T_1}
and the free ends of rods 2 and 3 are maintained at a lower
temperature \texttip{T_{\rm 0}}{T_0}. You may assume that
there is no heat loss from the surfaces of the rods.
Part A
What is \texttip{T_{\rm j}}{T_j}, the temperature of the junction point?
Express your answer in terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}.
Hint 1. Find a relationship among the heat currents
Heat is flowing from the free end of rod 1 to the free ends of rods 2 and 3, breaking into two flows at the
junction point. What is the relationship between the heat current in rod 1 ( \texttip{H_{\rm 1}}{H_1}) to the
heat currents in rods 2 and 3 ( \texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3} respectively)?
Express \texttip{H_{\rm 1}}{H_1} in terms of \texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3}.
Hint 1. Use an analogy to electric currents
If you are familiar with electrical current, you can solve this problem by using the analog of Kirchoff's
junction rule, which states that the algebraic sum of the currents into any junction is zero (any current
leaving the junction must be subtracted from the total). Similarly, in steady state, the algebraic sum of
all heat currents into a junction is zero (any heat current leaving the junction must be subtracted from
the total).
ANSWER:
\texttip{H_{\rm 1}}{H_1} = H_{2} + H_{3}
Hint 2. Find \texttip{H_{\rm 1}}{H_1} in terms of the junction temperature
Now, if you can find expressions for the heat currents in the rods in terms of their physical properties and
temperature difference across them, you can substitute these into the formula H_1=H_2+H_3 and solve
algebraically for \texttip{T_{\rm j}}{T_j}.
Derive a general expression for the heat current through rod 1 in terms of the temperature of the junction,
\texttip{T_{\rm j}}{T_j}, and other quantities given in the problem introduction.
ANSWER:
\texttip{H_{\rm 1}}{H_1} = \large{\frac{k A \left(T_{1}T_{j}\right)}{L}}
ANSWER:
\texttip{T_{\rm j}}{T_j} = \large{\frac{T_{1}+2T_{0}}{3}}
Correct
This is a weighted average of the temperatures at the two ends of the system and so always lies between them
as you would expect. Also, it is closer to \texttip{T_{\rm 0}}{T_0} than to \texttip{T_{\rm 1}}{T_1}. This is
because the heat current in rods 2 and 3 separately must be less than the heat current in rod 1 (since H_2 +
H_3 = H_1).
Part B
What is the heat current \texttip{H_{\rm 1}}{H_1} in rod 1?
Express the heat current in terms of any or all of \texttip{k}{k}, \texttip{L}{L}, \texttip{A}{A}, and the
temperatures \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}.
Hint 1. Determine the general formula for heat current
Given the difference in temperature between the ends of the rod ( \texttip{\Delta T}{DeltaT}), the rod's
thermal conductivity \texttip{k}{k}, and its length \texttip{L}{L} and crosssectional area \texttip{A}{A}, what is
the general formula for the heat current \texttip{H}{H} in the rod?
Express your answer in terms of \texttip{k}{k}, \texttip{L}{L}, \texttip{A}{A}, and \texttip{\Delta T}
{DeltaT} (typed as DeltaT).
ANSWER:
\texttip{H}{H} = \large{\frac{k A {\Delta{T}}}{L}}
Hint 2. Find the temperature difference
What is the temperature difference \texttip{\Delta T}{DeltaT} between the ends of rod 1?
Express your answer in terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}. Your answer
should be positive.
Hint 1. Use your answer to Part A
The temperature difference between the ends of rod 1 is \Delta T = T_1 T_{\rm j}. To express this in
terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}, use the relationship you found in Part A
of this problem.
ANSWER:
\texttip{\Delta T}{DeltaT} = \large{\frac{2}{3} \left(T_{1}T_{0}\right)}
ANSWER:
\texttip{H_{\rm 1}}{H_1} = \large{{\frac{2}{3}}{\frac{kA}{L}}\left(T_{1}T_{0}\right)}
Correct
This is greater than the heat current due to only 2 rods. You would expect this to be the case since the
effective surface area is increased by having a third rod. You can check that if you wanted to remove the third
rod while keeping the same heat current, you must increase the crosssectional area of the remaining two rods
by a factor of 1.33.
Part C
By symmetry, the heat current in rods 2 and 3 will be the same. Find this heat current.
Express your answer in terms of \texttip{k}{k}, \texttip{L}{L}, \texttip{A}{A}, and the temperatures
\texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}.
Hint 1. How to approach this problem
Recall that H_1 = H_2 + H_3. Also, by symmetry, H_2 = H_3. This information is sufficient to find
\texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3}.
ANSWER:
H_2 = H_3 = \large{kA{\frac{T_{1}T_{0}}{3L}}}
Correct
Problem 17.92
You cool a 110.0 {\rm g} slug of redhot iron (temperature 745 {\rm ^\circ C}) by dropping it into an insulated cup of
negligible mass containing 75.0 {\rm g} of water at 20.0 {\rm ^\circ C}. Assume no heat exchange with the surroundings.
Part A
What is the final temperature of the water?
Express your answer using three significant figures.
ANSWER:
T_{\rm final} = 100 {\rm ^\circ C}
Correct
Part B
What is the final mass of the iron and the remaining water?
Express your answer using three significant figures.
ANSWER:
m_{\rm final} = 181 {\rm g}
Correct
Problem 17.104
Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200 {\rm
W/(m \cdot K)} ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus)
as a sphere 1.60 {\rm {\rm m}} in diameter having a layer of fat 4.00 {\rm {\rm cm}} thick. (Actually, the thickness varies
with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was
found that the outer surface layer of the fur is at 2.90 {\rm ^\circ C} during hibernation, while the inner surface of the fat
layer is at 30.9 {\rm ^\circ C}.
Assume the surface area of each layer is constant and given by the surface area of the spherical model constructed for
the black bear.
Part A
What should the temperature at the fatinner fur boundary be so that the bear loses heat at a rate of 51.5 {\rm {\rm
W}} ?
ANSWER:
T = 29.6 {\rm ^\circ C}
Correct
Should you be required to use your answer to this part in a subsequent part, use your full precision answer,
only rounding as a final step before submitting your answer.
Part B
How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51.5 {\rm {\rm W}}
?
ANSWER:
L_{\rm air} = 10.0 {\rm cm}
Correct
Score Summary:
Your score on this assignment is 101%.
You received 15.22 out of a possible total of 16 points.
Your instructor adjusted your score by awarding 1 additional point. This adjustment is reflected in your score above.
Due: 11:59pm on Tuesday, September 8, 2015
To understand how points are awarded, read the Grading Policy for this assignment.
Heat Radiated by a Person
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.
Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a
human body may be considered to be that of the sides of a cylinder of length L = 2.0 m and circumference
C = 0.8 m.
For the StefanBoltzmann constant use σ
= 5.67 × 10
−8
W/m
2
/K
4
.
Part A
If the surface temperature of the skin is taken to be T body
described in the introduction radiate?
Take the emissivity to be e
∘
, how much thermal power P rb does the body
= 30 C
.
= 0.6
Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.
Hint 1. Area of person
Find the area A of the person, ignoring the ends of the cylinder that represents the person's area.
Express the area in terms of L, C , and any constants.
ANSWER:
A
=
LC
Hint 2. Symbolic expression
Find P rb , the total power radiated into the room by the body of the person.
Express the total power radiated in terms of the StefanBoltzmann constant σ, the body area A , the
body temperature T body , and the emissivity e .
ANSWER:
P rb
=
eσAT body
ANSWER:
P rb
= 460 W
Correct
Typesetting math: 38%
4
Part B
The basal metabolism of a human adult is the total rate of energy production when a person is not performing
significant physical activity. It has a value around 125 W, most of which is lost by heat conduction to the surrounding
air and especially to the exhaled air that was warmed while inside the lungs. Given this energy production rate, it
would seem impossible for a human body to radiate 460 W as you calculated in the previous part.
Which of the following alternatives seems to best explain this conundrum?
ANSWER:
The human body is quite reflective in the infrared part of the spectrum (where it radiates) so e is in fact
less than 0.1.
The surrounding room is near the temperature of the body and radiates nearly the same power into the
body.
Correct
The human body contains significant amounts of water and organic compounds. These typically have many
absorption bands in the infrared part of the spectrum where room temperature objects radiate thermal energy.
Consequently, human skin has a high emissivity in this part of the spectrum regardless of the emissivity of the
skin in the visible part of the spectrum.
Part C
Now calculate P br , the thermal power absorbed by the person from the thermal radiation field in the room, which is
assumed to be at T room = 20∘ C. If you do not understand the role played by the emissivities of room and person,
be sure to open the hint on that topic.
Express the thermal power numerically, giving your answer to the nearest 10 W.
Hint 1. Role of emissivities
The emissivity e is the complement of the reflectivity. The power incident on the person is partly absorbed
and partly reflected. The fraction absorbed is proportional to the emissivity of the person and thus is less
than or equal to the thermal power striking the person by the amount of this power that is reflected.
Because the room is closed, no allowance has to be made for its emissivity. If the room has an emissivity of
0.7, then when you look at the wall (with an infrared viewer) 0.7 of what you see is the power radiated by that
spot on the wall; the remaining 0.3 of what you see is what is reflected by the wall. But in a closed container
like a room, this reflection is simply another part of the wall (since the room is closed). Being at the same
temperature, this other part of the wall radiates power at the same rate as the spot you are looking at. When
all the emissions and reflections are added together, the net effect is that the radiation from any part of the
wall has the same total thermal power as if it were perfectly black. This accounts for the remarkable fact that
the thermal radiation field in any closed cavity is independent of the material of the cavity and depends only
on the temperature.
Such radiation is called cavity radiation or blackbody radiation. The fact that the intensity at each color
depended only on the temperature was so striking to physicists that the explanation of this phenomenom
was assiduously studied until it unlocked the key to quantum mechanics and revealed a fundamental
quantum of nature: Planck's constant h .
A good example of this is a kiln. Just after a part is placed in a kiln, the heaters are much hotter than the cool
part, and the part is visible through the peephole. As thermal equilibrium is reached everything glows a
uniform red or orange, and the part disappears into the background color, becomming invisible.
Hint 2. Area of person
Find the area A of the person.
Express the area in terms of L, C , and any constants.
ANSWER:
A
=
LC
Hint 3. Symbolic expression
Find P br , the total power absorbed by the person.
Express your answer in terms of the StefanBoltzmann constant σ, the body area A , the room
temperature T room , and the emissivity e .
ANSWER:
P br
=
eσAT room
4
ANSWER:
P br
= 400 W
Correct
Part D
Find P net , the net power radiated by the person when in a room with temperature T room
Express the net radiated power numerically, to the nearest 10 W.
Hint 1. How to set up the problem
The net power radiated is simply the power radiated minus the power absorbed.
ANSWER:
P net
= 60 W
∘
.
= 20 C
Correct
This net thermal power radiated is less than half of the basal metabolic rate. Under these circumstances a
person would feel cool, but not cold (if we neglect loss of heat by conduction). However, if the surrounding
temperature is much cooler than this, the net radiative loss will become a significant loss mechanism for body
heat. This explains why lightweight survival blankets have a shiny side: The low emissivity dramatically
reduces radiative heat loss (and the air trapped between blanket and body reduces heat loss by conduction).
Exercise 17.6
Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales.
Part A
The midday temperature at the surface of the moon (400 K ).
ANSWER:
T
= 127 ∘ C
Correct
Part B
The midday temperature at the surface of the moon (400 K ).
ANSWER:
T
= 260 ∘ F
Correct
Part C
The temperature at the tops of the clouds in the atmosphere of Saturn (95 K ).
ANSWER:
T
= 178 ∘ C
Correct
Part D
K
The temperature at the tops of the clouds in the atmosphere of Saturn (95 K ).
ANSWER:
T
= 289 ∘ F
Correct
Part E
The temperature at the center of the sun (1.55 × 107 K) .
ANSWER:
T
= 1.55×107 ∘ C
Correct
Part F
The temperature at the center of the sun (1.55 × 107 K) .
ANSWER:
T
= 2.79×107 ∘ F
Correct
Rankine Temperature Scale
Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0
∘
∘
R). However, the units of this scale are the same size as those on the Fahrenheit scale ( F ) rather than the Celsius
∘
scale ( C).
Part A
Given that water at standard pressure freezes at 0∘ C, which corresponds to 32∘ F , and that it boils at 100∘ C,
which corresponds to 212∘ F , calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a
temperature difference of 1 K on the Kelvin scale.
Give your answer to two significant figures.
Hint 1. Relation of Celsius and Kelvin temperature scales
A temperature increase of one kelvin corresponds to a temperature increase of one degree also on the
Celsius scale. The Kelvin temperature scale and the Celsius temperature scale differ only in their zero point.
ANSWER:
= 1.8 ∘ F
ΔT = 1 K
Correct
Part B
What is the numerical value of the triplepoint temperature T triple of water on the Rankine scale?
Give your answer to three significant figures.
Hint 1. Triplepoint temperature
On the Kelvin temperature scale, water freezes and coexists in three phases (solid, liquid, and vapor) at
273.16 K at standard pressure. This temperature is known as the triple point.
ANSWER:
T triple
= 492 ∘ R
Correct
Exercise 17.18
A steel tank is completely filled with 2.50 m 3 of ethanol when both the tank and the ethanol are at a temperature of 35.0
∘
C .
Part A
When the tank and its contents have cooled to 16.0 ∘ C , what additional volume of ethanol can be put into the
tank?
Express your answer using two significant figures.
ANSWER:
ΔV
= 3.4×10−2 m 3
Correct
Hot Rods
Two circular rods, both of length L and having the same diameter, are placed end to end between rigid supports with no
initial stress in the rods.
The coefficient of linear expansion and Young's modulus for rod A are α A and Y A respectively; those for rod B are α B
and Y B respectively. Both rods are "normal" materials with α > 0.
The temperature of the rods is now raised by ΔT .
Part A
After the rods have been heated, which of the following statements is true?
Choose the best answer.
ANSWER:
The length of each rod is still L.
The length of each rod changes but the combined length of the rods is still 2L.
Correct
The length of the combined rod remains the same, but because the rods have different expansion coefficients,
the lengths of the individual rods change. In other words, ΔLA + ΔLB = 0 even though ΔLA ≠ 0 and
ΔLB ≠ 0 .
Part B
After the rods have been heated, which of the following statements is true?
Choose the best answer.
ANSWER:
The stress in each rod remains zero.
A compressive stress arises that is the same for both rods.
A compressive stress arises that is different for the two rods.
A tensile stress arises that is the same for both rods.
A tensile stress arises that is different for the two rods.
A compressive stress arises in one rod and a tensile stress arises in the other rod.
Correct
Stress is a force per unit area. By Newton's 3rd law, the force on rod A due to rod B is the same as that on rod
B due to rod A. Since the rods have the same diameter, their crosssectional area is the same. Therefore, the
stress on each rod must be the same.
Part C
What is the stress F /A in the rods after heating?
Y
Y
ΔT
Express the stress in terms of α A , α B , Y A , Y B , and ΔT .
Hint 1. How to approach the problem
You know that the length of the two rods remains equal to 2L during the heating process. Find the change in
length of each rod as a function of the stress on the rods, then set the total change in length to zero and
solve for the stress.
Hint 2. Change in length for each rod
The change in length for each rod is due to both thermal as well as compressive stresses. The
corresponding formulas are
ΔLthermal = αLΔT
and
ΔL compressive
stress
The sum gives the net change in length for each rod.
=L
F
AY
.
ANSWER:
F
A
=
−(α A +α B )ΔT
1
Y
A
+
1
Y
B
Correct
Another way of thinking about this is that the combination of rods has a net thermal expansion coefficient
and a net Young's modulus given by
α = α1 + α2
Y =
Y1Y2
Y 1 +Y 2
.
Video Tutor: Heating Water and Aluminum
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an
experiment. Then, close the video window and answer the question on the right. You can watch the video again at any
point.
Part A
Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the
video, the mass of the metal is the same as that of the water. Room temperature is about 20∘ C before the start of
the experiment. The water heats up to 40∘ C, and the mystery metal heats up to 80∘ C. Compared to that of water,
the heat capacity of our mystery metal is
Hint 1. How to approach the problem
Recall that the heat Q delivered to a substance can be written Q = mc ΔT , where m is the mass of the
substance and c is its heat capacity.
ANSWER:
half as great.
three times greater.
onethird as great.
two times greater.
the same.
Correct
Given the same input of energy, the temperature of the metal increased three times as much as the
temperature of the water. Therefore, the metal has onethird the heat capacity of water. (Recall that the heat Q
delivered to a substance can be written Q = mc ΔT , where m is the mass of the substance and c is its heat
capacity.)
Exercise 17.27
An aluminum tea kettle with mass 1.55 kg and containing 1.70 kg of water is placed on a stove.
Part A
If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 21.0 ∘ C to 80.0
∘
C
∘
C
?
ANSWER:
Q
= 5.03×105 J
Answer Requested
Thermal Energy from Friction on a Rope
A capstan is a rotating drum or cylinder over which a rope or cord slides to provide a great amplification of the rope's
tension while keeping both ends free . Since the added
tension in the rope is due to friction, the capstan generates
thermal energy.
Part A
If the difference in tension (T 0 − T ′ ) between the two ends of the rope is 520 N and the capstan has a diameter
of 10.0 cm and turns once in 0.90 \;{\rm s}, find the rate \texttip{P_{\rm thermal}}{P_thermal} at which thermal
energy is being generated.
Give a numerical answer, in watts, rounded to the nearest 10 W.
Hint 1. Torque applied to the rope
In this case, the contact force between the capstan and the rope is not localized at a single point but is
distributed along the rope. (This is more realistic than assuming that forces act at a single point.) However,
we can still say that the net torque is the product of the difference between the tensions (T_0T') and the
radius. (Finding a numerical value for this torque can certainly be done as an intermediate calculation, but it
doesn't help much in reaching the answer.)
Hint 2. Power due to the torque
If you've found the torque, either symbolically or numerically, the power is given by P=\tau\omega, where
\omega=2\pi/t_0, with \texttip{t_{\rm 0}}{t_0} as the time for one revolution of the capstan.
Hint 3. Power from force
We should be able to find the power as the product of a force and a relative speed, and of course we can.
The net force on the rope is T_0T' and the relative speed is the circumference of the capstan divided by the
period of the capstan's rotation.
Hint 4. How both methods work equally well
If you've done this problem symbolically using torques, you've found that
P_{\rm thermal}=\tau\omega=(Fr)\left({2\pi\over t_0}\right)={2\pi r\left(T_0T'\right)\over t_0}\;.
If you've done this problem symbolically using forces, you've found that
P_{\rm thermal}= Fv=\left(T_0T'\right){2\pi r\over t_0}.
The fact that these expressions are the same should not be surprising.
ANSWER:
\texttip{P_{\rm thermal}}{P_thermal} = 180 W
Correct
The answer does not depend on the number of turns, only on the difference in tensions! A larger number of
turns might well create a larger tension difference, but once this difference has been achieved, the work done
per rotation of the capstan will not depend on the number of turns.
Part B
If the capstan is made of iron (with a specific heat capacity C_{\rm iron}=470\;{\rm J/(kg}\cdot ^\circ{\rm C}) and has
a mass of 6.00 \;{\rm kg}, at what rate d\Theta/ dt does its temperature rise? Assume that the temperature in the
capstan is uniform and that all the thermal energy generated flows into it.
Note that \texttip{\Theta }{Theta} is a temperature.
Give a numerical answer, in degrees Celsius per second, rounded to two significant figures.
Hint 1. Use the chain rule
In Part A, you found the rate at which heat is added to the capstan. This added heat will of course cause a
temperature rise, and this rate of temperature rise will be proportional to \texttip{P_{\rm thermal}}{P_thermal},
the proportionality being the reciprocal of the heat capacity. Symbolically,
P_{\rm thermal}={dQ\over dt}={dQ\over d\Theta}{d\Theta\over dt}=(mC){d\Theta\over dt},
where \texttip{C}{C} is the specific heat capacity, so that
{d\Theta\over dt}={P_{\rm thermal}\over mC}.
ANSWER:
{d\Theta\over dt} = 0.064 ^\circ {\rm C/s}
Correct
Video Tutor: Water Balloon Held over Candle Flame
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an
experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.
Part A
Which of the following changes would make the water balloon more likely to pop? (Ignore effects of convection
within the fluid.)
Select all that apply.
Hint 1. How to approach the problem.
Recall that in the video, the water balloon did not pop because most of the candle's thermal energy went into
heating the water. Let's think about the process in more detail.
The high heat capacity of water means that the water in the balloon warmed only slowly. (Recall that the
higher a substance's heat capacity, the less the substance's temperature will rise in response to a given
input of heat.) Contact with this relatively cool water kept the balloon's thin rubber skin from getting hot
enough to melt.
Rubber conducts heat more slowly than water. If you make the rubber skin of a water balloon thicker, will the
skin be more likely or less likely to melt? Think about the conduction of heat across the skin between the hot
side facing the candle and the cooler side in contact with water.
ANSWER:
Use a liquid that has a lower heat capacity than water.
Use a thinner balloon.
Use a thicker balloon.
Use a liquid that has a higher heat capacity than water.
Correct
Using a liquid with a lower heat capacity than water means that the candle flame will raise the liquid's
temperature more quickly than for water. A thicker balloon will conduct heat through to the liquid more slowly,
so that the outer part of the balloon wall will heat up more quickly to its melting point.
Exercise 17.41
A copper pot with a mass of 0.500 {\rm {\rm kg}} contains 0.170 {\rm {\rm kg}} of water, and both are at a temperature of
20.0 {\rm ^\circ C} . A 0.250 {\rm {\rm kg}} block of iron at 85.0 {\rm ^\circ C} is dropped into the pot.
Part A
Find the final temperature of the system, assuming no heat loss to the surroundings.
ANSWER:
T = 27.5 {\rm ^\circ C}
Correct
Steam vs. HotWater Burns
Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat
input to your skin from steam as opposed to hot water at the same temperature.
Assume that water and steam, initially at 100^\circ{\rm C}, are cooled down to skin temperature, 34^\circ{\rm C}, when
they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a
constant specific heat capacity c = 4190\; {\rm J/(kg \cdot K)} for both liquid water and steam.
Part A
Under these conditions, which of the following statements is true?
ANSWER:
Steam burns the skin worse than hot water because the thermal conductivity of steam is much higher than
that of liquid water.
Steam burns the skin worse than hot water because the latent heat of vaporization is released as well.
Hot water burns the skin worse than steam because the thermal conductivity of hot water is much higher
than that of steam.
Hot water and steam both burn skin about equally badly.
Correct
The key point is that the latent heat of vaporization has to be taken into account for the steam.
Part B
How much heat \texttip{H_{\rm 1}}{H_1} is transferred to the skin by 25.0 {\rm g} of steam onto the skin? The latent
heat of vaporization for steam is L = 2.256 \times 10^6 \; {\rm J/kg}.
Express the heat transferred, in joules, to three significant figures.
Hint 1. Determine the heat transferred from steam to skin
Find an expression for the heat \texttip{H_{\rm 1}}{H_1} transferred from the vapor to the skin.
Express your answer in terms of \texttip{c}{c}, the mass of steam \texttip{m}{m}, the latent heat of
vaporization \texttip{L}{L}, and the temperature difference \texttip{\Delta T}{DeltaT} between the
initial temperature of the steam and the skin temperature.
ANSWER:
\texttip{H_{\rm 1}}{H_1} = m \left(c {\Delta{T}}+L\right)
ANSWER:
\texttip{H_{\rm 1}}{H_1} = 6.33×104 {\rm J}
Correct
Here we assumed that the skin continues to remain at 34^{\circ}{\rm C}. Actually the local temperature in the
area where the steam condenses can be raised quite significantly.
Part C
How much heat \texttip{H_{\rm 2}}{H_2} is transferred by 25.0 {\rm g} of water onto the skin? To compare this to the
result in the previous part, continue to assume that the skin temperature does not change.
Express the heat transferred, in joules, to three significant figures.
Hint 1. Determine the heat transferred from water to skin
Find an expression for the heat \texttip{H_{\rm 2}}{H_2} transferred from the hot water to the skin.
Express your answer in terms of \texttip{c}{c}, the mass of water \texttip{m}{m}, and the temperature
difference \texttip{\Delta T}{DeltaT} between the initial temperature of the water and the skin
temperature.
ANSWER:
\texttip{H_{\rm 2}}{H_2} = m c {\Delta{T}}
ANSWER:
\texttip{H_{\rm 2}}{H_2} = 6910 {\rm J}
Correct
The amount of heat transferred to your skin is almost 10 times greater when you are burned by steam versus
hot water. The temperature of steam can also potentially be much greater than 100^\circ{\rm C}. For these
reasons, steam burns are often far more severe than hotwater burns.
An Electric Water Heater
An engineer is developing an electric water heater to provide a continuous ("on demand") supply of hot water. One trial
design is shown in the figure. Water is flowing at the rate
\texttip{F}{F}, the inlet thermometer registers \texttip{T_{\rm
1}}{T_1}, the voltmeter reads \texttip{V}{V}, and the ammeter
reads current \texttip{I}{I}. Then the power (i.e., the heat
generated per unit time by the heating element) is VI.
Assume that the heat capacity of water is \texttip{C}{C} and
that the heat capacity of the heater apparatus is
\texttip{C_{\rm h}}{C_h}.
Part A
When a steady state is finally reached, what is the temperature reading \texttip{T_{\rm 2}}{T_2} of the outlet
thermometer?
Express the outlet temperature in terms of \texttip{T_{\rm 1}}{T_1}, \texttip{F}{F}, \texttip{C}{C}, and any
other given quantities.
Hint 1. How to approach the problem
Determine the amount of heat input per unit mass of water that flows through the heater. Then use the
definition of heat capacity to compute how much the temperature of a unit mass of water will increase for a
given quantity of heat. Combine these two calculations to determine the increase in temperature of the water
as it flows through the heater.
Hint 2. Find heat input per unit mass of water
How much heat \texttip{Q}{Q} is delivered by the resistor per unit mass \texttip{m}{m} of water flowing
through the heater?
Express the heat per unit mass in terms of quanitities given in the problem introduction.
Hint 1. What relationships are involved
The power \texttip{P}{P} dissipated in the resistor is equal to the energy dissipated per unit time. The
flow rate \texttip{F}{F} is equal to the mass of the water that flows through the heater per unit time.
Thus, if all of the energy dissipated by the resistor goes into heating the water, then
\large{\frac{P}{F}=\frac{\frac{\mbox{heat}}{\mbox{time}}}{\frac{\mbox{mass}}
{\mbox{time}}}=\frac{\mbox{heat}}{\mbox{mass}}}.
Hint 2. Electrical power
The electrical power disspated in a device is given by the product of the voltage and the current:
P = V I.
ANSWER:
Q/m = \large{\frac{V I}{F}}
Hint 3. Find the change in water temperature
If an amount of heat \texttip{Q}{Q} is added uniformly to a substance of mass \texttip{m}{m} and heat
capacity \texttip{C}{C}, by how much will the temperature of that substance increase?
Express the temperature increase \texttip{\Delta T}{DeltaT} in terms of \texttip{Q}{Q}, \texttip{m}{m},
and \texttip{C}{C}.
ANSWER:
\texttip{\Delta T}{DeltaT} = \large{\frac{Q}{m} \frac{1}{C}}
ANSWER:
\texttip{T_{\rm 2}}{T_2} = \large{\frac{VI}{FC}+T_{1}}
Correct
Part B
Why is it unnecessary to take into account the heat capacity of the apparatus itself?
Choose the best explanation.
ANSWER:
In steady state the temperature of the apparatus doesn't change. Hence its heat capacity is irrelevant.
The heat capacity of any of the materials used in a water heater is much smaller than that of water.
Since the actual heating unit is immersed in the flow all the heat goes directly into the water.
Correct
Imagine that the input temperature of the water is T_1 = 18^\circ {\rm C}, the ammeter reads I = 15.0 \;{\rm A}, the
voltmeter reads V = 120\; {\rm V}, and the flow rate is F= 0.500\; {\rm kg/min}. The heat capacity of water C = 4200\; {\rm
J/(kg \cdot K)}.
Part C
What is the power \texttip{P}{P} at which the heater operates?
Express your answer numerically, in watts.
Hint 1. Electrical power
The electrical power disspated in a device is given by the product of the voltage and the current:
P = V I.
ANSWER:
\texttip{P}{P} = 1800 W
Correct
This is a large, but reasonable, power requirement for a household appliance.
Part D
Calculate the temperature \texttip{T_{\rm 2}}{T_2} of the water leaving the heater. Note that \texttip{F}{F} is defined
as the number of kilograms of water flowing through the heater per minute, whereas the power is measured in watts
(joules/second).
Express your answer numerically, in degrees Celsius, to the nearest integer.
ANSWER:
\texttip{T_{\rm 2}}{T_2} = 69 ^\circC
Correct
Part E
Consider using this heater to generate "on demand" hot water for a home shower. The input water temperature
(during the winter) can be as low as T_1=5.00^\circ {\rm C} and the output temperature should be at least
T_2=40.0^\circ {\rm C} for a moderately warm shower. Assume a conservative flow rate of F=8.00\; {\rm kg/min}
(corresponding to about 2.5 gallons per minute).
If the heater is to operate on a US standard V = 120 \;{\rm V} wall plug, how much current \texttip{I}{I} would it need
to draw to meet the design requirements?
Express the current numerically, in amperes, to the nearest integer.
ANSWER:
\texttip{I}{I} = 163 A
Correct
163 A is a tremendous amount of current for a home appliance. Most homes in the United States are wired for
a maximum of 200 A for the entire house, and only 20 A on any particular outlet. For the electric heater to meet
the modest design requirements given, it would consume almost 20 kW of power!
For this reason, most "on demand" water heaters generate heat by burning a fuel such as propane, rather than
using electricity.
Exercise 17.71
A picture window has dimensions of 1.40 {\rm m} \times2.50 {\rm m} and is made of glass 5.60 {\rm {\rm mm}} thick. On
a winter day, the outside temperature is 17.0 {\rm {\rm ^\circ C}} , while the inside temperature is a comfortable 20.5
{\rm {\rm ^\circ C}} .
Part A
At what rate is heat being lost through the window by conduction?
Express your answer using three significant figures.
ANSWER:
H = 1.88×104 {\rm W}
Correct
Part B
At what rate would heat be lost through the window if you covered it with a 0.750 {\rm mm}thick layer of paper
(thermal conductivity 0.0500 {\rm W/m \cdot K})?
Express your answer using three significant figures.
ANSWER:
H = 5970 {\rm W}
Correct
Heat Flow through Three Rods
Three identical rods are welded together to form a Yshaped figure. The crosssectional area of each rod is \texttip{A}
{A}, and they have length \texttip{L}{L} and thermal conductivity \texttip{k}{k}.
The free end of rod 1 is maintained at \texttip{T_{\rm 1}}{T_1}
and the free ends of rods 2 and 3 are maintained at a lower
temperature \texttip{T_{\rm 0}}{T_0}. You may assume that
there is no heat loss from the surfaces of the rods.
Part A
What is \texttip{T_{\rm j}}{T_j}, the temperature of the junction point?
Express your answer in terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}.
Hint 1. Find a relationship among the heat currents
Heat is flowing from the free end of rod 1 to the free ends of rods 2 and 3, breaking into two flows at the
junction point. What is the relationship between the heat current in rod 1 ( \texttip{H_{\rm 1}}{H_1}) to the
heat currents in rods 2 and 3 ( \texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3} respectively)?
Express \texttip{H_{\rm 1}}{H_1} in terms of \texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3}.
Hint 1. Use an analogy to electric currents
If you are familiar with electrical current, you can solve this problem by using the analog of Kirchoff's
junction rule, which states that the algebraic sum of the currents into any junction is zero (any current
leaving the junction must be subtracted from the total). Similarly, in steady state, the algebraic sum of
all heat currents into a junction is zero (any heat current leaving the junction must be subtracted from
the total).
ANSWER:
\texttip{H_{\rm 1}}{H_1} = H_{2} + H_{3}
Hint 2. Find \texttip{H_{\rm 1}}{H_1} in terms of the junction temperature
Now, if you can find expressions for the heat currents in the rods in terms of their physical properties and
temperature difference across them, you can substitute these into the formula H_1=H_2+H_3 and solve
algebraically for \texttip{T_{\rm j}}{T_j}.
Derive a general expression for the heat current through rod 1 in terms of the temperature of the junction,
\texttip{T_{\rm j}}{T_j}, and other quantities given in the problem introduction.
ANSWER:
\texttip{H_{\rm 1}}{H_1} = \large{\frac{k A \left(T_{1}T_{j}\right)}{L}}
ANSWER:
\texttip{T_{\rm j}}{T_j} = \large{\frac{T_{1}+2T_{0}}{3}}
Correct
This is a weighted average of the temperatures at the two ends of the system and so always lies between them
as you would expect. Also, it is closer to \texttip{T_{\rm 0}}{T_0} than to \texttip{T_{\rm 1}}{T_1}. This is
because the heat current in rods 2 and 3 separately must be less than the heat current in rod 1 (since H_2 +
H_3 = H_1).
Part B
What is the heat current \texttip{H_{\rm 1}}{H_1} in rod 1?
Express the heat current in terms of any or all of \texttip{k}{k}, \texttip{L}{L}, \texttip{A}{A}, and the
temperatures \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}.
Hint 1. Determine the general formula for heat current
Given the difference in temperature between the ends of the rod ( \texttip{\Delta T}{DeltaT}), the rod's
thermal conductivity \texttip{k}{k}, and its length \texttip{L}{L} and crosssectional area \texttip{A}{A}, what is
the general formula for the heat current \texttip{H}{H} in the rod?
Express your answer in terms of \texttip{k}{k}, \texttip{L}{L}, \texttip{A}{A}, and \texttip{\Delta T}
{DeltaT} (typed as DeltaT).
ANSWER:
\texttip{H}{H} = \large{\frac{k A {\Delta{T}}}{L}}
Hint 2. Find the temperature difference
What is the temperature difference \texttip{\Delta T}{DeltaT} between the ends of rod 1?
Express your answer in terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}. Your answer
should be positive.
Hint 1. Use your answer to Part A
The temperature difference between the ends of rod 1 is \Delta T = T_1 T_{\rm j}. To express this in
terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}, use the relationship you found in Part A
of this problem.
ANSWER:
\texttip{\Delta T}{DeltaT} = \large{\frac{2}{3} \left(T_{1}T_{0}\right)}
ANSWER:
\texttip{H_{\rm 1}}{H_1} = \large{{\frac{2}{3}}{\frac{kA}{L}}\left(T_{1}T_{0}\right)}
Correct
This is greater than the heat current due to only 2 rods. You would expect this to be the case since the
effective surface area is increased by having a third rod. You can check that if you wanted to remove the third
rod while keeping the same heat current, you must increase the crosssectional area of the remaining two rods
by a factor of 1.33.
Part C
By symmetry, the heat current in rods 2 and 3 will be the same. Find this heat current.
Express your answer in terms of \texttip{k}{k}, \texttip{L}{L}, \texttip{A}{A}, and the temperatures
\texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}.
Hint 1. How to approach this problem
Recall that H_1 = H_2 + H_3. Also, by symmetry, H_2 = H_3. This information is sufficient to find
\texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3}.
ANSWER:
H_2 = H_3 = \large{kA{\frac{T_{1}T_{0}}{3L}}}
Correct
Problem 17.92
You cool a 110.0 {\rm g} slug of redhot iron (temperature 745 {\rm ^\circ C}) by dropping it into an insulated cup of
negligible mass containing 75.0 {\rm g} of water at 20.0 {\rm ^\circ C}. Assume no heat exchange with the surroundings.
Part A
What is the final temperature of the water?
Express your answer using three significant figures.
ANSWER:
T_{\rm final} = 100 {\rm ^\circ C}
Correct
Part B
What is the final mass of the iron and the remaining water?
Express your answer using three significant figures.
ANSWER:
m_{\rm final} = 181 {\rm g}
Correct
Problem 17.104
Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200 {\rm
W/(m \cdot K)} ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus)
as a sphere 1.60 {\rm {\rm m}} in diameter having a layer of fat 4.00 {\rm {\rm cm}} thick. (Actually, the thickness varies
with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was
found that the outer surface layer of the fur is at 2.90 {\rm ^\circ C} during hibernation, while the inner surface of the fat
layer is at 30.9 {\rm ^\circ C}.
Assume the surface area of each layer is constant and given by the surface area of the spherical model constructed for
the black bear.
Part A
What should the temperature at the fatinner fur boundary be so that the bear loses heat at a rate of 51.5 {\rm {\rm
W}} ?
ANSWER:
T = 29.6 {\rm ^\circ C}
Correct
Should you be required to use your answer to this part in a subsequent part, use your full precision answer,
only rounding as a final step before submitting your answer.
Part B
How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51.5 {\rm {\rm W}}
?
ANSWER:
L_{\rm air} = 10.0 {\rm cm}
Correct
Score Summary:
Your score on this assignment is 101%.
You received 15.22 out of a possible total of 16 points.
Your instructor adjusted your score by awarding 1 additional point. This adjustment is reflected in your score above.
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