ques- heat
Content
The length of the mercury column in an uncalibrated mercury
thermometer is 2cm when its bulb is immerse in melting ice and 20cm
when the bulb is in steam above boiling water. The thermometer is used
to measure the temperature of a water in a glass. The length of the
mercury column is 11cm? What is the temperature of the water in the
glass?
Answer:
The temperature of the water, T
×100 =50 o C
=
lθ −l0 ×100 = (11)−(2 )
l 100 −l0
(20)−(2)
The length of the alcohol column in a thermometer is 2.5cm and 17.5cm
when the thermometer is dipped into a melting ice and a boiling water
respectively. Find the distance between every 10°C of the scale on the
thermometer.
Answer:
Distance for 100oC = 17.5cm – 2.5 cm = 15.0cm
Distance for 10oC = 15.0cm ÷ 10 = 1.50 cm
How much heat energy is required to change 2 kg of ice at 0°C into
water at 20°C? [Specific latent heat of fusion of water = 334 000 J/kg;
specific heat capacity of water = 4200 J/(kg K).]
Answer:
Energy needed to melt 2kg of ice,
Q1 = mL = (2)(334000) = 668000J
Energy needed to change the temperature from 0°C to 20°C.
Q2 = mcθ = (2)(4200)(20 - 0) = 168000J
Total energy needed = Q1 + Q2 = 668000 + 168000 = 836000J
Starting at 20°C, how much heat is required to heat 0.3 kg of aluminum
to its melting point and then to convert it all to liquid? [Specific heat
capacity of aluminium = 900J kg-1 °C-1; Specific latent heat of aluminium
= 321,000 Jkg-1, Melting point of aluminium = 660°C]
Answer:
Energy needed to increase the temperature from 20°C to 660°C
Q1 = mcθ = (0.3)(900)(660 - 20) = 172,800J
Energy needed to melt 0.3kg of aluminium,
Q2 = mL = (0.3)(321000) = 96,300J
Total energy needed = Q1 + Q2 = 172,800 + 96,300 = 269,100J
How much heat must be removed by a refrigerator from 2 kg of water at
70 °C to convert it to ice cubes at -11°C? [Specific latent heat of fusion of
ice = 334,000 Jkg-1 ; Specific heat capacity of water = 4200J kg-1 °C-1;
Specific heat capacity of ice = 2100 J/(kg K)]
Answer:
Energy to be removed to reduce the temperature from 70°C to 0°C
(Freezing point of water)
Q1 = mcθ = (2)(4200)(70 - 0) = 588,000J
Energy needed to freeze 2kg of water,
Q2 = mL = (2)(334,000) = 668,000J
Energy to be removed to reduce the temperature from 0°C to -11°C
Q3 = mcθ = (2)(2100)(0 - (-11)) = 46,200J
Total energy needed = Q1 + Q2 + Q3
= 588,000 + 668,000 + 46,200J = 1,302,200J
A fish releases a bubble of air of volume 1cm³ at the bottom of a lake.
The depth of the lake is 10m. Find the volume of the bubble when it
reaches the surface of the pond. (Assume that the atmospheric pressure
is equal to 10m of water)
V1 = 1cm³
P1 = 20m water
V2 = ?
P2 = 10m water
P1 V1 = P 2 V2
(20)(1) = (10)V2
V2 = 2cm3
An iron cylinder containing gas with pressure 200kPa when it is kept is a
room of temperature 27°C. What is the pressure of the gas when the
cylinder is located outdoor where the temperature is 35°C.
P1 = 200kPa
T1 = 273 + 27 = 300K
P1 = P2
T1
T2
P2 = ?
T2 = 273 + 35 = 308K
(200000) = P 2
(300)
(308)
P2 = (308)(200000) = 205kPa
(300)
The figure shows some air trapped in a capillary tube. Given that the
temperature of the air is 27°C. Find the length of the air column when
the temperature of the air is increased to 87°C.
V1 = 6cm
T1 = 273 + 27 = 300K
V1 = V2
T1 T2
(6) = V 2
(300) (360)
V2 = ?
T2 = 273 + 87 = 360K
V2 = (360)(6) =7.2cm
(300)
thermometer is 2cm when its bulb is immerse in melting ice and 20cm
when the bulb is in steam above boiling water. The thermometer is used
to measure the temperature of a water in a glass. The length of the
mercury column is 11cm? What is the temperature of the water in the
glass?
Answer:
The temperature of the water, T
×100 =50 o C
=
lθ −l0 ×100 = (11)−(2 )
l 100 −l0
(20)−(2)
The length of the alcohol column in a thermometer is 2.5cm and 17.5cm
when the thermometer is dipped into a melting ice and a boiling water
respectively. Find the distance between every 10°C of the scale on the
thermometer.
Answer:
Distance for 100oC = 17.5cm – 2.5 cm = 15.0cm
Distance for 10oC = 15.0cm ÷ 10 = 1.50 cm
How much heat energy is required to change 2 kg of ice at 0°C into
water at 20°C? [Specific latent heat of fusion of water = 334 000 J/kg;
specific heat capacity of water = 4200 J/(kg K).]
Answer:
Energy needed to melt 2kg of ice,
Q1 = mL = (2)(334000) = 668000J
Energy needed to change the temperature from 0°C to 20°C.
Q2 = mcθ = (2)(4200)(20 - 0) = 168000J
Total energy needed = Q1 + Q2 = 668000 + 168000 = 836000J
Starting at 20°C, how much heat is required to heat 0.3 kg of aluminum
to its melting point and then to convert it all to liquid? [Specific heat
capacity of aluminium = 900J kg-1 °C-1; Specific latent heat of aluminium
= 321,000 Jkg-1, Melting point of aluminium = 660°C]
Answer:
Energy needed to increase the temperature from 20°C to 660°C
Q1 = mcθ = (0.3)(900)(660 - 20) = 172,800J
Energy needed to melt 0.3kg of aluminium,
Q2 = mL = (0.3)(321000) = 96,300J
Total energy needed = Q1 + Q2 = 172,800 + 96,300 = 269,100J
How much heat must be removed by a refrigerator from 2 kg of water at
70 °C to convert it to ice cubes at -11°C? [Specific latent heat of fusion of
ice = 334,000 Jkg-1 ; Specific heat capacity of water = 4200J kg-1 °C-1;
Specific heat capacity of ice = 2100 J/(kg K)]
Answer:
Energy to be removed to reduce the temperature from 70°C to 0°C
(Freezing point of water)
Q1 = mcθ = (2)(4200)(70 - 0) = 588,000J
Energy needed to freeze 2kg of water,
Q2 = mL = (2)(334,000) = 668,000J
Energy to be removed to reduce the temperature from 0°C to -11°C
Q3 = mcθ = (2)(2100)(0 - (-11)) = 46,200J
Total energy needed = Q1 + Q2 + Q3
= 588,000 + 668,000 + 46,200J = 1,302,200J
A fish releases a bubble of air of volume 1cm³ at the bottom of a lake.
The depth of the lake is 10m. Find the volume of the bubble when it
reaches the surface of the pond. (Assume that the atmospheric pressure
is equal to 10m of water)
V1 = 1cm³
P1 = 20m water
V2 = ?
P2 = 10m water
P1 V1 = P 2 V2
(20)(1) = (10)V2
V2 = 2cm3
An iron cylinder containing gas with pressure 200kPa when it is kept is a
room of temperature 27°C. What is the pressure of the gas when the
cylinder is located outdoor where the temperature is 35°C.
P1 = 200kPa
T1 = 273 + 27 = 300K
P1 = P2
T1
T2
P2 = ?
T2 = 273 + 35 = 308K
(200000) = P 2
(300)
(308)
P2 = (308)(200000) = 205kPa
(300)
The figure shows some air trapped in a capillary tube. Given that the
temperature of the air is 27°C. Find the length of the air column when
the temperature of the air is increased to 87°C.
V1 = 6cm
T1 = 273 + 27 = 300K
V1 = V2
T1 T2
(6) = V 2
(300) (360)
V2 = ?
T2 = 273 + 87 = 360K
V2 = (360)(6) =7.2cm
(300)
