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Heat Capacities

Notes


Terms

  • Heat capacity is the amount of heat required to raise the temperature of a body by 1 K (or 1°C)
  • Specific heat capacity is the amount of heat required to raise the temperature of 1kg of the substance by 1 K (or 1°C) 
  • Latent heat of fusion of a solid substance is the heat energy needed to change it from solid to liquid state without any change in temperature 
  • Specific latent heat of fusion of a solid substance is the heat energy needed to change 1kg of it from solid to liquid state without any change in temperature 
    • SI unit: J/kg 
  • Latent heat of vaporisation of a substance is the heat energy needed to change it from liquid vapour state without any change in temperature 
  • Specific latent heat of vaporisation of a substance is the heat energy needed to change 1kg of it from liquid to vapour state without any change in temperature 
    • SI unit: J/kg

Heat Change Formula

  • Q = mcθ
    • m = mass (kg)
      c = specific heat capacity (J kg-1 oC-1)
      θ = temperature change (o)

Electric heater

Energy Supply, E = Pt
Energy Received, Q = mcθ

Energy Supplied, E = Energy Receive, Q
Pt = mcθ

E = electrical Energy (J or Nm)
P = Power of the electric heater (W)
t = time (in second) (s)
Q = Heat Change (J or Nm)
m = mass (kg)
c = specific heat capacity (J kg-1 oC-1)
θ = temperature change (o)

Mixing 2 liquids

Heat Gain by Liquid 1 = Heat Loss by Liquid 2
m1c1θ1 = m2c2θ2


m1 = mass of liquid 1
c1 = specific heat capacity of liquid 1
θ1 = temperature change of liquid 1

m2 = mass of liquid 2
c2 = specific heat capacity of liquid 2
θ2 = temperature change of liquid 2

Specific Latent Heat

Q = mL

Q = Heat Change (J or Nm)
m = mass (kg)
L = specific latent heat (J kg-1)

Example


1. A mass of 0.20kg of water at 0°C is placed in a vessel of negligible heat capacity. An electric heater with an output of 24 W is placed in the water and switched on. When the temperature of the water reaches 12°C, the heater is switched off.

a) Calculate the time for which the heater is switched on. Assume that the heat capacity of water is 4200J/kgK

b) An ice cube of mass 0.020kg is added to the 0.20kg of water at 0°C in the same vessel and the heater is switched on. Assuming that all the ice is at 0°C, calculate how long it will take for the water to reach 12°C. Assume that the specific latent heat of fusion on ice is 0.34 MJ/kg

Solutions

a) Heat supplied by heater = heat absorbed by water
E = mcθ
E = Pt
24 × t = 0.20 × 4200 × 12
t = 420
--> The heater is switched on for 420 s.

 

b) Heat absorbed by ice = Heat used to melt ice + Heat used to raise temperature of ice water from 0°C to 12°C
= ml + mcθ
= (0.020 × 340000) + (0.040 × 4200 × 12)
= 8816 J
Time t = 8816/24
= 367 s                                    

MCQ

1. When the temperature of a body increases, its

a. internal energy decreases

b. internal energy remains constant

c. internal energy increases

d. heat capacity increases

 

2. The internal energy of a body is measured in

a. kg

b. °C

c. J

d. JK-1

 

3. The heat capacity of a bottle of water is 2100 J°C-1. What is the amount of heat required to heat the water from 30°C to 50°C?

a. 2100J

b. 4200J

c. 42000J

d. 63000J

 

4. If the same amount of heat is supplied to 2 metal rods, A and B, rod B shows a smaller rise in temperature. Which of the following statements is true about the heat capacity of rods A and B?

a. The heat capacity of A is less than that of B

b. The heat capacity of B is less than that of A

c. The heat capacity of A is zero

d. The heat capacity of B is zero

 

5. The heat capacities of 10g of water and 1kg of water are in the ratio

a. 1 : 10

b. 10 : 1

c. 1 : 100

d. 100 : 1

 

6. 1 kg of substance X of specific heat capacity 2 kJkg-1°C-1 is heated from 30°C to 90°C. Assuming no heat loss, the heat required is

a. 7.5 kJ

b. 18 kJ

c. 80 kJ

d. 120 kJ

 

7. How much heat is required to raise the temperature of 20g of water from 10°C to 20°C if the specific heat capacity of water is 4.2 Jg-1°C-1?

a. 1.68 J

b. 84 J

c. 840 J

d. 1680 J

 

8. 4000 J of energy are given out when 2kg of  a metal is cooled from 50°C t0 40°C. The specific heat capacity of the metal, in Jg-1°C-1, is

a. 40

b. 50

c. 200

d. 400

 

9. What is the temperature rise when 42 kJ of energy is supplied to 5kg of water? (specific heat capacity of water is 4200 Jkg-1°C-1

a. 2°C

b. 5°C

c. 8.4°C

d. 10°C

 

10. A piece of copper of mass 2kg is cooled from 150°C to 50°C. The specific heat capacity of copper is 400 Jkg-1°C-1. The heat loss is

a. 800J

b. 4000J

c. 40000J

d. 80000J


11. 2 kg of oil is heated from 30°C to 40°C in 20s. The specific heat capacity of oil is 8 kJkg-1°C-1. The power of the heater is

a. 8 W

b. 8 kW

c. 24 kW

d. 32 kW

 

12. An immersion heater rated at 200 W is fitted into a large block of ice at 0°C. The latent heat of fusion of ice is 300J/g. How long does it take to melt 20g of ice?

a. 13s

b. 15s

c. 30s

d. 60s


13. An immersion heater rated at 150 W is fitted into a large block of ice at 0°C. The specific latent heat of fusion of ice is 300J/g. How long does it take to melt 10g of ice?

a. 2s

b. 5s

c. 20s

d. 150s

e. 4500 s


14. Aniline melts at -6°C and boils at 184°C. At which temperature would aniline not be a liquid?

a. -9°C

b. -3°C

c. 25°C

d. 100°C

e. 102°C


15. A 2 kg mass of copper is heated for 40 s by a heater that produces 100 J/s. The specific heat capacity of copper is 400 J/kgK. What is the rise in temperature?

a. 5 K

b. 10 K

c. 20 K

d. 50 K


16. When bubbles are seen forming rapidly in water and the temperature of the water remains constant,

a. the particles of the water are moving further apart

b. the particles of the water are moving faster

c. the particles of the water are moving faster and further apart

d. the particles of the water are moving slower and closer together


17. A 2 kW kettle containing boiling water is placed on a balance. It is left there and continues to boil for 5 minutes. The balance reading changes by 0.2kg. What does this information give as an estimate for the specific latent heat of vaporisation of water?

a. 2000 J/kg

b. 3000 J/kg

c. 50 000J/kg

d. 3 000 000J/kg


18. The specific heat capacity of copper is 400J/(kgºC). A 2kg mass of copper is heated for 40s by a 100W heater. What is the maximum possible rise in temperature?

a. 5ºC

b. 10ºC

c. 20ºC

d. 50ºC


MCQ Answers

1. c
2. c
3. c
4. a
5. c
6. d
7. c
8. c
9. a
10. d
11. b
12. c
13. c
14. a
15. a
16. a
17. d
18. a

Structured Question Worked Solutions


1. A 12-kW electric heater, working at its stated power, is found to heat 5kg of water from 20°C to 35°C in half a minute. The specific heat capacity of water is 4200 Jkg-1°C-1

a. Calculate
i) the heat produced by the heater in half a minute
ii) the heat absorbed by the water in the half minute


Solution

1i. 12000 x 30 = 360 kJ
1ii. 5 x 42000 x 15 = 315 kJ
1b. There is heat lost to the surroundings

b. Account for the difference in the answers to ai and ii. [1]

2. A lead cube of mass 0.25kg falls from rest from a height of 12m to the ground. Calculate, neglecting frictional loss,

a. the loss of potential energy of the cube
b. the gain in kinetic energy of the cube
c. the speed the cube has when it hits the ground
d. the rise of the temperature of the cube after it hits the ground, assuming that all the kinetic energy is converted into internal energy of the cube. [8]

The gravitational force on the mass of 1kg=10N The specific heat capacity of lead=0.13 kJ/(kgK)


Solution

2a. loss of p.e. of cube = mgh = 0.25 x 10 x 12 = 30 J

2b. gain in k.e. of cube = loss of p.e. of cube = 30 J

2c. 1/2 mv2 = 30
1/2 x 0.25 x v2 = 30
v = 15.5
speed of cube when it hits the ground = 15.5 m/s

2d. internal energy of cube = gain in k.e. of cube
mcθ = 30
0.25 x 130 x
θ = 30
θ = 0.923oC

3. Lemonade can be cooled by adding lumps of ice to it. A student discovers that 70g of ice at a temperature of 0°C cools 0.30kg of lemonade from 28°C to 7°C.

The latent heat of fusion of ice is 0.33 MJ/kg.
The specific heat capacity of water is 4.2 kJ/kgK.

Determine

a. the energy gained by the ice in melting
b. the energy gained by the melted ice
c. the enegy lost by the lemonade
d. a value for the specific heat capacity of the lemonade

Solution

3a. energy gained by ice in melting = ml = 0.07 x (3.3 x 105) = 23100 J

3b. energy gained by melted ice =
mcθ = 0.07 x 4200 x 7 = 2058 J

3c. energy lost by lemonade = 23100 + 2100 = 25200 J

3d. energy lost by lemonade = 25200 J
mc
θ = 25200
0.3 x c x 21 = 25200
c = 4000 J/kgK

4. A gas burner is used to heat 0.50kg of water in a beaker. The temperature of the water rises from 15oC to 60oC in 60s. Assuming that the specific heat capacity of water is 4200J/kgK, calculate the average rate at which heat is transferred to the water.

Solution

4. Heat gained by water = 0.5 x 4200 x (60 - 15) = 94500J
Average rate of heat transfer = heat gained / time taken = 94500 / 60 = 1575 J/s

5. The heater of an electric kettle is rated at 2.0kW.
ai. Calculate how long it would take to raise the temperature of 1.5kg of water in the kettle iron from 15oC to 100oC.

The specific heat capacity of water is 4200 J/kgK. Ignore heat losses and the heat needed to raise the temperature of the material of the kettle

aii. Calculate the cost of heating the water assuming that 1kWh of energy costs 6.0p

b. The heating element works from a 250 V a.c. supply. Calculate
i. the current through the heating element
ii. the resistance of the heating element


Solution

5ai. heat required = 1.5 x 4200 x (100 - 15) = 535500 J
Power = 2000W
Power = Energy / Time
Time = 535500 / 2000 = 267.75s

aii. energy consumed = power x time = 2 x (267.75/3600) = 0.14875
Cost = 2 x 0.14875 x 6 = 0.89p

bi. current in the heating element = power / voltage = 2000 / 250 = 8A

bii. resistance = voltage / current = 250 / 8 = 31.25 ohm

6a. What is meant by the term latent heat of fusion of a solid?

6b. Thermal energy is supplied to a melting solid at a constant rate of 2000W. Calculate the mass of the solid changed to liquid in 2.0 minutes. Assume that the specific latent heat of fusion of the solid is 95 000 J/kg and that heat exchange with the surroundings may be neglected.


Solution

6a. It is the heat required to change 1g of the solid at its melting point to liquid state at the same temperature.

6b. heat supplied by thermal energy = heat absorbed to convert solid to liquid
heat supplied in 2 minutes = ml
2000 x 2 x 60 = 95 000 x l
l = 2.526 kg

7. 2.0 kg of ice is placed in a vacuum flask, both ice and flask being at 0°C. It is found that exactly 14 hours elapse before the contents of the flask are entirely water at °C. Given that the specific latent heat of fusion of ice is 3.4 x 105 J/kg, calculate the average rate at which the contents gain heat from the surroundings.

Suggest a reason why the rate of gain of heat gradually decreases after all the ice has melted.

Solution


quantity of heat required to melt the ice = ml = 2 x 3.4 x 105 = 6.8 x 105 J

rate of heat gain = total heat gain / time = (6.8 x 105) / (14 x 60 x 60) = 13.49 J/s

After all the ice has melted, the temperature of water rises. The gap of difference in temperature between the water and the surroundings reduces and hence the rate of heat gain decreases.

8. 200g of ice at -10ºC was placed in a 300ºC copper cup. The ice in the copper cup eventually turned to water and reached a constant temperature of 50ºC.
Use the data below to answer the following questions.
a. How much thermal energy is needed to increase the temperature of the ice from -20ºC to 0ºC?
b. How much thermal energy is needed for the ice at 0ºC to melt to water at 0ºC
c. How much thermal energy is needed to increase the temperature of the water from 0ºC to 50ºC?
d. What is the final temperature of the copper cup when the water is at a constant temperature of 50ºC? Explain your answer.
e. Calculate the mass of the copper cup.
f. In real life, the mass of copper cup is different from the calculated value in (e). Should the actual mass of the copper cup be higher or lower than the calculated value? Explain your answer. 

Solution

8a. Q = mcø = 0.2 x 2100 x (0-(-20)) = 8400J
8b. Q = ml = 0.2 x 340,000 = 68,000J
8c. Q = mcø = 0.2 x 4200 x (50-0) = 42,000J
8d. 50ºC. Thermal equilibrium is reached between the copper cup and the water.
8e. Thermal energy lost by copper cup = thermal energy gained by ice/water
m x 400 x (300 - 50) = 8400 + 68,000 + 42,000
m = 1.18kg
8f. The actual mass of the copper cup should be higher than 1.18kg. In real life, thermal energy transfers from the copper cup to the surrounding at high rate due to its high temperature above the room temperature of 30ºC. Although ice is also absorbing thermal energy from the surrounding, the rate of absorption is not as high as what is lost by the copper cup to the surrounding due to the small temperature difference. When the copper cup has a higher mass, it can store more thermal energy and so have enough thermal energy to transfer to the ice/water while losing some energy to the surrounding.


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