# Heat Mass Specific heat The change in temperature – Problems and Solutions

9 Heat Mass Specific heat The change in temperature – Problems and Solutions

1. A 2 kg lead is heated from 50oC to 100oC. The specific heat of lead is 130 J.kg-1 oC-1. How much heat is absorbed by the lead?

Known :

Mass (m) = 2 kg

The specific heat (c) = 130 J.kg-1C-1

The change in temperature (ΔT) = 100oC – 50oC = 50oC

Wanted : Heat (Q)

Solution :

Q = m c ΔT

Q = heat, m = mass, c = the specific heat, ΔT = the change in temperature

The heat absorbed by lead :

Q = (2 kg)(130 J.kg-1C-1)(50oC)

Q = (100)(130)

Q = 13,000 Joule

Q = 1.3 x 104 Joule

2. The specific heat of copper is 390 J/kg oC, the change in temperature is 40oC. If the copper absorbs 40 Joule of heat, what is the copper’s mass!

Known :

The specific heat of copper (c) = 390 J/kgoC

The change in temperature (ΔT) = 40oC

Heat (Q) = 40 J

Wanted : Mass (m) of copper

Solution :

Q = m c ΔT

40 J = (m)(390 J/kg oC)(40oC)

40 = (m)(390 /kg)(40)

40 = (m)(390 /kg)(4)

40 = (m)(1560 /kg)

m = 40 / 1560

m = 0.026 kg

m = 26 gram

3. The initial temperature of 20 gram water is 30oC. The specific heat of water is 1 cal g-1 oC-1. If water absorbs 300 calories of heat, determine the final temperature!

Known :

Mass (m) = 20 gr

The initial temperature (T1) = 30oC

The specific heat of water (c) = 1 cal gr-1 oC-1

Heat (Q) = 300 cal

Wanted : The final temperature of water

Solution :

Q = m c ΔT

300 cal = (20 gr)(1 cal gr-1 oC-1)(T2-30)

300 = (20)(1)(T2-30)

300 = 20 (T2-30)

300 = 20T2 – 600

300 + 600 = 20T2

900 = 20T2

T2 = 900 / 20

T2 = 45

The change in temperature is 45oC – 30oC = 15oC.

4. The change in temperature of the sea water is 1oC when water absorbs 3900 Joule of heat. The specific heat of the sea water is 3.9 × 103 J/kg°C, what is the mass of the sea water.

Known :

The change in temperature (ΔT) = 1oC

Heat (Q) = 3900 Joule

The specific heat of the sea water (c) = 3.9 x 103 J/kg°C = 3900 J/Kg°C

Wanted : Mass (m)

Solution :

Q = m c ΔT

Q = heat, m = mass, c = specific heat, ΔT = the change in temperature

m = Q / c ΔT = 3900 / (3900)(1) = 3900 / 3900 = 1 kg

5. A 2-kg copper absorbs 39,000 J of heat at 30°C. If the specific heat of copper is 390 J/kg °C, what is the final temperature of the copper…

Known :

Mass (m) = 2 kg

Initial temperature (T1) = 30oC

Heat (Q) = 39,000 Joule

Specific heat (c) of copper = 390 J/kg oC

Wanted : The final temperature (T2)

Solution :

Q = m c ΔT

Q = heat, m = mass, c = specific heat, ΔT = the change in temperature

Q = m c ΔT = m c (T2 – T1)

39,000 = (2)(390)(T2 – 30)

100 = (2)(1)(T2 – 30)

100 = (2)(T2 – 30)

50 = T230

T2 = 50 + 30

T2 = 80oC

6. A 5-kg water is heated from 15°C to 40°C. What is the heat is absorbed by water. Specific heat of water is 4.2 × 103 J/Kg° C.

Known :

Mass (m) = 5 kg

Initial temperature (T1) = 15°C

Final temperature (T2) = 40°C

Specific heat of water (c) = 4.2 × 103 J/kg° C

Wanted : Heat (Q)

Solution :

Q = m c ΔT

Q = (5 kg)(4.2 × 103 J/kg°C)(40°C – 15°C)

Q = (5)(4.2 × 103 J)(25)

Q = 525 x 103 J

Q = 525,000 Joule

7. A 2-kg water is heated from 24°C to 90°C. What is the heat is absorbed by water. Specific heat of water is 4.2 × 103 J/Kg° C.

Known :

Mass (m) = 2 kg

Initial temperature (T1) = 24°C

Final temperature (T2) = 90°C

Specific heat of water (c) = 4,200 Joule/kg°C

Wanted :: Heat (Q)

Solution :

Q = m c ΔT

Q = (2 kg)(4,200 Joule/kg°C)(90°C – 24°C)

Q = (2 kg)(4,200 Joule/kg°C)(66°C)

Q = (132)(4,200 Joule)

Q = 554,400 Joule

8. A 5-gram water is heated from 10°C to 40°C. What is the heat is absorbed by water. Specific heat of water is 1 × 103 cal/gr° C.

Known :

Mass (m) = 5 gram

Initial temperature (T1) = 10oC

Final temperature (T2) = 40oC

Specific heat of water (c) = 1 cal/ gr°C

Wanted : Heat

Solution :

Q = m c ΔT

Q = (5 gram)(1 cal/ gr°C)(40oC – 10oC)

Q = (5)(1 cal)(30)

Q = 150 calories

9. A 0.2-kg water absorbs 42,000 Joule of heat at 25oC. The specific heat of water is 4200 J/kg oC, what is the final temperature of water.

Known :

Mass of water (m) = 0.2 kg

Heat (Q) = 42,000 Joule

Specific heat of water (c) = 4200 J/kg oC

Initial temperature (T1) = 25oC

Wanted : Final temperature (T2)

Solution :

Q = m c ΔT = m c (T2 – T1)

Q = heat, m = mass, c = specific heat, ΔT = the change in temperature, T1 = the initial temperature, T2 = the final temperature

Q = m c (T2 – T1)

42,000 = (0.2)(4200)(T2 – 25)

42,000 = 840 (T2 – 25)

42,000 = 840 T2 – 21,000

42,000 + 21,000 = 840 T2

63,000 = 840 T2

T2 = 63,000 / 840

T2 = 75oC

1. Converting temperature scales
2. Linear expansion
3. Area expansion
4. Volume expansion
5. Heat
6. Mechanical equivalent of heat
7. Specific heat and heat capacity
8. Latent heat, heat of fusion, heat of vaporization
9. Energy conservation for heat transfer

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