# Heat, mass, specific heat, the change in temperature – problems and solutions

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

__Known :__

Mass (m) = 2 kg

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

The change in temperature (ΔT) = 100^{o}C – 50^{o}C = 50^{o}C

__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^{-1}C^{-1})(50^{o}C)

Q = (100)(130)

Q = 13,000 Joule

Q = 1.3 x 10^{4} Joule

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

__Known :__

The specific heat of copper (c) = 390 J/kg^{o}C

The change in temperature (ΔT) = 40^{o}C

Heat (Q) = 40 J

__Wanted :__ Mass (m) of copper

__Solution :__

Q = m c ΔT

40 J = (m)(390 J/kg ^{o}C)(40^{o}C)

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 30^{o}C. The specific heat of water is 1 cal g^{-1} ^{o}C^{-1}. If water absorbs 300 calories of heat, determine the final temperature!

__Known :__

Mass (m) = 20 gr

The initial temperature (T_{1}) = 30^{o}C

The specific heat of water (c) = 1 cal gr^{-1} ^{o}C^{-1}

Heat (Q) = 300 cal

__Wanted :__ The final temperature of water

__Solution :__

Q = m c ΔT

300 cal = (20 gr)(1 cal gr^{-1} ^{o}C^{-1})(T_{2}-30)

300 = (20)(1)(T_{2}-30)

300 = 20 (T_{2}-30)

300 = 20T_{2 }– 600

300 + 600 = 20T_{2}

900 = 20T_{2}

T_{2} = 900 / 20

T_{2} = 45

The change in temperature is 45^{o}C – 30^{o}C = 15^{o}C.

4. The change in temperature of the sea water is 1^{o}C when water absorbs 3900 Joule of heat. The specific heat of the sea water is 3.9 × 10^{3} J/kg°C, what is the mass of the sea water.

__Known :__

The change in temperature (ΔT) = 1^{o}C

Heat (Q) = 3900 Joule

The specific heat of the sea water (c) = 3.9 x 10^{3} 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 (T_{1}) = 30^{o}C

Heat (Q) = 39,000 Joule

Specific heat (c) of copper = 390 J/kg ^{o}C

__Wanted ____:__ The final temperature (T_{2})

__Solution :__

Q = m c ΔT

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

Q = m c ΔT = m c (T_{2 }– T_{1})

39,000 = (2)(390)(T_{2 }– 30)

100 = (2)(1)(T_{2 }– 30)

100 = (2)(T_{2 }– 30)

50 = T_{2} – 30

T_{2 }= 50 + 30

T_{2 }= 80^{o}C

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 × 10^{3 }J/Kg° C.

__Known :__

Mass (m) = 5 kg

Initial temperature (T_{1}) = 15°C

Final temperature (T_{2}) = 40°C

Specific heat of water (c) = 4.2 × 10^{3 }J/kg° C

__Wanted :__ Heat (Q)

__Solution :__

Q = m c ΔT

Q = (5 kg)(4.2 × 10^{3 }J/kg°C)(40°C – 15°C)

Q = (5)(4.2 × 10^{3 }J)(25)

Q = 525 x 10^{3 }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 × 10^{3 }J/Kg° C.

__Known :__

Mass (m) = 2 kg

Initial temperature (T_{1}) = 24°C

Final temperature (T_{2}) = 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 × 10^{3 }cal/gr° C.

__Known :__

Mass (m) = 5 gram

Initial temperature (T_{1}) = 10^{o}C

Final temperature (T_{2}) = 40^{o}C

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

__Wanted__ : Heat

__Solution :__

Q = m c ΔT

Q = (5 gram)(1 cal/ gr°C)(40^{o}C – 10^{o}C)

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

Q = 150 calories

9. A 0.2-kg water absorbs 42,000 Joule of heat at 25^{o}C. The specific heat of water is 4200 J/kg ^{o}C, 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 ^{o}C

Initial temperature (T_{1}) = 25^{o}C

__Wanted :__ Final temperature (T_{2})

__Solution :__

Q = m c ΔT = m c (T_{2} – T_{1})

*Q = **heat**, m = mass, c = **specific heat**, **ΔT = **the change in temperature**, T*_{1 }*= **the initial temperature**, T*_{2 }*= **the final temperature*

Q = m c (T_{2} – T_{1})

42,000 = (0.2)(4200)(T_{2} – 25)

42,000 = 840 (T_{2} – 25)

42,000 = 840 T_{2} – 21,000

42,000 + 21,000 = 840 T_{2}

63,000 = 840 T_{2}

T_{2} = 63,000 / 840

T_{2} = 75^{o}C

### Ebook PDF Heat, mass, specific heat, the change in temperature sample problems with solutions

1 file(s) 90.16 KB- Converting temperature scales
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