Basic Physics

# 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

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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.

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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

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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

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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

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