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 50^oC to 100^oC. 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^-1C^-1

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

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)(50^oC)

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 40^oC. If the copper absorbs 40 Joule of heat, what is the copper’s mass!

Known :

The specific heat of copper (c) = 390 J/kg^oC

The change in temperature (ΔT) = 40^oC

Heat (Q) = 40 J

Wanted : Mass (m) of copper

Solution :

Q = m c ΔT

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

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^oC. 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 (T₁) = 30^oC

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)(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 45^oC – 30^oC = 15^oC.

4. The change in temperature of the sea water is 1^oC 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) = 1^oC

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) = 30^oC

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 = T₂ – 30

T₂ = 50 + 30

T₂ = 80^oC

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) = 10^oC

Final temperature (T₂) = 40^oC

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

Wanted : Heat

Solution :

Q = m c ΔT

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

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

Q = 150 calories

9. A 0.2-kg water absorbs 42,000 Joule of heat at 25^oC. 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) = 25^oC

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 = 75^oC

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