Energy conservation for heat transfer – problems and solutions

1. 1-kg water at 100^oC mixed with 1-kg water at 10^oC in an isolated system. The specific heat of water is 4200 J/kg ^oC. Determine the final temperature of the mixture!

Known :

Mass of hot water (m₁) = 1 kg

Temperature of hot water (T₁) = 100^oC

Mass of cold water (m₂) = 1 kg

Temperature of cold water (T₂) = 10^oC

Wanted : The final temperature (T)

Solution :

Heat lost = Heat gained (isolated system)

m c ΔT = m c ΔT

m ΔT = m ΔT

m₁ (T₁ – T) = m₂ (T – T₂)

(1)(100 – T) = (1)(T – 10)

100 – T = T – 10

100 + 10 = T + T

110 = 2T

T = 110 / 2

T = 55

The final temperature is 55^oC.

2. A 3 kg block of lead at 80^o placed in 10 kg of water. The specific heat of lead is 1400 J.kg^-1C^-1 and the specific heat of water is 4200 J.kg^-1C^-1. The final temperature in thermal equilibrium is 20^oC. Determine the initial temperature of water!

Known :

Mass (m₁) = 3 kg

The specific heat of lead (c₁) = 1400 J.kg^-1C^-1

The temperature of lead (T₁) = 80 ^oC

Mass of water (m₂) = 10 kg

The specific heat of water (c₂) = 4200 J.kg^-1C^-1

The temperature of thermal equilibrium (T) = 20 ^oC

Wanted : The initial temperature of water (T₂)

Solution :

Heat lost = Heat gained

Q lead = Q water

m₁ c₁ ΔT = m₂ c₂ ΔT

(3)(1400)(80-20) = (10)(4200)(20-T)

(4200)(60) = (42,000)(20-T)

252,000 = 840,000 – 42,000 T

42,000 T = 840,000 – 252,000

42,000 T = 588,000

T = 588,000 / 42,000

T = 14

The initial temperature of water is 14^oC.

3. A block of copper at 100^oC placed in 128 gram water at 30 ^oC. The specific heat of water is 1 cal.g^-1oC^-1 and the specific heat of copper is 0.1 cal.g^-1oC^-1. If the temperature of the thermal equilibrium is 36 ^oC, determine the mass of the copper!

Known :

The temperature of copper (T₁) = 100 ^oC

The specific heat of copper (c₁) = 0.1 cal.g^-1oC^-1

Mass of water (m₂) = 128 gram

Temperature of water (T₂) = 30 ^oC

The specific heat of water (c₂) = 1 cal.g^-1oC^-1

The temperature of thermal equilibrium (T) = 36 ^oC

Wanted : Mass of copper (m₁)

Solution :

Heat lost = Heat gained

Q copper = Q water

m₁ c₁ ΔT = m₂ c₂ ΔT

(m₁)(0.1)(100-36) = (128)(1)(36-30)

(m₁)(0.1)(64) = (128)(1)(6)

(m₁)(6.4) = 768

m₁ = 768 / 6.4

m₁ = 120

The mass of copper is 120 gram.

4. A M-kg block of ice at 0^oC placed in 340-gram water at 20^oC in a vat. If the heat of fusion for water = 80 cal g^-1, the specific heat of water is 1 cal g^{-1 o}C^-1. All ice melts and the temperature of thermal equilibrium is 5^oC, determine the mass of ice!

Known :

Mass of water (m) = 340 gram

The temperature of ice (T _ice) = 0^oC

The temperature of water (T _water) = 20^oC

The temperature of thermal equilibrium (T) = 5^oC

The heat of fusion for water (L) = 80 cal g^-1

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

Wanted : Mass of ice (M)

Solution :

Heat lost = Heat gained

Q water = Q ice

m c (ΔT) = m_es L_es + m c (ΔT)

(340)(1)(20-5) = M (80) + M (1)(5-0)

(340)(15) = 80M + 5M

5100 = 85M

M = 5100/85

M = 60 gram

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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