Energy conservation for heat transfer – problems and solutions

1. 1-kg water at 100oC mixed with 1-kg water at 10oC 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) = 100oC

Mass of cold water (m2) = 1 kg

Temperature of cold water (T2) = 10oC

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 55oC.

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2. A 3 kg block of lead at 80o 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 20oC. 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 14oC.

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3. A block of copper at 100oC 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.

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

Known :

Mass of water (m) = 340 gram

The temperature of ice (T ice) = 0oC

The temperature of water (T water) = 20oC

The temperature of thermal equilibrium (T) = 5oC

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

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

Wanted : Mass of ice (M)

Solution :

Heat lost = Heat gained

Q water = Q ice

m c (ΔT) = mes Les + 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

  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