Basic Physics

# 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

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

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