# Energy conservation for heat transfer – problems and solutions

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

__Known :__

Mass of hot water (m_{1}) = 1 kg

Temperature of hot water (T_{1}) = 100^{o}C

Mass of cold water (m_{2}) = 1 kg

Temperature of cold water (T_{2}) = 10^{o}C

__Wanted :__ The final temperature (T)

__Solution :__

Heat lost = Heat gained (isolated system)

m c ΔT = m c ΔT

m ΔT = m ΔT

m_{1} (T_{1 }– T) = m_{2} (T – T_{2})

(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^{o}C.

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^{-1}C^{-1} and the specific heat of water is 4200 J.kg^{-1}C^{-1}. The final temperature in thermal equilibrium is 20^{o}C. Determine the initial temperature of water!

__Known :__

Mass (m_{1}) = 3 kg

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

The temperature of lead (T_{1}) = 80 ^{o}C

Mass of water (m_{2}) = 10 kg

The specific heat of water (c_{2}) = 4200 J.kg^{-1}C^{-1}

The temperature of thermal equilibrium (T) = 20 ^{o}C

__Wanted :__ The initial temperature of water (T_{2})

__Solution :__

Heat lost = Heat gained

Q lead = Q water

m_{1} c_{1} ΔT = m_{2} c_{2} 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^{o}C.

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

__Known :__

The temperature of copper (T_{1}) = 100 ^{o}C

The specific heat of copper (c_{1}) = 0.1 cal.g^{-1o}C^{-1}

Mass of water (m_{2}) = 128 gram

Temperature of water (T_{2}) = 30 ^{o}C

The specific heat of water (c_{2}) = 1 cal.g^{-1o}C^{-1}

The temperature of thermal equilibrium (T) = 36 ^{o}C

__Wanted : __Mass of copper (m_{1})

__Solution :__

Heat lost = Heat gained

Q copper = Q water

m_{1} c_{1} ΔT = m_{2} c_{2} ΔT

(m_{1})(0.1)(100-36) = (128)(1)(36-30)

(m_{1})(0.1)(64) = (128)(1)(6)

(m_{1})(6.4) = 768

m_{1} = 768 / 6.4

m_{1} = 120

The mass of copper is 120 gram.

4. A M-kg block of ice at 0^{o}C placed in 340-gram water at 20^{o}C 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^{o}C, determine the mass of ice!

__Known :__

Mass of water (m) = 340 gram

The temperature of ice (T _{ice}) = 0^{o}C

The temperature of water (T _{water}) = 20^{o}C

The temperature of thermal equilibrium (T) = 5^{o}C

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

### Ebook PDF Energy conservation for heat transfer sample problems with solutions

1 file(s) 56.88 KB- Converting temperature scales
- Linear expansion
- Area expansion
- Volume expansion
- Heat
- Mechanical equivalent of heat
- Specific heat and heat capacity
- Latent heat, heat of fusion, heat of vaporization
- Energy conservation for heat transfer