# Black principle – problems and solutions

1. 200-gram water at 30°C mixed with 100-gram water at 90°C. The specific heat of water = 1 cal.gram^{−1}°C^{−1}. Determine the final temperature of the mixture!

__Known :__

Mass of water 2 (m_{1}) = 200 gram

Temperature of water 1 (T_{1}) = 30^{o}C

Mass of water 2 (m_{2}) = 100 gram

Temperature of water 2 (T_{2}) = 90^{o}C

The specific heat for water (c) = 1 cal.gram^{−1}°C^{−1}

__Wanted :__ The final temperature

__Solution :__

Heat released by hooter water (Q_{2}) = heat absorbed by cooler water (Q_{1})

m_{2} c (ΔT) = m_{1} c (ΔT)

(100)(1)(90 – T) = (200)(1)(T – 30)

(100)(90 – T) = (200)(T – 30)

9000 – 100T = 200T – 6000

9000 + 6000 = 200T + 100T

15000 = 300T

T = 15000/300

T = 50^{o}C

2. 60-gram at 90^{o}C mixed with 40-gram water at 25^{o}C. What is the final temperature of the mixture. The specific heat for water = 1 cal.g^{-1}.^{o}C^{-1}.

__Known :__

Mass of water 1 (m_{1}) = 60 gram

Temperature of water 1 (T_{1}) = 90^{o}C

Mass of water 2 (m_{2}) = 40 gram

Temperature of water 2 (T_{2}) = 25^{o}C

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

__Wanted :__ The final temperature

__Solution :__

Heat released by hooter water (Q_{2}) = heat absorbed by cooler water (Q_{1})

m_{1} c (ΔT) = m_{2} c (ΔT)

(60)(1)(90 – T) = (40)(1)(T – 25)

(60)(90 – T) = (40)(T – 25)

5400 – 60T = 40T – 1000

5400 + 1000 = 40T + 60T

6400 = 100T

T = 6400/100

T = 64^{o}C

3. M-gram ice at 0^{o}C placed in 340-gram water at 20^{o}C in a specific container. Latent heat of fusion (L_{ice}) = 80 cal g^{-1}, the specific heat of water (c_{water}) = 1 cal g^{-1} ^{o}C^{-1}. All the ice melts and the thermal equilibrium = 5^{o}C. Find mass of ice.

__Known :__

Mass of water (m) = 340 gram

Temperature of ice (T_{ice}) = 0^{o}C

Temperature of water (T_{water}) = 20^{o}C

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

Latent heat of fusion for ice (L_{ice}) = 80 cal g^{-1}

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

__Wanted :__ Mass of ice (M)

__Solution :__

Heat released by water (Q_{2}) = heat absorbed by ice (Q_{1})

m_{water} c_{water} (ΔT) = m_{ice} L_{ice} + m_{ice} c_{water} (ΔT)

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

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

5100 = 85M

M = 5100/85

M = 60 gram

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

__Known :__

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

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

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

__Wanted :__ Mass of copper (m_{1})

__Solution :__

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 gram

5. A 3-kg lead with the specific heat for lead is 1400 J.kg^{-1}C^{-1} at 80^{o}C placed in 10-kg water with the specific heat for water is 4200 J.kg^{-1}C^{-1}. The temperature of the thermal equilibrium is 20^{o}C. What is the initial temperature of water.

__Known :__

Mass of lead (m_{1}) = 3 kg

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

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 the thermal equilibrium (T) = 20 ^{o}C

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

__Solution :__

Q release = Q absorb

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.

6. 75-gram water at 0^{o}C mixed with 50-gram water so the temperature of mixture is 40^{o}C. What is the initial temperature of 50-gram water.

__Known :__

Mass of water 1 (m_{1}) = 75 gram

The initial temperature of water 1 (T_{1}) = 0^{o}C

Mass of water 2 (m_{2}) = 50 gram

The temperature of mixture (T) = 40^{o}C

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

__Solution :__

*Heat released by hooter water (Q*_{release}*) = heat absorbed by cooler water (Q*_{absorbs}*)*

m_{1 }c (ΔT_{1}) = m_{2} c (ΔT_{2})

m_{1 }(ΔT_{1}) = m_{2} (ΔT_{2})

(75)(40 – 0) = (50)(T_{2 }– 40)

(75)(40) = (50)(T_{2 }– 40)

3000 = 50 T_{2 }– 2000

3000 + 2000 = 50 T_{2}

5000 = 50 T_{2}

T_{2 }= 100 ^{o}C

7. 200-gram metal heated to 120^{o}C, then placed in 100-gram water at 30^{o}C. The temperature of mixture at thermal equilibrium is 60^{o}C. If the specific heat of water is 4200 J.kg^{-1 o}C^{-1}, what is the specific heat of the metal.

Solution :

*Convert mass unit from gram to kilogram (International unit)*

__Known :__

Mass of metal (m_{1}) = 200 gram = 0.2 kg

Temperature of metal (T_{1}) = 120^{o}C

Mass of water (m_{2}) = 100 gram = 0.1 kg

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

The temperature of mixture (T) = 60^{o}C

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

__Wanted :__ The specific heat of metal (c_{1})

__Solution :__

Q_{release} = Q_{absorb}

m_{1 }c_{1} (ΔT_{1}) = m_{2} c_{2} (ΔT_{2})

(0.2)(c_{1})(120 – 60) = (0.1)(4200)(60 – 30)

(0.2)(c_{1})(60) = (0.1)(4200)(30)

12 c_{1} = 12600

c_{1} = 12600 / 12

c_{1} = 1050