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

**1. Question:** Who was Joseph Black? **Answer:** Joseph Black was an 18th-century Scottish physician and chemist, known for his discoveries of latent heat, specific heat, and carbon dioxide.

**2. Question:** What is latent heat? **Answer:** Latent heat is the amount of energy absorbed or released by a substance during a phase change, like melting or boiling, without changing its temperature.

**3. Question:** How does latent heat differ from sensible heat? **Answer:** While sensible heat causes a change in temperature of a substance without a phase change, latent heat causes a phase change without a change in temperature.

**4. Question:** What significance did Black’s experiments on latent heat have on the study of thermodynamics? **Answer:** Black’s work laid the foundational understanding of heat energy and its conservation, paving the way for further advancements in thermodynamics and calorimetry.

**5. Question:** Why does ice at 0°C and water at 0°C have different amounts of heat energy? **Answer:** Ice at 0°C needs to absorb the latent heat of fusion to become water at 0°C. Thus, water at 0°C has more heat energy than ice at the same temperature due to this absorbed latent heat.

**6. Question:** What is specific heat? **Answer:** Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin.

**7. Question:** How did Black’s work on specific heat lead to the understanding of different heat capacities of substances? **Answer:** Black discovered that equal masses of different substances required different amounts of heat to achieve the same change in temperature. This observation introduced the concept of specific heat capacities of substances.

**8. Question:** Why is the concept of latent heat crucial in weather systems? **Answer:** Latent heat plays a significant role in weather systems because of the energy involved in the phase changes of water, like evaporation and condensation. This energy transfer affects temperature, cloud formation, and weather patterns.

**9. Question:** What apparatus did Joseph Black use to measure heat changes in his experiments? **Answer:** Black used a simple calorimeter, an instrument for measuring heat change, in his experiments to quantify the amount of heat absorbed or released.

**10. Question:** How did Black’s discovery of carbon dioxide (which he called “fixed air”) contribute to chemistry? **Answer:** Black’s identification of carbon dioxide as a distinct gas different from atmospheric air was pivotal in the advancement of gas chemistry and paved the way for understanding chemical reactions involving gases.