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A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during the process is:
- a)12.65 kJ/K
- b)16.97 kJ/K
- c)4.32 kJ/K
- d)29.62 kJ/K
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A 50-Kg block of iron casting at 500K is thrown into a large lake that...
Specific heat of iron block:
Cv = m cv = 22.5 kJ/K
Entropy of iron:
S(T) = ∫dS = ∫ 1/T CvdT = Cv lnT
Entropy change of iron:
∆Si = Cv ln(To/T) = -12.65 kJ/K
Entropy change of water:
∆Sw = ∆Q/To = Cv(T/To -1) = +16.97 kJ/K
Total entropy change:
∆S = ∆SI + ∆Sw = +4.3 kJ/K
Most Upvoted Answer
A 50-Kg block of iron casting at 500K is thrown into a large lake that...
Given data:
Mass of iron block, m = 50 kg
Initial temperature of iron block, Ti = 500 K
Temperature of lake water, Tl = 285 K
Specific heat of iron, Cp = 0.45 kJ/kg-K
Calculation:
1. The heat lost by the iron block is equal to the heat gained by the lake water, i.e., Qlost = Qgain.
2. The heat lost by the iron block can be calculated using the formula: Qlost = m*Cp*(Ti-Tf), where Tf is the final temperature of the iron block after reaching thermal equilibrium with the lake water.
3. The heat gained by the lake water can be calculated using the formula: Qgain = m*Cp*(Tf-Tl).
4. Equating Qlost and Qgain, we get: m*Cp*(Ti-Tf) = m*Cp*(Tf-Tl).
5. Solving for Tf, we get: Tf = (Ti+Tl)/2 = (500+285)/2 = 392.5 K.
6. The change in entropy of the iron block can be calculated using the formula: ΔS = m*Cp*ln(Tf/Ti).
7. Substituting the values, we get: ΔS = 50*0.45*ln(392.5/500) = -1.26 kJ/K.
8. The change in entropy of the lake water can be calculated using the formula: ΔS = m*Cp*ln(Tf/Tl).
9. Substituting the values, we get: ΔS = 50*0.45*ln(392.5/285) = 5.58 kJ/K.
10. The total entropy generation can be calculated by adding the change in entropy of the iron block and the lake water, i.e., ΔStotal = ΔSiron + ΔSlake = -1.26 + 5.58 = 4.32 kJ/K.
Therefore, the entropy generation during the process is 4.32 kJ/K (option C).
Mass of iron block, m = 50 kg
Initial temperature of iron block, Ti = 500 K
Temperature of lake water, Tl = 285 K
Specific heat of iron, Cp = 0.45 kJ/kg-K
Calculation:
1. The heat lost by the iron block is equal to the heat gained by the lake water, i.e., Qlost = Qgain.
2. The heat lost by the iron block can be calculated using the formula: Qlost = m*Cp*(Ti-Tf), where Tf is the final temperature of the iron block after reaching thermal equilibrium with the lake water.
3. The heat gained by the lake water can be calculated using the formula: Qgain = m*Cp*(Tf-Tl).
4. Equating Qlost and Qgain, we get: m*Cp*(Ti-Tf) = m*Cp*(Tf-Tl).
5. Solving for Tf, we get: Tf = (Ti+Tl)/2 = (500+285)/2 = 392.5 K.
6. The change in entropy of the iron block can be calculated using the formula: ΔS = m*Cp*ln(Tf/Ti).
7. Substituting the values, we get: ΔS = 50*0.45*ln(392.5/500) = -1.26 kJ/K.
8. The change in entropy of the lake water can be calculated using the formula: ΔS = m*Cp*ln(Tf/Tl).
9. Substituting the values, we get: ΔS = 50*0.45*ln(392.5/285) = 5.58 kJ/K.
10. The total entropy generation can be calculated by adding the change in entropy of the iron block and the lake water, i.e., ΔStotal = ΔSiron + ΔSlake = -1.26 + 5.58 = 4.32 kJ/K.
Therefore, the entropy generation during the process is 4.32 kJ/K (option C).
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A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during theprocess is:a)12.65 kJ/Kb)16.97 kJ/Kc)4.32 kJ/Kd)29.62 kJ/KCorrect answer is option 'C'. Can you explain this answer?
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A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during theprocess is:a)12.65 kJ/Kb)16.97 kJ/Kc)4.32 kJ/Kd)29.62 kJ/KCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during theprocess is:a)12.65 kJ/Kb)16.97 kJ/Kc)4.32 kJ/Kd)29.62 kJ/KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during theprocess is:a)12.65 kJ/Kb)16.97 kJ/Kc)4.32 kJ/Kd)29.62 kJ/KCorrect answer is option 'C'. Can you explain this answer?.
A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during theprocess is:a)12.65 kJ/Kb)16.97 kJ/Kc)4.32 kJ/Kd)29.62 kJ/KCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during theprocess is:a)12.65 kJ/Kb)16.97 kJ/Kc)4.32 kJ/Kd)29.62 kJ/KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50-Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/Kg-K for the iron, the entropy generation during theprocess is:a)12.65 kJ/Kb)16.97 kJ/Kc)4.32 kJ/Kd)29.62 kJ/KCorrect answer is option 'C'. Can you explain this answer?.
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