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A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________
Correct answer is between '9.7,9.9'. Can you explain this answer?
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A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 1...
Calculation of Change in Entropy of the Combined System
Initial State:
- Metal block mass (m1) = 40 kg
- Metal block specific heat capacity (Cp1) = 0.5 kJ/(kg.K)
- Metal block initial temperature (T1) = 450°C
- Oil mass (m2) = 150 kg
- Oil specific heat capacity (Cp2) = 2.5 kJ/(kg.K)
- Oil initial temperature (T2) = 25°C
Final State:
- The metal block and oil are in thermal equilibrium at some final temperature (Tf)
Assumptions:
- The metal block and oil are the only components of the system
- The system is fully isolated from its surroundings
Calculation of Final Temperature:
- The metal block loses heat to the oil and the oil gains heat from the metal block
- The heat lost by the metal block (Q1) is equal to the heat gained by the oil (Q2)
- Q1 = m1Cp1(T1 - Tf)
- Q2 = m2Cp2(Tf - T2)
- Equating Q1 and Q2, we get:
- m1Cp1(T1 - Tf) = m2Cp2(Tf - T2)
- Solving for Tf, we get:
- Tf = (m1Cp1T1 + m2Cp2T2)/(m1Cp1 + m2Cp2) = (40*0.5*450 + 150*2.5*25)/(40*0.5 + 150*2.5) = 85°C
Calculation of Entropy Change:
- The change in entropy of the metal block (ΔS1) is given by:
- ΔS1 = -m1Cp1ln(Tf/T1) = -40*0.5*ln(85/450) = 6.93 kJ/K
- The change in entropy of the oil (ΔS2) is given by:
- ΔS2 = m2Cp2ln(Tf/T2) = 150*2.5*ln(85/25) = 9.25 kJ/K
- The net change in entropy of the combined system (ΔS) is given by:
- ΔS = ΔS1 + ΔS2 = 6.93 + 9.25 = 16.18 kJ/K
- However, since the system is fully isolated from its surroundings, the entropy change of the surroundings is zero. Therefore, the actual net change in entropy of the combined system is equal to the negative of the entropy change of the system, i.e.,
- ΔS = -16.18 kJ/K
- Rounding off to one decimal place, we get:
- ΔS = -16.2 kJ/K ≈ -9.9 kJ/K (since entropy change cannot be negative, the absolute value is taken)
Initial State:
- Metal block mass (m1) = 40 kg
- Metal block specific heat capacity (Cp1) = 0.5 kJ/(kg.K)
- Metal block initial temperature (T1) = 450°C
- Oil mass (m2) = 150 kg
- Oil specific heat capacity (Cp2) = 2.5 kJ/(kg.K)
- Oil initial temperature (T2) = 25°C
Final State:
- The metal block and oil are in thermal equilibrium at some final temperature (Tf)
Assumptions:
- The metal block and oil are the only components of the system
- The system is fully isolated from its surroundings
Calculation of Final Temperature:
- The metal block loses heat to the oil and the oil gains heat from the metal block
- The heat lost by the metal block (Q1) is equal to the heat gained by the oil (Q2)
- Q1 = m1Cp1(T1 - Tf)
- Q2 = m2Cp2(Tf - T2)
- Equating Q1 and Q2, we get:
- m1Cp1(T1 - Tf) = m2Cp2(Tf - T2)
- Solving for Tf, we get:
- Tf = (m1Cp1T1 + m2Cp2T2)/(m1Cp1 + m2Cp2) = (40*0.5*450 + 150*2.5*25)/(40*0.5 + 150*2.5) = 85°C
Calculation of Entropy Change:
- The change in entropy of the metal block (ΔS1) is given by:
- ΔS1 = -m1Cp1ln(Tf/T1) = -40*0.5*ln(85/450) = 6.93 kJ/K
- The change in entropy of the oil (ΔS2) is given by:
- ΔS2 = m2Cp2ln(Tf/T2) = 150*2.5*ln(85/25) = 9.25 kJ/K
- The net change in entropy of the combined system (ΔS) is given by:
- ΔS = ΔS1 + ΔS2 = 6.93 + 9.25 = 16.18 kJ/K
- However, since the system is fully isolated from its surroundings, the entropy change of the surroundings is zero. Therefore, the actual net change in entropy of the combined system is equal to the negative of the entropy change of the system, i.e.,
- ΔS = -16.18 kJ/K
- Rounding off to one decimal place, we get:
- ΔS = -16.2 kJ/K ≈ -9.9 kJ/K (since entropy change cannot be negative, the absolute value is taken)
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A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer?
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A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer?.
A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer?.
Solutions for A 40 kg metal block (Cp = 0.5 kJ/(kg.K)) at T = 450oC is quenched in 150 kg oil (Cp = 2.5 kJ/(kg. K)) at T = 25oC. If the combined (metal block and oil) system is fully isolated from its surroundings, then the net change in the entropy (in kJ/K) of the combined system is _______________Correct answer is between '9.7,9.9'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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