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A Carnot cycle works between 27 ºC and 327 ºC. If engine produces 100 kJ of work the entropy change during heat rejection is .
- a)0.33 kJ/kJk
- b)1 kJ/ kgk
- c)0.5 kJ/kgk
- d)1.5 kK/kgk
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A Carnot cycle works between 27 C and 327 C. If engine produces 100 kJ...
**Explanation:**
To find the entropy change during heat rejection in a Carnot cycle, we can use the formula:
\(\Delta S = \frac{Q}{T}\)
Where:
- \(\Delta S\) is the entropy change
- \(Q\) is the heat transfer during the process
- \(T\) is the temperature at which the heat transfer occurs
In the Carnot cycle, the heat transfer during the heat rejection process is equal to the work output of the engine. So, \(Q = 100 \, kJ\).
The temperature at which the heat transfer occurs during the heat rejection process is the lower temperature of the cycle, which is 27°C. We need to convert this temperature to Kelvin by adding 273 to it.
\(T = 27°C + 273 = 300 \, K\)
Now we can plug in the values into the formula:
\(\Delta S = \frac{100 \, kJ}{300 \, K}\)
To simplify the units, we can convert kJ to J:
\(\Delta S = \frac{100 \, kJ \times 1000}{300 \, K}\)
\(\Delta S = \frac{100000 \, J}{300 \, K}\)
\(\Delta S = 333.33 \, J/K\)
To convert the units to kJ/kJ·K, we divide by 1000:
\(\Delta S = 0.33 \, kJ/kJ·K\)
Therefore, the entropy change during heat rejection in the Carnot cycle is 0.33 kJ/kJ·K, which corresponds to option A.
To find the entropy change during heat rejection in a Carnot cycle, we can use the formula:
\(\Delta S = \frac{Q}{T}\)
Where:
- \(\Delta S\) is the entropy change
- \(Q\) is the heat transfer during the process
- \(T\) is the temperature at which the heat transfer occurs
In the Carnot cycle, the heat transfer during the heat rejection process is equal to the work output of the engine. So, \(Q = 100 \, kJ\).
The temperature at which the heat transfer occurs during the heat rejection process is the lower temperature of the cycle, which is 27°C. We need to convert this temperature to Kelvin by adding 273 to it.
\(T = 27°C + 273 = 300 \, K\)
Now we can plug in the values into the formula:
\(\Delta S = \frac{100 \, kJ}{300 \, K}\)
To simplify the units, we can convert kJ to J:
\(\Delta S = \frac{100 \, kJ \times 1000}{300 \, K}\)
\(\Delta S = \frac{100000 \, J}{300 \, K}\)
\(\Delta S = 333.33 \, J/K\)
To convert the units to kJ/kJ·K, we divide by 1000:
\(\Delta S = 0.33 \, kJ/kJ·K\)
Therefore, the entropy change during heat rejection in the Carnot cycle is 0.33 kJ/kJ·K, which corresponds to option A.
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A Carnot cycle works between 27 C and 327 C. If engine produces 100 kJ of work the entropy change during heat rejection is .a)0.33 kJ/kJkb)1 kJ/ kgkc)0.5 kJ/kgkd)1.5 kK/kgkCorrect answer is option 'A'. 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 Carnot cycle works between 27 C and 327 C. If engine produces 100 kJ of work the entropy change during heat rejection is .a)0.33 kJ/kJkb)1 kJ/ kgkc)0.5 kJ/kgkd)1.5 kK/kgkCorrect answer is option 'A'. 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 Carnot cycle works between 27 C and 327 C. If engine produces 100 kJ of work the entropy change during heat rejection is .a)0.33 kJ/kJkb)1 kJ/ kgkc)0.5 kJ/kgkd)1.5 kK/kgkCorrect answer is option 'A'. Can you explain this answer?.
A Carnot cycle works between 27 C and 327 C. If engine produces 100 kJ of work the entropy change during heat rejection is .a)0.33 kJ/kJkb)1 kJ/ kgkc)0.5 kJ/kgkd)1.5 kK/kgkCorrect answer is option 'A'. 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 Carnot cycle works between 27 C and 327 C. If engine produces 100 kJ of work the entropy change during heat rejection is .a)0.33 kJ/kJkb)1 kJ/ kgkc)0.5 kJ/kgkd)1.5 kK/kgkCorrect answer is option 'A'. 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 Carnot cycle works between 27 C and 327 C. If engine produces 100 kJ of work the entropy change during heat rejection is .a)0.33 kJ/kJkb)1 kJ/ kgkc)0.5 kJ/kgkd)1.5 kK/kgkCorrect answer is option 'A'. Can you explain this answer?.
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