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A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will be
- a)372 K, 310 K
- b)181 K, 150 K
- c)472 K, 410 K
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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A Carnot reversible engine converts 1/6 of heat input work. When the t...
To solve this problem, we will use the equations for the efficiency of a Carnot engine and the relationship between temperature and efficiency when the temperature of the sink is changed.
1. Efficiency of a Carnot engine:
The efficiency (η) of a Carnot engine is given by the equation:
η = 1 - (T_sink / T_source)
where T_sink is the temperature of the sink and T_source is the temperature of the source.
2. Heat input work conversion:
Given that the Carnot engine converts 1/6 of the heat input work, we can write the equation:
η = 1/6
3. Change in efficiency with temperature:
When the temperature of the sink is reduced by ΔT, the new efficiency (η') can be calculated using the equation:
η' = 1 - ((T_sink - ΔT) / T_source)
4. Calculate the temperatures:
We need to find the temperatures of the source and sink. Substituting the given values into the equations above, we can solve for T_sink and T_source.
First, using equation 2, we have:
1/6 = 1 - (T_sink / T_source)
Simplifying and rearranging the equation, we get:
5/6 = T_sink / T_source
Next, using equation 3, we have:
1/3 = 1 - ((T_sink - 62) / T_source)
Simplifying and rearranging the equation, we get:
2/3 = (T_sink - 62) / T_source
5. Solve the equations:
We can solve the above two equations simultaneously to find the values of T_sink and T_source.
From equation 5, we can rewrite it as:
2/3 = (5/6 - 62/T_source)
Simplifying and rearranging the equation, we get:
2/3 = (5T_source - 372) / (6T_source)
Cross-multiplying, we get:
4T_source = 5T_source - 372
Rearranging the equation, we get:
T_source = 372
Substituting this value into equation 4, we get:
5/6 = T_sink / 372
Rearranging the equation, we get:
T_sink = 310
Therefore, the temperatures of the source and sink are 372 K and 310 K, respectively. Thus, the correct answer is option A: 372 K, 310 K.
1. Efficiency of a Carnot engine:
The efficiency (η) of a Carnot engine is given by the equation:
η = 1 - (T_sink / T_source)
where T_sink is the temperature of the sink and T_source is the temperature of the source.
2. Heat input work conversion:
Given that the Carnot engine converts 1/6 of the heat input work, we can write the equation:
η = 1/6
3. Change in efficiency with temperature:
When the temperature of the sink is reduced by ΔT, the new efficiency (η') can be calculated using the equation:
η' = 1 - ((T_sink - ΔT) / T_source)
4. Calculate the temperatures:
We need to find the temperatures of the source and sink. Substituting the given values into the equations above, we can solve for T_sink and T_source.
First, using equation 2, we have:
1/6 = 1 - (T_sink / T_source)
Simplifying and rearranging the equation, we get:
5/6 = T_sink / T_source
Next, using equation 3, we have:
1/3 = 1 - ((T_sink - 62) / T_source)
Simplifying and rearranging the equation, we get:
2/3 = (T_sink - 62) / T_source
5. Solve the equations:
We can solve the above two equations simultaneously to find the values of T_sink and T_source.
From equation 5, we can rewrite it as:
2/3 = (5/6 - 62/T_source)
Simplifying and rearranging the equation, we get:
2/3 = (5T_source - 372) / (6T_source)
Cross-multiplying, we get:
4T_source = 5T_source - 372
Rearranging the equation, we get:
T_source = 372
Substituting this value into equation 4, we get:
5/6 = T_sink / 372
Rearranging the equation, we get:
T_sink = 310
Therefore, the temperatures of the source and sink are 372 K and 310 K, respectively. Thus, the correct answer is option A: 372 K, 310 K.
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A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will bea)372 K, 310 K b) 181 K, 150 Kc)472 K, 410 Kd)none of theseCorrect answer is option 'A'. Can you explain this answer?
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A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will bea)372 K, 310 K b) 181 K, 150 Kc)472 K, 410 Kd)none of theseCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will bea)372 K, 310 K b) 181 K, 150 Kc)472 K, 410 Kd)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will bea)372 K, 310 K b) 181 K, 150 Kc)472 K, 410 Kd)none of theseCorrect answer is option 'A'. Can you explain this answer?.
A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will bea)372 K, 310 K b) 181 K, 150 Kc)472 K, 410 Kd)none of theseCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will bea)372 K, 310 K b) 181 K, 150 Kc)472 K, 410 Kd)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot reversible engine converts 1/6 of heat input work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes1/3. The temperature of the source and sink will bea)372 K, 310 K b) 181 K, 150 Kc)472 K, 410 Kd)none of theseCorrect answer is option 'A'. Can you explain this answer?.
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