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A heat engine has an efficiency η. Temperatures of source and sink are each decreased by 100 K. Then, the efficiency of the engine.
- a)Increases
- b)Decreases
- c)Remains constant
- d)Becomes 1
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A heat engine has an efficiencyη. Temperatures of source and sink ...
Effect of Decreasing Temperatures on Efficiency of Heat Engine
Increasing the efficiency of a heat engine is always desirable as it indicates a better conversion of heat into work. Let's analyze how the efficiency of a heat engine changes when the temperatures of the source and sink are each decreased by 100 K.
Explanation:
1. Efficiency of a Heat Engine:
Efficiency of a heat engine is given by the formula:
\[\eta = 1 - \frac{T_C}{T_H}\]
where \(\eta\) is the efficiency, \(T_C\) is the temperature of the cold reservoir (sink), and \(T_H\) is the temperature of the hot reservoir (source).
2. Impact of Decreasing Temperatures:
When both the source and sink temperatures are decreased by 100 K, the new efficiency of the engine can be calculated using the formula mentioned above. Let's denote the new temperatures as \(T_C' = T_C - 100\) and \(T_H' = T_H - 100\).
Substitute the new temperatures in the efficiency formula:
\[\eta' = 1 - \frac{T_C'}{T_H'} = 1 - \frac{T_C - 100}{T_H - 100}\]
3. Comparison with Initial Efficiency:
Now, compare the new efficiency \(\eta'\) with the initial efficiency \(\eta\):
\[\eta' = 1 - \frac{T_C - 100}{T_H - 100}\]
\[\eta = 1 - \frac{T_C}{T_H}\]
Since the numerator of \(\eta'\) is less than the numerator of \(\eta\) (due to the subtraction of 100 from both temperatures), the fraction \(\frac{T_C - 100}{T_H - 100}\) is greater than \(\frac{T_C}{T_H\). This implies that \(\eta' > \eta\), meaning the efficiency of the engine increases when the temperatures of the source and sink are each decreased by 100 K.
Therefore, the correct answer is option 'A) Increases'.
Increasing the efficiency of a heat engine is always desirable as it indicates a better conversion of heat into work. Let's analyze how the efficiency of a heat engine changes when the temperatures of the source and sink are each decreased by 100 K.
Explanation:
1. Efficiency of a Heat Engine:
Efficiency of a heat engine is given by the formula:
\[\eta = 1 - \frac{T_C}{T_H}\]
where \(\eta\) is the efficiency, \(T_C\) is the temperature of the cold reservoir (sink), and \(T_H\) is the temperature of the hot reservoir (source).
2. Impact of Decreasing Temperatures:
When both the source and sink temperatures are decreased by 100 K, the new efficiency of the engine can be calculated using the formula mentioned above. Let's denote the new temperatures as \(T_C' = T_C - 100\) and \(T_H' = T_H - 100\).
Substitute the new temperatures in the efficiency formula:
\[\eta' = 1 - \frac{T_C'}{T_H'} = 1 - \frac{T_C - 100}{T_H - 100}\]
3. Comparison with Initial Efficiency:
Now, compare the new efficiency \(\eta'\) with the initial efficiency \(\eta\):
\[\eta' = 1 - \frac{T_C - 100}{T_H - 100}\]
\[\eta = 1 - \frac{T_C}{T_H}\]
Since the numerator of \(\eta'\) is less than the numerator of \(\eta\) (due to the subtraction of 100 from both temperatures), the fraction \(\frac{T_C - 100}{T_H - 100}\) is greater than \(\frac{T_C}{T_H\). This implies that \(\eta' > \eta\), meaning the efficiency of the engine increases when the temperatures of the source and sink are each decreased by 100 K.
Therefore, the correct answer is option 'A) Increases'.
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Community Answer
A heat engine has an efficiencyη. Temperatures of source and sink ...
where T1 and T2 are the temperatures of a source and sink respectively.
When T1 and T2 both are decreased by 100 K each, (T1 - T2) stays constant. T1 decreases.
∴ η increases.
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A heat engine has an efficiencyη. Temperatures of source and sink are each decreased by 100 K. Then, the efficiency of the engine.a)Increasesb)Decreasesc)Remains constantd)Becomes 1Correct answer is option 'A'. Can you explain this answer?
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