Question:
A heat engine operates between a cold reservoir at temperature 300 K and a hot reservoir at temperature \(T_1\). It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to the cold reservoir in a cycle. The minimum temperature \(T_1\) of the hot reservoir is:
A heat engine operates between a cold reservoir at temperature 300 K and a hot reservoir at temperature \(T_1\). It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to the cold reservoir in a cycle. The minimum temperature \(T_1\) of the hot reservoir is:
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For heat engine problems, use efficiency \(\eta = \frac{W}{Q_H}\) and compare with Carnot efficiency for limits.
Updated On: Jun 19, 2026
- 450 K
- 400 K
- 500 K
- 350 K
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The Correct Option is C
Solution and Explanation
Step 1: Find work done by engine.
\[ W = Q_H - Q_C = 200 - 120 = 80\,J \]
Step 2: Efficiency of engine.
\[ \eta = \frac{W}{Q_H} = \frac{80}{200} = 0.4 \]
Step 3: Carnot efficiency relation.
For maximum efficiency (minimum \(T_1\)): \[ \eta = 1 - \frac{T_C}{T_H} \]
Step 4: Substitute values.
\[ 0.4 = 1 - \frac{300}{T_1} \]
Step 5: Solve for \(T_1\).
\[ \frac{300}{T_1} = 0.6 \Rightarrow T_1 = 500\,K \]
Step 6: Final conclusion.
Thus, minimum temperature of hot reservoir is \(500\,K\).
Final Answer: \[ \boxed{500\,K} \]
\[ W = Q_H - Q_C = 200 - 120 = 80\,J \]
Step 2: Efficiency of engine.
\[ \eta = \frac{W}{Q_H} = \frac{80}{200} = 0.4 \]
Step 3: Carnot efficiency relation.
For maximum efficiency (minimum \(T_1\)): \[ \eta = 1 - \frac{T_C}{T_H} \]
Step 4: Substitute values.
\[ 0.4 = 1 - \frac{300}{T_1} \]
Step 5: Solve for \(T_1\).
\[ \frac{300}{T_1} = 0.6 \Rightarrow T_1 = 500\,K \]
Step 6: Final conclusion.
Thus, minimum temperature of hot reservoir is \(500\,K\).
Final Answer: \[ \boxed{500\,K} \]
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