Consider a Carnot engine with a hot source kept at 500 K. From the hot source, 100 J of energy (heat) is withdrawn at 500 K. The cold sink is kept at 300 K. The efficiency of the Carnot engine is ___________ (rounded off to one decimal place).
Consider a Carnot engine with a hot source kept at 500 K. From the hot source, 100 J of energy (heat) is withdrawn at 500 K. The cold sink is kept at 300 K. The efficiency of the Carnot engine is ___________ (rounded off to one decimal place).
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Solution and Explanation
The efficiency (\( \eta \)) of a Carnot engine is determined solely by the temperatures of the hot and cold reservoirs: \[ \eta = 1 - \frac{T_C}{T_H} \] where \( T_H \) is the absolute temperature of the hot source and \( T_C \) is the absolute temperature of the cold sink. Given: Temperature of the hot source, \( T_H = 500 \) K Temperature of the cold sink, \( T_C = 300 \) K Substitute these values into the formula: \[ \eta = 1 - \frac{300 \, \text{K}}{500 \, \text{K}} = 1 - 0.6 = 0.4 \] The efficiency of the Carnot engine is 0.4. Rounded off to one decimal place, the answer remains 0.4.
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