GATE ME 2014 Set 4 — Question 23
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Thermodynamics → Laws & Properties → Carnot Cycle & Entropy
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Question
A reversed Carnot cycle refrigerator maintains a temperature of –5° C. The ambient air temperature is 35°C. The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watt) required to pump this heat out continuously is __________
Correct answer
370 to 375
Solution
The refrigerator operates on a reversed Carnot cycle.Given:
Lower temperature (refrigerator temperature), K.
Higher temperature (ambient temperature), K.
Heat gained by the refrigerator (refrigeration effect), kJ/s = 2500 W.The Coefficient of Performance (COP) of a refrigerator is defined as:For a reversible cycle like the Carnot cycle, the COP can also be expressed in terms of the absolute temperatures of the heat reservoirs:Equating the two expressions for the COP:We need to find the power required, which is the work input . Rearranging the formula to solve for :Now, substitute the given values into the equation:The power required to pump the heat out is approximately 373.13 W. This value falls within the official answer range of 370 to 375.
Lower temperature (refrigerator temperature), K.
Higher temperature (ambient temperature), K.
Heat gained by the refrigerator (refrigeration effect), kJ/s = 2500 W.The Coefficient of Performance (COP) of a refrigerator is defined as:For a reversible cycle like the Carnot cycle, the COP can also be expressed in terms of the absolute temperatures of the heat reservoirs:Equating the two expressions for the COP:We need to find the power required, which is the work input . Rearranging the formula to solve for :Now, substitute the given values into the equation:The power required to pump the heat out is approximately 373.13 W. This value falls within the official answer range of 370 to 375.
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