Question:

A heat pump, operating in reversed Carnot cycle, maintains a steady air temperature of 300 K inside an auditorium. The heat pump receives heat from the ambient air. The ambient air temperature is 280 K. Heat loss from the auditorium is 15 kW. The power consumption of the heat pump is _________ kW (rounded off to 2 decimal places).

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For a Carnot heat pump, the coefficient of performance (COP) is the ratio of the heat transferred to the work input. To calculate the power consumption, divide the heat extracted by the COP.
Updated On: Apr 18, 2025
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Solution and Explanation

Given: \[ T_H = 300 \ {K}, \quad T_L = 280 \ {K}, \quad Q_H = 15 \ {kW} \] For a reversed Carnot cycle, the Coefficient of Performance (COP) of the heat pump is: \[ {COP}_{{HP}} = \frac{T_H}{T_H - T_L} = \frac{300}{300 - 280} = \frac{300}{20} = 15 \] Power consumption of the heat pump is given by: \[ {COP}_{{HP}} = \frac{Q_H}{W} \Rightarrow W = \frac{Q_H}{{COP}_{{HP}}} = \frac{15}{15} = 1.00 \ {kW} \] 
Correct Answer: 1.00 kW

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