Question

A heat pump is used to heat a building. The external temperature is less than the internal temperature. The pump’s coefficient of performance is 3.8, and the heat pump delivers 7.54 MJ as heat to the building each hour. If the heat pump is a Carnot engine working in reverse, at what rate must work be done to run it?

Short answer

The rate of work that must be done to run the reverse Carnot engine is 436.4 W

Step-by-step solution

The given data

Pump performance is${\mathrm{K}}_{\mathrm{performance}}=3.8$

Heat delivered by the pump is$Q_H=7.54\mathrm{MJ}\left(\frac{{10}^6\mathrm{J}}{1\mathrm{MJ}}\right)=7.54\times {10}^6\mathrm{J}$

Understanding the concept of the Carnot engine

We use the concept of coefficient of performance of the heat pump and coefficient of performance of Carnot’s engine. Using, the coefficient of performance of the heat pump, we can get the work done by the engine to run it properly.

Formula:

The formula of coefficient of performance,${\mathrm{K}}_{\mathrm{heatpump}}=\frac{{\mathrm{Q}}_{\mathrm{H}}}{\mathrm{W}}$ (1)

are heat energy and work done respectively.

If the heat pump works in a reverse manner to Carnot’s engine, then

$K_{heatpump}=1+K_{engine}$ (2)

Calculation of the rate of work to be done by the engine

Heat Pump works in a reverse manner to Carnot’s engine, so using equation (2) in equation (1), we get the work that must be done by the Carnot engine can be given as:

$$\begin{gathered} \frac{{\mathrm{Q}}_{\mathrm{H}}}{\mathrm{W}}=1+{\mathrm{K}}_{\mathrm{engine}} \\ \frac{{\mathrm{Q}}_{\mathrm{H}}/\mathrm{t}}{\mathrm{W}/\mathrm{T}}=1+{\mathrm{K}}_{\mathrm{engine}} \\ \frac{{\mathrm{Q}}_{\mathrm{H}}/\mathrm{t}}{\left(1+{\mathrm{K}}_{\mathrm{engine}}\right)}\mathrm{W}/\mathrm{t} \\ \frac{\mathrm{W}}{\mathrm{t}}=\frac{7.54\times {10}^6\mathrm{J}\times {\displaystyle \frac{1}{3600\mathrm{s}}}}{1+3.8} \\ \frac{\mathrm{W}}{\mathrm{t}}=436.4\mathrm{W} \end{gathered}$$

Hence, the rate of work done is 436.4 W