In a very mild winter climate, a heat pump has heat transfer from an environment at 5.00 ºC to one at 35.0 ºC. What is the best possible coefficient of performance for these temperatures? Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics.

The reciprocal of efficiency of a pump is called the Coefficient of performance.

The temperature of a cold environment\({T_c} = 5\;^\circ {\rm{C}} = {\rm{278}}\;{\rm{K}}\)

The temperature of the warm environment \({T_h} = 35\;^\circ {\rm{C}} = {\rm{308}}\;{\rm{K}}\)

The efficiency of the Carnot engine \(\eta = 1 - \frac{{{T_c}}}{{{T_h}}}\)

Here,

\(\eta \)- efficiency of the engine.

\({T_c}\)- the temperature of the cold environment

\({T_h}\)- temperature of the warmenvironment

\({\beta _{hp}}\) - heat pump’s performance coefficient

Maximum efficiency can be calculated by calculating Carnot efficiency as

\(\begin{array}{}\eta & = 1 - \frac{{{T_c}}}{{{T_h}}}\\ &= 1 - \frac{{278}}{{308}}\\ &= 0.097\end{array}\)

Heat pump’s performance coefficient is

\(\begin{array}{c}{\beta _{hp}} &= 1/\eta \\ &= 1/0.097\\ &= 10.31\end{array}\)

Therefore, the heat pump’s performance coefficient is \(10.31\).