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The definition of a refrigerator's coefficient of performance can be written as

$K=\frac{Q_c}{W_{in}}=\frac{Q_c/t}{W_{in}/t}=\frac{P_c}{P_{in}}$

Here $P_c$is power of cooling and $P_{in}$is electrical power used. As a result, we can calculate the pace at which the refrigerator takes out energy as

$P_c=P_{in}K$

The motor's power input, on the other hand, is given as

$P_{in}=畏P_e.$

$P_e$ is electrical power consumed.

where $P_e$is the electric power of the motor and$畏$ is its efficiency. So, we put that in our previous result and find

$P_c=P_{in}K=畏K{P}_e$

Numerically, we will have

$$\begin{gathered} P_c=0.95鈰�6鈰�115 \\ {P}_c鈮�660\mathrm{W}. \end{gathered}$$

The best explanation is that the total heating power discharged into the environment equals the sum of the cooling and motor power. Because the loss in an electric motor is likewise transformed to heat, we shall analyse its electric power for the motor.. That is,

$$\begin{gathered} P_h=P_c+P_e \\ {P}_h鈮�660+115 \\ {P}_h鈮�780\mathrm{W}. \end{gathered}$$

The differences are due to the fact that we are rounding to only have one meaningful figure after the decimal point.