Question

An ideal refrigerator has a freezer at a temperature of $-13°\mathrm{C}$, The coefficient of performance of the engine is $5$. The temperature of the air to which heat is rejected will be

• A. $325°\mathrm{C}$
• B. $39°\mathrm{C}$
• C. $325\mathrm{K}$
• D. $320°\mathrm{C}$

Answer 1

The correct option is B

$39°\mathrm{C}$

Explanation for the correct option

Option B: $39°\mathrm{C}$

Step 1: Given data

The temperature of the refrigerator = $-13°\mathrm{C}=-13+273=260\mathrm{K}$

The coefficient of performance of the engine = $\mathrm{\eta }=5$

Step 2: Formula used

The formula for the coefficient of performance of the engine is: $\mathrm{\eta }=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1-{\mathrm{T}}_2}$where

$$\begin{gathered} {\mathrm{T}}_2=\mathrm{low}\mathrm{temperature} \\ {\mathrm{T}}_1=\mathrm{high}\mathrm{temperature} \end{gathered}$$

Step 3: Calculations

The temperature of the air outside will be higher than the temperature of the freezer.

Hence, ${\mathrm{T}}_1=\mathrm{temperature}\mathrm{of}\mathrm{air}$ and ${\mathrm{T}}_2=\mathrm{temperature}\mathrm{of}\mathrm{freezer}=260\mathrm{K}$

Putting the values in the formula for the coefficient of performance

$$\begin{gathered} \Rightarrow 5=\frac{260}{{\mathrm{T}}_1-260} \\ \Rightarrow 5{\mathrm{T}}_1-1300=260 \\ \Rightarrow 5{\mathrm{T}}_1=1560 \\ \Rightarrow {\mathrm{T}}_1=312\mathrm{K} \\ \Rightarrow {\mathrm{T}}_1=312-273=39°\mathrm{C} \end{gathered}$$

Hence the temperature of the air to which heat will be rejected is $39°\mathrm{C}$. Hence this option is correct.