Question

An ideal refrigerator has freezer at a temperature of $-13^{\circ}\text{C}$. The coefficient of performance of the engine is $5$. The temperature of the air (to which heat is rejected) is

• A. 320^{\circ}\text{C}
• B. 39^{\circ}\text{C}
• C. 325\text{ K}
• D. 325^{\circ}\text{C}

Hint:

Always perform thermodynamic calculations using absolute temperature in Kelvin.

Answer 1

The Correct Option is C

textbf{Step 1:} Identify the formula for the coefficient of performance ($\beta$) of an ideal refrigerator: \[ \beta = \frac{T_2}{T_1 - T_2} \] textbf{Step 2:} Convert the given freezer temperature ($T_2$) to Kelvin: \[ T_2 = -13 + 273 = 260\text{ K} \] textbf{Step 3:} Substitute $\beta = 5$ and $T_2 = 260$ into the formula: \[ 5 = \frac{260}{T_1 - 260} \] textbf{Step 4:} Solve for $T_1$: \[ 5(T_1 - 260) = 260 \] \[ T_1 - 260 = \frac{260}{5} = 52 \] \[ T_1 = 312\text{ K} \] textbf{Step 5:} Convert $T_1$ back to Celsius: \[ T_1 = 312 - 273 = 39^{\circ}\text{C} \]