Question

If the coefficient of performance of a refrigerator is 5 and the temperature inside it is -20 °C, then the temperature of its surroundings is

• A. 21.6 °C
• B. 30.6 °C
• C. 40.6 °C
• D. 10.6 °C

Hint:

When working with thermodynamic formulas like the coefficient of performance of refrigerators or heat engines, always remember to convert temperatures to the absolute Kelvin scale. A common mistake is using Celsius directly, which leads to incorrect results. The Kelvin scale starts from absolute zero (0 K = -273.15 °C), and temperature differences are the same in both Celsius and Kelvin, but the absolute values are crucial for ratios and products in formulas.

Answer 1

The Correct Option is B

Step 1: Identify the given information and the quantity to be calculated.
Given: \begin{itemize} \item Coefficient of performance of the refrigerator ($\beta$) = 5. \item Temperature inside the refrigerator (cold reservoir temperature, $T_c$) = -20 °C. \end{itemize} We need to calculate the temperature of its surroundings (hot reservoir temperature, $T_h$). Step 2: Convert the given temperature to the Kelvin scale.
The formula for the coefficient of performance uses absolute temperatures (Kelvin). $T_c = -20 \text{ °C} = (-20 + 273.15) \text{ K} = 253.15 \text{ K}$ Step 3: Recall the formula for the coefficient of performance of a refrigerator.
The coefficient of performance ($\beta$) of a refrigerator is given by: $\beta = \frac{T_c}{T_h - T_c}$ where $T_c$ is the absolute temperature of the cold reservoir and $T_h$ is the absolute temperature of the hot reservoir. Step 4: Substitute the known values into the formula and solve for $T_h$.
$5 = \frac{253.15 \text{ K}}{T_h - 253.15 \text{ K}}$ Rearrange the equation to solve for $T_h$: $5 (T_h - 253.15) = 253.15$ $5T_h - 5 \times 253.15 = 253.15$ $5T_h - 1265.75 = 253.15$ $5T_h = 253.15 + 1265.75$ $5T_h = 1518.9$ $T_h = \frac{1518.9}{5}$ $T_h = 303.78 \text{ K}$ Step 5: Convert the calculated temperature back to Celsius.
$T_h \text{ (in °C)} = T_h \text{ (in K)} - 273.15$
$T_h \text{ (in °C)} = 303.78 - 273.15$
$T_h \text{ (in °C)} = 30.63 \text{ °C}$
Step 6: Compare the calculated temperature with the given options.
The calculated temperature of the surroundings is approximately 30.6 °C, which matches option (2). The final answer is $\boxed{\text{30.6 °C}}$.