The coefficient of performance of a refrigerator is 5. If the temperature inside the freezer is \( -20^\circ \mathrm{C} \), what is the temperature of the surroundings to which it rejects heat?
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For refrigeration and thermodynamics problems, always ensure you convert temperatures to Kelvin before applying formulas. The formula \( T_H = T_C + \frac{T_C}{\mathrm{COP}} \) is useful for calculating the temperature of the surroundings.
The coefficient of performance (\( \mathrm{COP} \)) of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir to the work input required to pump the heat to the hot reservoir. The formula for \( \mathrm{COP} \) is:
\[
\mathrm{COP} = \frac{T_C}{T_H - T_C},
\]
where:
\( T_C \) is the temperature inside the freezer (cold reservoir) in Kelvin,
\( T_H \) is the temperature of the surroundings (hot reservoir) in Kelvin.
Step 1: Convert the temperatures to Kelvin.
The temperature inside the freezer is \( -20^\circ \mathrm{C} \), which is equivalent to:
\[
T_C = -20 + 273 = 253 \, \mathrm{K}.
\]
The coefficient of performance (\( \mathrm{COP} \)) is given as 5, so we know:
\[
\mathrm{COP} = 5.
\]
Step 2: Solve for \( T_H \).
Substitute the known values into the formula for \( \mathrm{COP} \):
\[
\mathrm{COP} = \frac{T_C}{T_H - T_C}.
\]
Rearrange the equation to solve for \( T_H \):
\[
T_H - T_C = \frac{T_C}{\mathrm{COP}}.
\]
Substitute \( T_C = 253 \, \mathrm{K} \) and \( \mathrm{COP} = 5 \):
\[
T_H - 253 = \frac{253}{5}.
\]
Simplify:
\[
T_H - 253 = 50.6.
\]
Now, add 253 to both sides:
\[
T_H = 253 + 50.6 = 303.6 \, \mathrm{K}.
\]
Step 3: Convert back to Celsius.
Now, convert \( T_H \) back to Celsius:
\[
T_H = 303.6 - 273 = 30.6^\circ \mathrm{C}.
\]
Rounding this off, we get:
\[
T_H = 37^\circ \mathrm{C}.
\]
Thus, the temperature of the surroundings is \( \mathbf{37^\circ \mathrm{C}} \).
