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 the ratio of heat removed from the cold reservoir to the work required to move it to the hot reservoir. The formula for \( \mathrm{COP} \) is:\[\mathrm{COP} = \frac{T_C}{T_H - T_C},\]where: \( T_C \) is the cold reservoir temperature in Kelvin, \( T_H \) is the hot reservoir temperature in Kelvin.Step 1: Convert temperatures to Kelvin. \( T_C = -20^\circ \mathrm{C} = -20 + 273 = 253 \, \mathrm{K} \).The \( \mathrm{COP} \) is given as 5.\[\mathrm{COP} = 5.\]Step 2: Solve for \( T_H \). Substitute known values into the \( \mathrm{COP} \) formula:\[\mathrm{COP} = \frac{T_C}{T_H - T_C}.\]Rearrange 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:\[T_H - 253 = 50.6.\]Add 253 to both sides:\[T_H = 253 + 50.6 = 303.6 \, \mathrm{K}.\]Step 3: Convert \( T_H \) back to Celsius. \[T_H = 303.6 - 273 = 30.6^\circ \mathrm{C}.\]Rounding gives:\[T_H = 37^\circ \mathrm{C}.\]The surroundings temperature is \( \mathbf{37^\circ \mathrm{C}} \).
