Question

During an experiment conducted in a room at \(25^{\circ} \mathrm{C},\) a laboratory assistant measures that a refrigerator that draws \(2 \mathrm{kW}\) of power has removed \(30,000 \mathrm{kJ}\) of heat from the refrigerated space, which is maintained at \(-30^{\circ} \mathrm{C} .\) The running time of the refrigerator during the experiment was 20 min. Determine if these measurements are reasonable.

Short answer

Answer: No, the measurements are not reasonable as the calculated Coefficient of Performance (COP) is 12.5, which is outside the typical range of 3 to 6 for a refrigerator.

Step-by-step solution

Convert all units to the same base unit

The power input is given in kW and the heat removed is given in kJ. Convert both values to joules to have the same base units. Additionally, convert the running time to seconds. Power input: \(2 \mathrm{kW} = 2000 \mathrm{W}\) Heat removed: \(30,000 \mathrm{kJ} = 30,000,000 \mathrm{J}\) Running time: \(20 \mathrm{min} = 1200 \mathrm{s}\)

Calculate the work done by the refrigerator through power input

Use the formula \(\mathrm{W}_{\mathrm{in}} = \mathrm{P}_{\mathrm{input}} \cdot \mathrm{t}\), where \(\mathrm{P}_{\mathrm{input}}\) is the power input and \(\mathrm{t}\) is the running time. \(\mathrm{W}_{\mathrm{in}} = (2000 \mathrm{W})(1200 \mathrm{s}) = 2,400,000 \mathrm{J}\)

Calculate the Coefficient of Performance (COP)

Use the COP formula, \(\mathrm{COP} = \frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{W}_{\mathrm{in}}}\) \(\mathrm{COP} = \frac{30,000,000 \mathrm{J}}{2,400,000 \mathrm{J}} = 12.5\)

Compare the calculated COP with typical values for a refrigerator

A common range for the Coefficient of Performance of a refrigerator is between 3 and 6. The calculated COP of 12.5 is outside of this range, which suggests that the given measurements may not be reasonable.