Question

A household refrigerator that has a power input of 450 \(\mathrm{W}\) and a COP of 1.5 is to cool 5 large watermelons, \(10 \mathrm{kg}\) each, to \(8^{\circ} \mathrm{C}\). If the watermelons are initially at \(28^{\circ} \mathrm{C}\), determine how long it will take for the refrigerator to cool them. The watermelons can be treated as water whose specific heat is \(4.2 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) Is your answer realistic or optimistic? Explain.

Short answer

Short Answer: To determine the time it takes for the refrigerator to cool the watermelons, first calculate the amount of heat to be removed from the watermelons using heat removed(Q) = mass(m) × specific heat(c) × temperature difference(ΔT). Next, calculate the heat removed by the refrigerator using Q_removed = COP × P_input. Finally, divide the total amount of heat to be removed by the heat removal rate to find the time required to cool the watermelons, t = Q_total / Q_removed. The calculated time might be optimistic since refrigerators are not perfectly efficient, and other variables could affect the cooling rate.

Step-by-step solution

Find the amount of heat to be removed from the watermelons

We are given the initial temperature, the final temperature, the mass of the watermelons, and the specific heat of water. We can use the following formula to find the amount of heat that needs to be removed from the watermelons: Heat removed (Q) = mass (m) × specific heat (c) × temperature difference (ΔT) The temperature difference (ΔT) is the difference between the initial temperature (Ti) and the final temperature (Tf): ΔT = Ti - Tf = 28°C - 8°C = 20°C Now, we can find the heat to be removed: Q = m × c × ΔT = (5 × 10 kg) × (4.2 kJ/kg°C) × (20°C)

Calculate the heat removed by the refrigerator

We are given the power input (P) of the refrigerator and its COP. We can use the formula: COP = Q_removed / P_input Where Q_removed is the heat removed by the refrigerator, and P_input is the power input. We can rearrange the formula for Q_removed: Q_removed = COP × P_input = 1.5 × 450 W Since 1 W = 1 J/s, Q_removed = 675 J/s

Calculate the time required to cool the watermelons

Now, we can find the time it takes for the refrigerator to remove the heat from the watermelons. We just need to divide the total amount of heat to be removed by the heat removal rate: time (t) = Q_total / Q_removed = (5 × 10 kg) × (4.2 kJ/kg°C) × (20°C) / (675 J/s) Make sure the units are consistent, converting kJ to J: t = (5 × 10 kg) × (4200 J/kg°C) × (20°C) / (675 J/s)

Evaluate the answer's realism

After calculating the time in Step 3, we need to evaluate whether the answer is realistic or optimistic. Since refrigerators are not perfectly efficient and there may be other variables that could affect the cooling rate, the calculated time might be optimistic.