Question

A heat pump with a COP of 3.2 is used to heat a perfectly sealed house (no air leaks). The entire mass within the house (air, furniture, etc.) is equivalent to \(1200 \mathrm{kg}\) of air. When running, the heat pump consumes electric power at a rate of \(5 \mathrm{kW}\). The temperature of the house was \(7^{\circ} \mathrm{C}\) when the heat pump was turned on. If heat transfer through the envelope of the house (walls, roof, etc.) is negligible, the length of time the heat pump must run to raise the temperature of the entire contents of the house to \(22^{\circ} \mathrm{C}\) is (a) \(13.5 \mathrm{min}\) (b) \(43.1 \mathrm{min}\) \((c) 138 \min\) \((d) 18.8 \mathrm{min}\) \((e) 808 \mathrm{min}\)

Short answer

#Short Answer# (a) 13.5 minutes

Step-by-step solution

Calculate the change in temperature of the air inside the house.

We are given that the initial temperature is \(7^{\circ} \mathrm{C}\) and the final temperature is \(22^{\circ} \mathrm{C}\). To find the change in temperature, we simply subtract the initial temperature from the final temperature: \(\Delta T = T_f-T_i = 22 - 7 = 15^{\circ} \mathrm{C}\).

Calculate the change in internal energy of the air.

Now, we need to find the change in internal energy of the air by using the formula: \(\Delta U = mC_v * \Delta T\) The mass of the air is given as \(1200 \mathrm{kg}\), and the specific heat capacity at constant volume for air is \(C_v = 0.718 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\). Using these values and the change in temperature calculated in Step 1, we can now find the change in internal energy: \(\Delta U = 1200 \cdot 0.718 \cdot 15 = 12954\, \mathrm{kJ}\).

Calculate the heat transferred to the house.

We can now find the heat transferred to the house (\(Q_H\)) using the COP formula: \(COP = \frac{Q_H}{W}\) We are given the COP value as \(3.2\) and the electric power consumed by the heat pump as \(5\, \mathrm{kW}\). We can rearrange the formula to find \(Q_H\): \(Q_H = COP \cdot W = 3.2 \cdot 5 = 16\, \mathrm{kW}\) We must convert this value to \(\mathrm{kJ/s}\): \(Q_H = 16,000\frac{\mathrm{J}}{\mathrm{s}} = 16\frac{\mathrm{kJ}}{\mathrm{s}}\)

Calculate the time required for the heat pump to raise the temperature.

Finally, we can find the time required for the heat pump to raise the temperature of the house by dividing the change in internal energy by the heat transfer rate: \(time = \frac{\Delta U}{Q_H}=\frac{12954}{16} = 809.625\, \mathrm{s}\) Let's convert this to \(\mathrm{min}\): \(time = \frac{809.625}{60} = 13.49375\, \mathrm{min}\) The closest option to our calculated value is (a) \(13.5\, \mathrm{min}\), which is the correct answer.