Question

A typical new household refrigerator consumes about \(680 \mathrm{kWh}\) of electricity per year and has a coefficient of performance of \(1.4 .\) The amount of heat removed by this refrigerator from the refrigerated space per year is \((a) 952 \mathrm{MJ} / \mathrm{yr}\) (b) 1749 MJ/yr \((c) 2448\) MJ/yr \((d) 3427 \mathrm{MJ} / \mathrm{yr}\) \((e) 4048 \mathrm{MJ} / \mathrm{yr}\)

Short answer

Answer: (d) 3427 MJ/yr.

Step-by-step solution

Recall the formula for Coefficient of Performance (COP)

The coefficient of performance (COP) for a refrigerator is given by: COP = \(\frac{Q_L}{W}\), where \(Q_L\) is the amount of heat removed from the refrigerated space (in J), and W is the work done by the refrigerator (in J).

Convert the energy consumed by the refrigerator to Joules

We are given the energy consumed by the fridge in kilowatt-hours (kWh) per year. First, we need to convert this to Joules to work with the formula above. 1 kWh = 3.6 * 10^6 J, so the energy consumed per year in Joules is: _energy_consumed_J = \(680 kWh \times 3.6 \times 10^6 J/kWh = 2.448 \times 10^9 J\).

Calculate the amount of heat removed from the refrigerated space

Now we can use the COP formula to find the amount of heat (\(Q_L\)) removed from the refrigerated space: COP = \(\frac{Q_L}{W}\) => \(Q_L = COP \times W\) _Q_removed = 1.4 × 2.448 × 10^9 J = 3.4272 × 10^9 J.

Convert the amount of heat removed to MJ and compare to the given options

We need to present the answer in Mega Joules per year (MJ/yr). To convert Joules to Mega Joules, divide by 10^6: _Q_removed_MJ = \(\frac{3.4272 \times 10^9 J}{10^6} = 3427.2 MJ/yr\). Comparing this value to the given options, it is closest to option (d) with 3427 MJ/yr. Therefore, the correct answer for this exercise is: (d) 3427 MJ/yr.