Question

The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)

Short answer

Answer: The maximum amount of ice that can be produced in 2 hours is approximately 30.4 kg.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin to work with them in thermodynamic calculations. To do this, we will add 273.15 to each temperature. \(T_H = 20.0^\circ C + 273.15 K = 293.15 K\) \(T_C = -5.0^\circ C + 273.15 K = 268.15 K\) \(T_W = 8.0^\circ C + 273.15 K = 281.15 K\)

Calculate the Coefficient of performance (COP) for the refrigerator

For a reversible refrigerator, the Coefficient of Performance (COP) is given by the following formula: \(\text{COP} = \frac{T_C}{T_H-T_C}\) Plug in the values of \(T_H\) and \(T_C\) to get the COP. \(\text{COP} = \frac{268.15}{293.15 - 268.15} = \frac{268.15}{25} = 10.726\)

Determine the heat removed from the cold reservoir

Since we know the power output of the refrigerator (148 W), we can calculate the heat removed from the cold reservoir using the formula: \(Q_C = \text{COP} \times W = 10.726 \times 148 W = 1587.448 W\) Now we need to find the heat removed in 2 hours, so we multiply this by the time: \(Q_C = 1587.448 W \times 2 h \times 3600 s/h = 11445504.64 J\)

Calculate the heat removed from the water to produce ice

To produce ice from the water, firstly, we need to cool down the water from \(281.15 K\) to \(273.15 K\), and then freeze it. The specific heat capacity of water (\(c_w\)) is \(4.18 \times 10^3 \mathrm{J/kg K}\) and the heat of fusion for water (\(L\)) is \(3.34 \times 10^5 \mathrm{J/kg}\). Let's denote the mass of water to be produced into ice by \(m\). Then, we have: \(\text{Heat removed to cool down water} = m \cdot c_w \cdot (T_W - 273.15 K )\) \(\text{Heat removed to freeze water} = m \cdot L \) The total heat removed can be found by summing these two terms, thus: \(Q_C = m \cdot (c_w \cdot 8 K + L)\)

Calculate the mass of ice produced

Now we can find the mass of ice produced by solving for \(m\): \(m = \frac{Q_C}{c_w \cdot 8 K + L} = \frac{11445504.64 J}{(4.18 \times 10^3 J/kgK) \cdot 8 K + 3.34 \times 10^5 J/kg} = 30.4 kg\) So, the maximum amount of ice that can be produced in 2 hours is approximately 30.4 kg.