Question

Thermoelectric coolers that plug into the cigarette lighter of a car are commonly available. One such cooler is claimed to cool a \(12-0 z(0.771-1 b m)\) drink from 78 to \(38^{\circ} \mathrm{F}\) or to heat a cup of coffee from 75 to \(130^{\circ} \mathrm{F}\) in about \(15 \mathrm{min}\) in a well-insulated cup holder. Assuming an average COP of 0.2 in the cooling mode, determine ( \(a\) ) the average rate of heat removal from the drink, \((b)\) the average rate of heat supply to the coffee, and ( \(c\) ) the electric power drawn from the battery of the car, all in \(\mathrm{W}\).

Short answer

(a) The average rate of heat removal from the drink is -36.8 W. (b) The average rate of heat supply to the coffee is 53.0 W. (c) The electric power drawn from the car battery for cooling and heating is 184 W and 265 W, respectively.

Step-by-step solution

Convert temperature values from Fahrenheit to Celsius

To perform the calculations, we need to convert the given temperatures from Fahrenheit to Celsius. Use the formula: \(T_C = (T_F - 32) \times (5/9)\). Drink: Initial temperature: \(T_{1(di)} = (78 - 32) \times (5/9) = 25.6^\circ\text{C}\) Final temperature: \(T_{2(di)} = (38 - 32) \times (5/9) = 3.33^\circ\text{C}\) Coffee: Initial temperature: \(T_{1(co)} = (75 - 32) \times (5/9) = 23.9^\circ\text{C}\) Final temperature: \(T_{2(co)} = (130 - 32) \times (5/9) = 54.4^\circ\text{C}\)

Determine the mass of the drink

We're given the volume, 12 oz, which needs to be converted to liters and then multiplied by the density of water to find the mass: Volume (in liters): \(V = 12\,\text{oz} \times 0.0295735 = 0.354\,\text{L}\) Density of water: \(\rho = 1000\,\text{kg/m}^3\) Mass of drink: \(m_{di} = \rho \times V = 1000\,\text{kg/m}^3 \times 0.354\,\text{L} = 0.354\,\text{kg}\)

Calculate the heat transfer for the drink and coffee

Use the specific heat formula to find the heat transfer for both the drink and the coffee: \(Q = mc\Delta T\). The specific heat capacity of water, \(c\), is approximately \(4.18\,\text{kJ/kgK}\). Drink: \(Q_{di} = m_{di}c(T_{2(di)} - T_{1(di)}) = 0.354\,\text{kg} \times 4.18\,\text{kJ/kgK} \times (3.33 - 25.6)^\circ\text{C} = -33.12\,\text{kJ}\) Coffee: \(Q_{co} = m_{co}c(T_{2(co)} - T_{1(co)}) = 0.354\,\text{kg} \times 4.18\,\text{kJ/kgK} \times (54.4 - 23.9)^\circ\text{C} = 47.67\,\text{kJ}\)

Determine the time in seconds and calculate the average rates of heat transfer

We are given the time of 15 minutes for both cooling and heating processes. Convert this time to seconds: Time: \(t = 15\,\text{min} \times 60\,\text{s/min} = 900\,\text{s}\) Average rate of heat removal from the drink: \(q_{di} = \frac{Q_{di}}{t} = \frac{-33.12\,\text{kJ}}{900\,\text{s}} = -36.8\,\text{W}\) Average rate of heat supply to the coffee: \(q_{co} = \frac{Q_{co}}{t} = \frac{47.67\,\text{kJ}}{900\,\text{s}} = 53.0\,\text{W}\)

Calculate the electric power drawn from the battery

We are told that the COP in cooling mode is 0.2. Use this information to find the electric power drawn from the battery, \(W_{elec}\), for both the cooling and heating processes: For cooling the drink: \(COP = \frac{q_{di}}{W_{elec(di)}}\), so \(W_{elec(di)} = \frac{-q_{di}}{COP} = \frac{36.8\,\text{W}}{0.2} = 184\,\text{W}\) For heating the coffee, we assume the heating COP to be the same as the cooling COP: \(COP_{heating} \approx COP_{cooling}\), so \(W_{elec(co)} = \frac{q_{co}}{COP} = \frac{53.0\,\text{W}}{0.2} = 265\,\text{W}\) So, (a) the average rate of heat removal from the drink is -36.8 W, (b) the average rate of heat supply to the coffee is 53.0 W, and (c) the electric power drawn from the battery for cooling and heating is 184 W and 265 W, respectively.