Question

A thermoelectric refrigerator is powered by a \(12-\mathrm{V}\) car battery that draws 3 A of current when running. The refrigerator resembles a small ice chest and is claimed to cool nine canned drinks, 0.350 -L each, from 25 to \(3^{\circ} \mathrm{C}\) in \(12 \mathrm{h}\). Determine the average COP of this refrigerator.

Short answer

Answer: The average Coefficient of Performance (COP) of the thermoelectric refrigerator is 0.1877.

Step-by-step solution

Find the Mass of Drinks

First, we'll find the mass of the drinks. We are given that there are nine canned drinks, each with a volume of 0.350 L. We'll assume that the drinks are mostly water and, thus, can use the mass density of water, which is \(\rho = 1000 \, \mathrm{kg/m^3}\). Total volume of drinks = 9 × 0.350 L = 3.15 L We'll convert the volume to cubic meters by multiplying by \(\frac{1}{1000}\) Total volume of drinks = 3.15 × \(\frac{1}{1000}\) m³ = 0.00315 m³ Now, we find the mass by using the mass density formula: \(m = \rho V\) \(m = 1000 \, \mathrm{kg/m^3} \times 0.00315 \, \mathrm{m^3}\) \(m = 3.15 \, \mathrm{kg}\)

Calculate the Heat extracted, Q_L

Now, we calculate the heat extracted from the drinks during the cooling process using the heat equation: \(Q_L = mc\Delta T\) Where: - \(m = 3.15 \, \mathrm{kg}\) (mass of the drinks) - \(c = 4190 \, \mathrm{J/(kg \cdot K)}\) (specific heat capacity of water) - \(\Delta T = (25 - 3)^\circ \mathrm{C} = 22^\circ \mathrm{C}\) (temperature change) \(Q_L = 3.15 \, \mathrm{kg} \times 4190 \, \mathrm{J/(kg \cdot K)} \times 22 \, \mathrm{K}\) \(Q_L = 292117 \, \mathrm{J}\)

Calculate the work done, W

Next, we find the electrical work done by the car battery using the following formula: \(W = VI\Delta t\) Where: - \(V = 12 \, \mathrm{V}\) (voltage of the car battery) - \(I = 3 \, \mathrm{A}\) (current drawn by the refrigerator) - \(\Delta t = 12 \, \mathrm{h} = 43200 \, \mathrm{s}\) (time for which the refrigerator is running) \(W = 12 \, \mathrm{V} \times 3 \, \mathrm{A} \times 43200 \, \mathrm{s}\) \(W = 1555200 \, \mathrm{J}\)

Calculate the Coefficient of Performance, COP

Finally, we find the average Coefficient of Performance (COP) by dividing the heat extracted by the work done: COP = \(\frac{Q_L}{W}\) COP = \(\frac{292117 \, \mathrm{J}}{1555200 \, \mathrm{J}}\) COP = 0.1877 The average Coefficient of Performance (COP) of the thermoelectric refrigerator is 0.1877.

Key concepts

Thermodynamics

At its core, thermodynamics is the study of energy, its transformations, and its relation to matter. It operates on a set of fundamental laws, which apply to all physical processes and provide the foundation upon which thermoelectric refrigeration systems operate. These laws govern the principles of heat energy transfer, efficiency, and the work requirements of thermal systems.

In the context of the given exercise, thermodynamics helps us understand how the electrical energy from the car battery is converted into the work necessary for the refrigeration process. This process involves extracting heat from the drinks (inside the refrigerator) to the surroundings, cooling them down. The efficiency of this transfer is quantified by the Coefficient of Performance (COP), which in simple terms indicates how effectively an appliance transforms input work into desired heat transfer.

Thermoelectric Refrigeration

Thermoelectric refrigeration is a unique technology that uses the Peltier effect to create a heat flux between two different types of materials. It operates on the principle of semiconductor physics where an electric current flows through the junction of two distinct conductive materials, causing one side to absorb heat and cool down (the interior of the refrigerator) while the other side releases heat (to the environment).

For our exercise, the thermoelectric refrigerator uses the battery's electrical power to facilitate this heat flow, cooling the drinks inside it. Thermoelectric coolers are known for their simplicity, lack of moving parts, and ability to maintain precise temperature control. However, they are generally not as efficient as traditional vapor-compression refrigerators, which is partially reflected in the lower COP value calculated from the exercise.

Heat Transfer

Heat transfer is a discipline of thermal engineering that involves the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. Three primary modes of heat transfer exist: conduction, convection, and radiation.

In this exercise, we're particularly interested in the heat removed from the drinks—a term known as 'heat load' or 'cooling load.' The calculation of heat extracted (\(Q_L\)) employs the mass of the drinks, their specific heat capacity, and the desired temperature drop. By understanding how heat is transferred from the drinks to the refrigeration device, one can gauge the energy efficiency of the process. Since heat transfer is the main function of any refrigeration system, including thermoelectric ones, the practical understanding of these concepts is vital for anyone studying the field.