Question

A refrigerator has a coefficient of performance of \(5.00 .\) If the refrigerator absorbs 40.0 cal of heat from the low-temperature reservoir in each cycle, what is the amount of heat expelled into the high-temperature reservoir?

Short answer

Answer: The amount of heat expelled into the high-temperature reservoir is 200.832 J.

Step-by-step solution

Write down the given information

We are given the following information: - Coefficient of performance (COP) = 5.00 - Heat absorbed from the low-temperature reservoir (Q_cold) = 40.0 cal We need to find the amount of heat expelled into the high-temperature reservoir (Q_hot).

Convert calories to Joules

Since calories and Joules are both units of energy, we need to convert the 40.0 cal to Joules using the conversion factor of 1 calorie = 4.184 Joules. Q_cold = 40.0 cal * 4.184 J/cal = 167.36 J Now, we have the heat absorbed from the low-temperature reservoir in Joules: Q_cold = 167.36 J.

Calculate the work done by the refrigerator

The coefficient of performance (COP) is defined as the ratio of the heat absorbed by the refrigerator from the low-temperature reservoir to the work done by the refrigerator (W). COP = \(\frac{Q_{cold}}{W}\) Now, we can solve for the work done (W) using the given values of COP and Q_cold. W = \(\frac{Q_{cold}}{COP}\) = \(\frac{167.36 \thinspace J}{5.00}\) = 33.472 J So, the work done by the refrigerator is 33.472 J.

Apply the energy conservation principle to find Q_hot

Now, we can apply the energy conservation principle. The heat absorbed from the low-temperature reservoir (Q_cold) and the work done by the refrigerator (W) together will be equal to the heat expelled into the high-temperature reservoir (Q_hot). Q_hot = Q_cold + W Substitute the values of Q_cold and W into the equation: Q_hot = 167.36 J + 33.472 J = 200.832 J

Present the final answer

The amount of heat expelled into the high-temperature reservoir is Q_hot = 200.832 J.

Key concepts

Thermodynamics

Understanding the principles of thermodynamics is essential when discussing concepts like the coefficient of performance (COP) of a refrigerator. Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In this context, we focus on the second law of thermodynamics, which states that energy is transferred from a warmer body to a cooler body.

Within a refrigerator, thermodynamics governs the refrigeration cycle, where the refrigerant absorbs heat from the inside of the refrigerator (the low-temperature reservoir) and releases it to the surroundings (the high-temperature reservoir). This process is guided by the concept of COP, which is a measure of a refrigerator's efficiency. A COP of 5.00, as given in the exercise, indicates that for every unit of work energy put in to run the refrigeration cycle, five units of heat energy are removed from the low-temperature reservoir.

Energy Conservation

The principle of energy conservation is pivotal in analyzing thermodynamic systems like refrigerators. It asserts that energy cannot be created or destroyed, only converted from one form to another. In our exercise, the energy conservation principle helps us understand the relationship between the work done by the refrigerator and the heat transferred.

When the refrigerator does work to transfer heat from the inside to the outside, it consumes energy in the form of work (W), which is added to the heat absorbed (Q_cold) to calculate the heat expelled (Q_hot). The calculation of Q_hot is a straightforward application of the energy conservation principle, showing that the sum of energy inputs into the system (work plus absorbed heat) must equal the heat output.

Heat Transfer

Heat transfer is a fundamental aspect of the refrigeration process, and it occurs when heat moves from a warmer area to a cooler one. In the context of our exercise, the refrigerator absorbs heat energy from the food compartment, which is cooler, and releases it to the room, which is warmer.

This transfer is not 100% efficient — some energy must be supplied, typically in the form of electrical work, to facilitate the movement of heat against the natural direction (from cold to hot). That's where the coefficient of performance comes in; it tells us how effectively a refrigerator transfers heat, using a minimal amount of work for a maximum amount of heat transfer. Knowing how to calculate energy transfer with the given COP can help students understand how refrigerators manage the heat transfer process to maintain low temperatures inside the compartment.