Question

To make some ice, a freezer extracts \(185 \mathrm{~kJ}\) of heat at \(-12.0^{\circ} \mathrm{C} .\) The freezer has a coefficient of performance of 5.70. The room temperature is \(26.0^{\circ} \mathrm{C}\). (a) How much heat is delivered to the room? ( \(b\) ) How much work is required to run the freezer?

Short answer

(a) The heat delivered to the room is \(217.46 \mathrm{~kJ}\). (b) The work required to run the freezer is \(32.46 \mathrm{~kJ}\).

Step-by-step solution

The process to solve for part (a)

The coefficient of performance (COP) is the ratio of the desired output (heat removed, or \(Q_c\)) to the required input (work done, or \(W\)). It is given by the equation COP = \(Q_c / W\). You can rearrange this equation to solve for \(W\): \(W = Q_c / COP\). The heat delivered to the room is the sum of the work done and the heat extracted, so it is given by \(Q_h = W + Q_c\).

Calculation for part (a)

First, calculate the work done using the equation \(W = Q_c / COP\). Use the given values: \(Q_c = 185 kJ\) and \(COP = 5.70\) to get \(W = (185 \mathrm{~kJ}) / 5.70 = 32.46 \mathrm{~kJ}\). Then calculate the heat delivered to the room using \(Q_h = W + Q_c = 32.46 \mathrm{~kJ} + 185 \mathrm{~kJ} = 217.46 \mathrm{~kJ}\).

The process to solve for part (b)

The work required to run the freezer was already calculated in part (a).

Conclusion for part (b)

The work required to run the freezer is \(32.46 \mathrm{~kJ}\)

Key concepts

Understanding Thermodynamics in Everyday Applications

The study of thermodynamics revolves around the principles governing the transfer of energy from one form to another, and the efficiency of its use. In the context of a home freezer, as presented in our exercise, thermodynamics is key to understanding how the appliance maintains its cold temperature. The freezer operates by extracting heat from its interior and expelling it to the surrounding room, thereby reducing the internal temperature and allowing ice to form. It's essential for students to grasp that even though the interior of the freezer gets colder, the overall energy within the system and environment increases due to the heat expelled into the room.

Moreover, thermodynamics introduces us to the concept of the 'coefficient of performance (COP)', which is a measure of a heat pump's or refrigerator's efficiency. A higher COP value indicates a more efficient device, as it requires less work (input energy) to transfer a certain amount of heat (output energy). In our exercise, the COP of the freezer helps us determine the work required to transfer heat from the cooler inside to the warmer room outside.

The Role of Heat Transfer in Cooling Systems

Heat Exchange Mechanisms

Heat transfer plays a crucial role in the functionality of cooling systems like freezers. The primary objective of a freezer is to transfer heat from inside the unit to its external environment. This process occurs through mechanisms of conduction, where heat moves through materials, and convection, where it is carried away by air or fluid particles. By better understanding heat transfer, students can appreciate how the insulation in a freezer's walls reduces the rate of heat intrusion, and how the coolant within the freezer's coils absorbs heat from the inside space.

Applicability to the Exercise

In the exercise we've discussed, the heat extracted by the freezer (185 kJ) is a direct application of heat transfer principles. The heat transferred to the room is a sum of the work done by the freezer and the heat removed from its interior, reflecting the incessant flow and conservation of energy. Recognizing this relationship between heat extraction and the work performed by the freezer is vital for students' understanding of thermal systems.

Applying the Work-Energy Principle

The work-energy principle is a fundamental concept in physics, asserting that work done on an object results in a change in energy. In terms of our freezer example, the work done by the freezer, through its compressor and other mechanical parts, leads to a removal of heat energy from the interior, demonstrating the transformation of work into thermal energy (heat). The use of work here aligns perfectly with the work-energy principle; the energy input into the freezer is converted into work done on the refrigerant, which is then used to transfer to heat out of the freezer.

This principle allows students to quantitatively connect the amount of work done by the freezer to the amount of heat transferred. As calculated in the exercise, the work required (32.46 kJ) is crucial for evaluating the energy efficiency of the freezer, helping one to infer that even though work is required to produce cooling, the process is governed by the fundamental conservation of energy, as laid out by the work-energy principle.