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Chapter 5: Problem 12

A refrigerator operates at steady state using \(500 \mathrm{~W}\) of electric power with a COP of \(2.5\). What is the net effect on the kitchen air?

Short Answer

Expert verified
The net effect on the kitchen air is a gain of 1750 W of heat.

Step by step solution

01

Understand the concept of COP

The Coefficient of Performance (COP) of a refrigerator is defined as the ratio of the heat removed from the refrigerated space to the work input. Mathematically, it is expressed as:\[ \text{COP} = \frac{Q_c}{W} \]where \(Q_c\) is the heat removed from the refrigerated space and \(W\) is the work input in watts.
02

Rearrange the COP formula

We know the COP and the work input \(W\). We can rearrange the formula to find \(Q_c\):\[ Q_c = \text{COP} \times W \]
03

Calculate the heat removed from the refrigerated space

Given that the refrigerator's work input \(W = 500 \text{ W}\) and COP = 2.5, substitute these values into the rearranged formula:\[ Q_c = 2.5 \times 500 \]\[ Q_c = 1250 \text{ W} \]This means the refrigerator removes 1250 W of heat from inside the refrigerated space.
04

Determine the net effect on kitchen air

The refrigerator vents this energy into the kitchen. Since the refrigerator uses 500 W to remove 1250 W from inside, the total energy released to the kitchen is the sum of the work input and the extracted heat:\[ Q_{kitchen} = Q_c + W \]\[ Q_{kitchen} = 1250 + 500 = 1750 \text{ W} \]
05

Conclude the net effect

The net effect on the kitchen air is that it receives 1750 W of thermal energy. This energy is dissipated as heat and could raise the temperature of the kitchen.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Performance (COP)
The Coefficient of Performance, often abbreviated as COP, is a crucial concept in understanding refrigeration systems. It is essentially a measure of a refrigerator's efficiency. Specifically, COP is defined as the ratio of the heat removed from the refrigerated space (\[ Q_c \]) to the input work required (\[ W \]).
  • A higher COP indicates a more efficient refrigerator, as it can remove more heat with less work input.
  • Mathematically, COP is expressed as follows:\[\text{COP} = \frac{Q_c}{W}\]
This formula shows that with a given work input, the heat removal depends directly on the COP. Thus, knowing the COP helps us calculate how much heat the refrigerator can transfer from the inner space using a certain amount of work. Understanding this ratio is essential for assessing and optimizing refrigerator performance.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, and energy. In the context of refrigeration, thermodynamics helps us understand how energy is transferred and how systems like refrigerators operate.
  • Key principles include the conservation of energy, which states that energy in a system is conserved.
  • For a refrigerator, thermodynamics explains how electric energy is converted into thermal energy and moved inside the system to the kitchen air.
The first law of thermodynamics, also known as the principle of energy conservation, is especially critical here. It mandates that the total energy added to the system must equal the sum of energy output. Hence, in a refrigerator, the energy removed from inside plus the work done by the fridge must balance the energy expelled into the kitchen. This ensures we are mindful of energy efficiency.
Energy Transfer
Energy transfer plays a central role in the functioning of any refrigeration cycle. In this context, energy is moved from inside the refrigerator to the surrounding kitchen air. Understanding Energy Transfer:
  • Two forms of energy transfers occur: the work done on the system and the heat transfer out of the system.
  • The refrigerator uses electrical power as work input, utilizing COP to remove more heat than the power used.
For example, in this exercise, with a work input of 500 W and a COP of 2.5, the system removes 1250 W of heat from the refrigerated space, demonstrating a more significant energy transfer due to its efficiency. Moreover, the refrigerator not only removes heat from the inside but also adds this heat plus the input energy to the kitchen air, increasing the thermal energy released.
Steady State
A steady state is a condition in which the variables (like energy, pressure, temperature) of a system remain constant over time. In terms of refrigeration, operating at a steady state means that the refrigerator maintains consistent performance, maintaining constant temperatures and energy removal rates.
  • Steady state indicates the system is in balance, consistently cycling the refrigerant without fluctuations.
  • This ensures the desired temperature inside the refrigerator is consistently maintained without abrupt variations in performance.
In the given exercise, the refrigerator, operating at a steady state, signifies that it consistently uses 500 W of power with a COP of 2.5, ensuring the refrigerator keeps removing 1250 W of heat continuously. This stability is essential for reliable and predictable cooling performance, ensuring the system effectively controls the internal climate without unexpected shifts.

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Most popular questions from this chapter

Air in a rigid \(1-\mathrm{m}^{3}\) box is at \(300 \mathrm{~K}, 200 \mathrm{kPa}\). It is heated to \(600 \mathrm{~K}\) by heat transfer from a reversible heat pump that receives energy from the ambient at \(300 \mathrm{~K}\) besides the work input. Use constant specific heat at \(300 \mathrm{~K}\). Since the COP changes, write \(d Q=m_{\text {air }} C_{v} d T\) and find \(d W\). Integrate \(d W\) with the temperature to find the required heat pump work.

On a cold \(\left(-10^{\circ} \mathrm{C}\right)\) winter day, a heat pump provides \(20 \mathrm{~kW}\) to heat a house maintained at \(20^{\circ} \mathrm{C}\), and it has a \(\mathrm{COP}_{H P}\) of 4 . How much power does the heat pump require? The next day, a storm brings the outside temperature to \(-15^{\circ} \mathrm{C}\), assuming the same COP and the same house heat transfer coefficient for the heat loss to the outside air. How much power does the heat pump require then?

We wish to produce refrigeration at \(-20 \mathrm{~F}\). A reservoir is available at \(400 \mathrm{~F}\) and the ambient temperature is \(80 \mathrm{~F}\), as shown in Fig. P5.75. Thus, work can be done by a cyclic heat engine operating between the 400 F reservoir and the ambient. This work is used to drive the refrigerator. Determine the ratio of the heat transferred from the 400 F reservoir to the heat transferred from the \(-20 \mathrm{~F}\) reservoir, assuming all processes are reversible.

An industrial machine is being cooled by \(0.8\) \(\mathrm{lbm} / \mathrm{s}\) water at \(60 \mathrm{~F}\) that is chilled from \(95 \mathrm{~F}\) by a refrigeration unit with a COP of 3 . Find the rate of cooling required and the power input to the unit.

A refrigerator maintaining a \(5^{\circ} \mathrm{C}\) inside temperature is located in a \(30^{\circ} \mathrm{C}\) room. It must have a high temperature \(\Delta T\) above room temperature and a low temperature \(\Delta T\) below the refrigerated space in the cycle to actually transfer the heat. For a \(\Delta T\) of \(0^{\circ}, 5^{\circ}\), and \(10^{\circ} \mathrm{C}\), respectively, calculate the \(\mathrm{COP}\), assuming a Carnot cycle.
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