/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Prove that \(if\) you had a refr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Prove that \(if\) you had a refrigerator whose COP was better than the ideal value \((4.9),\) you could hook it up to an ordinary Carnot engine to make an engine that produces no waste heat.

Short Answer

Expert verified
A refrigerator with a COP > 4.9 theoretically allows an engine that produces no waste heat.

Step by step solution

01

Understanding the Problem

We need to prove whether a refrigerator with a COP greater than 4.9 can operate with a Carnot engine to produce an engine without waste heat. COP (Coefficient of Performance) is the ratio of the heat output from the cold reservoir to work input, and for an ideal refrigerator, the maximum COP is given by the Carnot formula.
02

Calculate Ideal COP

The ideal COP for a refrigerator operating between two temperatures is calculated by the formula: \[COP_{ideal} = \frac{T_c}{T_h - T_c}\]where \(T_c\) is the temperature of the cold reservoir and \(T_h\) is the temperature of the hot reservoir, both in Kelvin.
03

Positive Net Work

For a system composed of a refrigerator and Carnot engine, the work input to the refrigerator can be less than or equal to the work output by the engine, allowing for zero net waste work. If ideal COP is 4.9, then any COP > 4.9 means the system could theoretically create a scenario where all input work is converted back, without generating net waste heat.
04

Assertion

Given that the actual COP is greater than 4.9, this implies that when these components are hooked together, they could theoretically provide enough efficiency such that there's no net waste heat.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Coefficient of Performance
The Coefficient of Performance (COP) is a measure of a refrigerator's efficiency. It tells you how much heat energy is removed from the cold region, compared to the work energy input. Think of it in terms of getting the most cooling for the least amount of work. In mathematical terms, the COP is given by the equation:\[COP = \frac{Q_c}{W} \]where:- \(Q_c\) is the heat removed from the cold reservoir.- \(W\) is the work input to the refrigerator.For an ideal refrigerator, the COP can be maximized using the Carnot formula. This happens only under specific conditions. A higher COP indicates a more efficient refrigeration cycle.
Carnot Engine
The Carnot engine is an idealized heat engine, considered to be the most efficient possible engine. It operates between two thermal reservoirs at constant temperatures. The great thing about the Carnot engine is that it sets a benchmark for efficiency. No real engine can exceed the Carnot efficiency.Its efficiency is calculated using the Carnot efficiency formula:\[\eta_{Carnot} = 1 - \frac{T_c}{T_h} \]where:- \(\eta_{Carnot}\) is the efficiency of the Carnot engine.- \(T_c\) is the temperature of the cold reservoir.- \(T_h\) is the temperature of the hot reservoir.The key takeaway is that the efficiency of any engine is fundamentally limited by the temperature difference between its heat reservoirs.
Ideal Refrigerator
An ideal refrigerator is a device that achieves the maximum possible efficiency for cooling. The ideal efficiency is determined by the same principles that govern the Carnot cycle. In practice, an ideal refrigerator would operate in a perfectly reversible cycle.The maximum theoretical COP for an ideal refrigerator is given by:\[COP_{ideal} = \frac{T_c}{T_h - T_c} \]This formula shows the theoretical limit for efficiency based on the temperatures of the hot and cold reservoirs. A refrigerator reaching this ideal COP is incredibly efficient, but in reality, various losses prevent achieving this perfect coefficient.
Net Work
Net Work refers to the balance between the work done by machines and the work input into them. In this context, if a refrigerator and a Carnot engine are coupled, they exchange work. The goal in such a setup is to achieve a net zero work requirement, meaning: - The work output by the Carnot engine can potentially match the work input required by the refrigerator. When work input and output are balanced in this way, it results in an efficient system with minimal waste. This scenario outlines a theoretical setup where no net waste heat is emitted.
Heat Reservoirs
Heat reservoirs are a critical concept in thermodynamics. They are bodies or systems with a large thermal capacity, which allows them to absorb or release heat without a significant temperature change. In any thermodynamic process, there are typically two reservoirs: - The hot reservoir that provides heat. - The cold reservoir that absorbs the rejected heat. The temperatures of these reservoirs heavily influence the efficiency of thermodynamic cycles, as seen with concepts like COP and Carnot efficiency. Understanding the role of heat reservoirs helps in designing more efficient thermodynamic systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Can you cool off your kitchen by leaving the refrigerator door open? Explain.

Estimate the maximum possible COP of a household air conditioner. Use any reasonable values for the reservoir temperatures.

Prove that if you had a heat engine whose efficiency was better than the ideal value \((4.5),\) you could hook it up to an ordinary Carnot refrigerator to make a refrigerator that requires no work input.

To get more than an infinitesimal amount of work out of a Carnot engine, we would have to keep the temperature of its working substance below that of the hot reservoir and above that of the cold reservoir by non-infinitesimal amounts. Consider, then, a Carnot cycle in which the working substance is at temperature \(T_{h w}\) as it absorbs heat from the hot reservoir, and at temperature \(T_{\mathrm{cw}}\) as it expels heat to the cold reservoir. Under most circumstances the rates of heat transfer will be directly proportional to the temperature differences: $$\begin{aligned} &\frac{Q_{h}}{\Delta t}=K\left(T_{h}-T_{h w}\right) \quad \text { and } &\frac{Q_{c}}{\Delta t}=K\left(T_{c w}-T_{c}\right) \end{aligned}$$ I've assumed here for simplicity that the constants of proportionality ( \(K\) ) are the same for both of these processes. Let us also assume that both processes take the same amount of time, so the \(\Delta t^{\prime}\) s are the same in both of these equations. (a) Assuming that no new entropy is created during the cycle except during the two heat transfer processes, derive an equation that relates the four temperatures \(T_{h}, T_{c}, T_{h w},\) and \(T_{c w}\) (b) Assuming that the time required for the two adiabatic steps is negligible, write down an expression for the power (work per unit time) output of this engine. Use the first and second laws to write the power entirely in terms of the four temperatures (and the constant \(K\) ), then eliminate \(T_{c w}\) using the result of part (a). (c) When the cost of building an engine is much greater than the cost of fuel (as is often the case), it is desirable to optimize the engine for maximum power output, not maximum efficiency. Show that, for fixed \(T_{h}\) and \(T_{c}\) the expression you found in part (b) has a maximum value at \(T_{h w}=\) \(\frac{1}{2}\left(T_{h}+\sqrt{T_{h} T_{c}}\right) .\) (Hint: You'll have to solve a quadratic equation.) Find the corresponding expression for \(T_{c w}\) (d) Show that the efficiency of this engine is \(1-\sqrt{T_{c} / T_{h}}\). Evaluate this efficiency numerically for a typical coal-fired steam turbine with \(T_{h}=600^{\circ} \mathrm{C}\) and \(T_{c}=25^{\circ} \mathrm{C},\) and compare to the ideal Carnot efficiency for this temperature range. Which value is closer to the actual efficiency, about \(40 \%\) of a real coal-burning power plant?

What is the maximum possible COP for a cyclic refrigerator operating between a high-temperature reservoir at \(1 \mathrm{K}\) and a low-temperature reservoir at \(0.01 \mathrm{K} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.