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(a) A Carnot engine operates between a hot reservoir at \(320 \mathrm{~K}\) and a cold one at \(260 \mathrm{~K}\). If the engine absorbs \(500 \mathrm{~J}\) as heat per cycle at the hot reservoir, how much work per cycle does it deliver? (b) If the engine working in reverse functions as a refrigerator between the same two reservoirs, how much work per cycle must be supplied to remove \(1000 \mathrm{~J}\) as heat from the cold reservoir?

Short Answer

Expert verified
The engine delivers 93.75 J of work; the refrigerator requires 230.76 J of work.

Step by step solution

01

Understanding Carnot engine efficiency

The efficiency of a Carnot engine is given by \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_h = 320 \mathrm{~K} \) (hot reservoir temperature) and \( T_c = 260 \mathrm{~K} \) (cold reservoir temperature). This is the maximum efficiency any engine operating between these two temperatures can achieve.
02

Calculate the efficiency

Plug the values into the efficiency formula: \[ \eta = 1 - \frac{260}{320} = 1 - 0.8125 = 0.1875 \]. The efficiency of the Carnot engine is 0.1875 or 18.75%.
03

Determine work done by the engine

The work done by the engine is calculated by multiplying the efficiency by the heat absorbed, which is given as \( Q_h = 500 \mathrm{~J} \). Therefore, \( W = \eta \times Q_h = 0.1875 \times 500 = 93.75 \mathrm{~J} \). The Carnot engine delivers 93.75 J of work per cycle.
04

Understanding refrigerator work input

When the Carnot engine operates as a refrigerator between the same temperatures, we use the formula for the coefficient of performance (COP): \( \text{COP} = \frac{T_c}{T_h - T_c} \). This will help us determine how much work needs to be supplied to remove a certain amount of heat (\( Q_c \)).
05

Calculate the COP

Use the temperatures to calculate the COP: \[ \text{COP} = \frac{260}{320 - 260} = \frac{260}{60} \approx 4.333 \]. The COP of the refrigerator is approximately 4.333.
06

Determine work supplied to the refrigerator

The work supplied \( W \) is found by rearranging the COP formula: \( W = \frac{Q_c}{\text{COP}} \). Given that \( Q_c = 1000 \mathrm{~J} \), \[ W = \frac{1000}{4.333} \approx 230.76 \mathrm{~J} \]. Thus, the work supplied per cycle is approximately 230.76 J.

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Key Concepts

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

Thermodynamic Efficiency
When discussing thermodynamic systems like a Carnot engine, understanding efficiency is crucial. Thermodynamic efficiency is a measure of how well an engine converts heat energy into work.

The efficiency of a Carnot engine is determined by the temperatures of the heat reservoirs it operates between. The formula for efficiency, denoted by \( \eta \), is given as:
  • \( \eta = 1 - \frac{T_c}{T_h} \)
Here, \( T_h \) is the temperature of the hot reservoir and \( T_c \) is the temperature of the cold reservoir, both in Kelvin.

This formula highlights that the higher the temperature difference between the reservoirs, the more efficient the engine is. Significantly, the Carnot cycle is considered to achieve the maximum efficiency possible for any system operating between these two temperatures. However, practical limitations often prevent real engines from achieving this theoretical efficiency.
Heat Reservoirs
In thermodynamics, a heat reservoir is a large body with the ability to absorb or supply heat without undergoing a significant change in temperature.

They are essential components of Carnot engines. There are generally two types of reservoirs involved:
  • Hot reservoir (supplies heat)
  • Cold reservoir (absorbs heat)
During a Carnot cycle, the engine absorbs heat from the hot reservoir, converts some of this energy into work, and releases the rest into the cold reservoir.

It is important for the reservoirs to maintain a constant temperature as this allows the cycle to repeat under identical conditions. The temperatures of these reservoirs directly influence the efficiency of the engine as seen in the efficiency formula.
Coefficient of Performance
The coefficient of performance (COP) is a measure used to evaluate the efficiency of refrigeration cycles, such as when a Carnot engine operates in reverse as a refrigerator.

COP is defined by the formula:
  • \( \text{COP} = \frac{T_c}{T_h - T_c} \)
This ratio helps determine how effectively a refrigerator can remove heat from the cold reservoir per unit of work input. The larger the COP, the more efficient the refrigerator.

Unlike a regular engine's efficiency, where we want it as high as possible, COP values for refrigerators mean that less work is required to transfer more heat. This aspect of thermodynamics is especially critical in designing systems that need to efficiently moderate temperatures, such as air conditioners and heat pumps.
Refrigeration Cycles
Refrigeration cycles are processes that remove heat from a low-temperature reservoir and transfer it to a high-temperature reservoir.

A Carnot cycle, when used as a refrigeration cycle, aims to maximize the efficiency of this process. Unlike power engines, the focus here is on transferring heat rather than generating work. To achieve this, work must be input into the system. The efficiency of refrigeration cycles is gauged using the coefficient of performance (COP), which helps determine the work-to-heat output ratio as outlined earlier.

The central idea of refrigeration cycles is fundamentally the same, irrespective of whether they are used for cooling buildings or preserving food: they extract heat from a cooler environment and discharge it into a warmer one. The performance of these cycles is vitally important to industries and processes that depend on temperature control.

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Most popular questions from this chapter

What is the entropy change for \(3.20 \mathrm{~mol}\) of an ideal monatomic gas undergoing a reversible increase in temperature from \(380 \mathrm{~K}\) to \(425 \mathrm{~K}\) at constant volume?

A box contains \(N\) identical gas molecules equally divided between its two halves. For \(N=50\), what are (a) the multiplicity \(W\) of the central configuration, (b) the total number of microstates, and (c) the percentage of the time the system spends in the central configuration? For \(N=100\), what are (d) \(W\) of the central configuration, (e) the total number of microstates, and (f) the percentage of the time the system spends in the central configuration? For \(N=200\), what are (g) \(W\) of the central configuration, (h) the total number of microstates, and (i) the percentage of the time the system spends in the central configuration? (j) Does the time spent in the central configuration increase or decrease with an increase in \(N\) ?

Suppose that a deep shaft were drilled in Earth's crust near one of the poles, where the surface temperature is \(-40^{\circ} \mathrm{C}\), to a depth where the temperature is \(800^{\circ} \mathrm{C}\). (a) What is the theoretical limit to the efficiency of an engine operating between these temperatures? (b) If all the energy released as heat into the lowtemperature reservoir were used to melt ice that was initially at \(-40^{\circ} \mathrm{C}\), at what rate could liquid water at \(0^{\circ} \mathrm{C}\) be produced by a 100 MW power plant (treat it as an engine)? The specific heat of ice is \(2220 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} ;\) water's heat of fusion is \(333 \mathrm{~kJ} / \mathrm{kg}\). (Note that the engine can operate only between \(0^{\circ} \mathrm{C}\) and \(800^{\circ} \mathrm{C}\) in this case. Energy exhausted at \(-40^{\circ} \mathrm{C}\) cannot warm anything above \(-40^{\circ} \mathrm{C}\).)

An apparatus that liquefies helium is in a room maintained a \(300 \mathrm{~K}\). If the helium in the apparatus is at \(4.0 \mathrm{~K}\), what is the minimum ratio \(Q_{\mathrm{to}} / Q_{\text {from }}\), where \(Q_{\mathrm{to}}\) is the energy delivered as heat to the room and \(Q_{\text {from }}\) is the energy removed as heat from the helium?

In a hypothetical nuclear fusion reactor, the fuel is deuterium gas at a temperature of \(7 \times 10^{8} \mathrm{~K}\). If this gas could be used to operate a Carnot engine with \(T_{\mathrm{L}}=100^{\circ} \mathrm{C}\), what would be the engine's efficiency? Take both temperatures to be exact and report your answer to seven significant figures.

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