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Chapter 13: Problem 117

A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by \(62^{\circ} \mathrm{C}\), the efficiency of the engine is doubled. The temperatures of the source and sink are (a) \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (b) \(80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (c) \(95^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (d) \(90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperatures of the source and sink are \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\). Option (a) is correct.

Step by step solution

01

Understand the Problem

We are given a reversible engine converting one-sixth of the heat input into work, which implies an efficiency of \( \frac{1}{6} \). The problem states that by reducing the sink temperature by \(62^{\circ} \mathrm{C}\), the efficiency doubles. We need to find the temperatures of the source and sink from the given options.
02

Establish Relationships Using Heat Engine Efficiency

The efficiency \( \eta \) of a reversible engine is given by \( \eta = 1 - \frac{T_{sink}}{T_{source}} \). Initially, the efficiency is \( \frac{1}{6} \). Therefore, we can write: \[ \frac{1}{6} = 1 - \frac{T_{sink}}{T_{source}} \] Solving this provides: \[ \frac{T_{sink}}{T_{source}} = \frac{5}{6} \]
03

Determine Temperature Change and New Efficiency

The new sink temperature is \( T_{sink} - 62 \). The new efficiency is double the initial efficiency: \( \frac{1}{3} \). So, we have the equation: \[ \frac{1}{3} = 1 - \frac{T_{sink} - 62}{T_{source}} \] Simplifying, we get: \[ \frac{T_{sink} - 62}{T_{source}} = \frac{2}{3} \]
04

Solve for Source and Sink Temperatures

We have two equations: \( \frac{T_{sink}}{T_{source}} = \frac{5}{6} \) and \( \frac{T_{sink} - 62}{T_{source}} = \frac{2}{3} \). Solve them simultaneously to find: \[ T_{sink} = 310 \ K \] \[ T_{source} = 372 \ K \] Converting to Celsius, the source temperature \( T_{source} = 372 - 273 = 99^{\circ} \mathrm{C} \) and sink temperature \( T_{sink} = 310 - 273 = 37^{\circ} \mathrm{C} \).
05

Verify Against Answer Choices

Comparing \( T_{source} = 99^{\circ} \mathrm{C} \) and \( T_{sink} = 37^{\circ} \mathrm{C} \) with the given options, the correct answer is option (a), which is \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\).

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Key Concepts

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

Reversible engine
A reversible engine is a theoretical model used in thermodynamics to maximize efficiency. Such engines aim to convert heat into work through a cycle that can be operated forwards or backwards without any increase in entropy.
The key feature of a reversible engine is that it operates in a perfectly efficient manner without energy loss, which is not possible in real-world scenarios.
  • It means every part of the cycle is run without any friction or unneeded energy dissipation.
  • This type of engine serves as a benchmark for comparing the performance of real engines.
Understanding reversible engines helps us grasp the theoretical limits of converting heat into work efficiently.
Heat input
In thermodynamics, heat input is the energy supplied to a system. For a heat engine, this energy is usually extracted or absorbed from a hot source.
The reversible engine mentioned in our problem receives a certain amount of heat, out of which only a portion gets converted into useful work.
  • The initial problem states that the heat input is divided such that one-sixth becomes work, signifying the engine's efficiency.
  • The remainder of the heat (five-sixths) is typically dumped or rejected to the cooler sink.
Calculating the efficiency of converting this heat input into work is crucial in assessing the engine's performance.
Efficiency
Efficiency in the context of a heat engine is a measure of how well it converts heat (input energy) into work (output energy). For reversible engines, the efficiency is expressed mathematically by the formula: \[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]
  • A higher efficiency means a greater proportion of input heat is converted to useful work.
  • In our problem, doubling the efficiency implies significantly more work is obtained from the same or similar input.
Efficiency calculations are foundational for designing engines and assessing how practical they are in real-world applications.
Source temperature
The source temperature is the temperature at the hot end or the reservoir supplying heat to the engine. It directly affects the engine's efficiency because it is part of the core efficiency formula.
  • Higher source temperatures tend to allow for greater efficiency, as reflected by "1 - \frac{T_{\text{sink}}}{T_{\text{source}}}" formula.
  • In our problem, the source temperature is eventually calculated and found to be \(99^{\circ} \mathrm{C}\), which is critical since it's the baseline for efficiency improvements.
Understanding the role of source temperature helps in optimizing engine performance through effective temperature management.
Sink temperature
Sink temperature is the temperature at the cold end or reservoir where rejected heat is dumped. It impacts the efficiency because keeping it lower tends to increase overall efficiency.
  • A lower sink temperature means a smaller value of \( \frac{T_{\text{sink}}}{T_{\text{source}}} \), thus improving efficiency.
  • In our scenario, decreasing the sink temperature by \(62^{\circ} \mathrm{C}\) doubled the engine's efficiency.
Knowing how the sink temperature affects an engine's efficiency is essential for making informed design and operational decisions.

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Most popular questions from this chapter

A Carnot engine uses first an ideal monoatomic gas \((\gamma=5 / 3)\) and then an ideal diatomic gas \((\gamma=7 / 5)\) as its working substance. The source and sink temperatures are \(411^{\circ} \mathrm{C}\) and \(69^{\circ} \mathrm{C}\) respectively and the engine extracts \(1000 \mathrm{~J}\) of heat from the source in each cycle. Then (a) the efficiencies of the engine in the two cases are in the ratio \(21: 25\) (b) the area enclosed by the \(P-V\) diagram in the first case only is \(500 \mathrm{~J}\) (c) the area enclosed by the \(P-V\) diagram in both cases is \(500 \mathrm{~J}\) (d) the heat energy rejected by the engine in the first case is \(600 \mathrm{~J}\) while that in the second case is \(714.3 \mathrm{~J}\)

When two bodies \(A\) and \(B\) are in thermal equilibrium (a) the KE of all the molecules of \(A\) and \(B\) will be equal (b) the PE of all the molecules of \(A\) and \(B\) will be equal (c) the internal energies of the two bodies will be equal (d) the average kinetic energy of the molecules of the two bodies will be equal

The maximum amount of mechanical energy that can be converted into heat in any process (a) is \(100 \%\) (b) depends upon the temperatures at intake and exhaust (c) depends upon the amount of friction present (d) depends upon the nature of mechanical energy

Of the following thermometers the one which is most useful for the measurement of a rapidly varying temperature is a (a) platinum resistance thermometer (b) gas thermometer (c) thermoelectric thermometer (d) saturation vapour pressure thermometer

A piece of metal weighs \(46 \mathrm{~g}\) in air. When immersed in a liquid of specific gravity \(1.24\) at \(27^{\circ} \mathrm{C}\) it weighs \(30 \mathrm{~g} .\) When the temperature of liquid is raised to \(42^{\circ} \mathrm{C}\) the metal piece weighs \(30.5 \mathrm{~g}\). Specific gravity of liquid at \(42^{\circ} \mathrm{C}\) is \(1.20 .\) Calculate the coefficient of linear expansion of metal (a) \(2.23 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) (b) \(6.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) (c) \(4.46 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) (d) none of these
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