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A Carnot's reversible engine converts \(\frac{1}{6}\) of heat input into work, When the temperature of the sink is reduced by \(62 \mathrm{~K}\). the efficiency of Carnot's cycle becomes \((1 / 3)\). The temperature of the source and the sink (in degree kelvin) are respectively (a) 372,310 (b) 472,410 (c) 310,372 (d) 744,682

Short Answer

Expert verified
The temperatures of the source and the sink are 372 K and 310 K, respectively.

Step by step solution

01

Understand Efficiency Formula

The efficiency () of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_2}{T_1} \), where \( T_1 \) is the temperature of the source and \( T_2 \) is the temperature of the sink, both in Kelvin. Initially, \( \eta = \frac{1}{6} \).
02

Express Initial Efficiency

Convert the initial efficiency condition \( \eta = \frac{1}{6} \) to the formula: \[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \] This implies that \[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \].
03

Use Sink Temperature Change in Efficiency Formula

When the sink temperature decreases by 62 K, the efficiency becomes \( \frac{1}{3} \). The new sink temperature is \( T_2 - 62 \). Substitute into the efficiency formula: \[ \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \].
04

Solve New Efficiency Equation

Rearrange the new efficiency equation: \[ \frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]. This implies \[ T_2 - 62 = \frac{2}{3} T_1 \].
05

Solve System of Equations

We now have two equations: \( \frac{T_2}{T_1} = \frac{5}{6} \) and \( T_2 - 62 = \frac{2}{3} T_1 \). Solve these to find \( T_1 \) and \( T_2 \).From the first equation, we have \( T_2 = \frac{5}{6} T_1 \). Substitute into the second equation:\[ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 \].
06

Simplify and Solve for T1

Simplify the equation: \[ \frac{5}{6} T_1 - \frac{2}{3} T_1 = 62 \]\[ \frac{5}{6} T_1 - \frac{4}{6} T_1 = 62 \]\[ \frac{1}{6} T_1 = 62 \]\[ T_1 = 372 \].
07

Calculate T2 from T1

Use \( T_2 = \frac{5}{6} T_1 \) to find \( T_2 \):\[ T_2 = \frac{5}{6} \times 372 = 310 \].

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Key Concepts

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

Understanding the Efficiency Formula
The efficiency of a Carnot engine is a fundamental concept that helps us understand how well the engine converts heat into work. In simple terms, efficiency is the measurement of how much useful work we can get from a given amount of heat. For a Carnot engine, the efficiency \( \eta \) is expressed as:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Where:
  • \( T_1 \) is the temperature of the source (higher temperature reservoir) in Kelvin.
  • \( T_2 \) is the temperature of the sink (lower temperature reservoir) in Kelvin.
This formula tells us the potential efficiency of a Carnot engine based on the temperatures of its source and sink. A higher efficiency indicates that more of the heat input is being converted into work. The closer the value of \( \frac{T_2}{T_1} \) is to one, the less efficient the engine becomes.
The Role of Temperature in Source and Sink
The temperatures of the source and sink play a critical role in determining the efficiency of a Carnot engine. The temperature of the source \( T_1 \) is where the engine absorbs heat. On the other hand, the temperature of the sink \( T_2 \) is where the engine rejects heat. Both of these temperatures must be measured in Kelvin for calculations to be accurate.
A crucial insight from the efficiency formula is that reducing the sink temperature \( T_2 \) while maintaining the same source temperature \( T_1 \) can increase the engine’s efficiency. Conversely, increasing \( T_1 \) can also improve efficiency, provided \( T_2 \) remains the same or is reduced. This sensitivity to temperatures is why real-world engines often aim to maximize the source temperature and minimize the sink temperature to achieve greater efficiency.
Heat Conversion in Engines
In engines, one of the primary goals is to convert as much heat input into useful work as possible. This process is essentially what the efficiency of a Carnot engine measures. The Carnot cycle is an idealized thermodynamic cycle, theorized by Sadi Carnot, that defines the maximum possible efficiency that any heat engine can achieve.
In a Carnot engine, the process of heat conversion involves four key stages:
  • Isothermal Expansion: The engine absorbs heat at a constant temperature from the source.
  • Adiabatic Expansion: The engine expands without exchanging heat, doing work on the surroundings.
  • Isothermal Compression: The engine releases heat to the sink at a constant temperature.
  • Adiabatic Compression: The working substance is compressed without heat exchange, moving the system back to its initial state.
These stages underline how the Carnot engine turns heat into work and why not all input heat can be converted into useful work due to intrinsic thermodynamic limitations. The stages essentially demonstrate the engine's need to expel some heat to the sink, which is an unavoidable aspect of real-world engines, though minimized in the idealized Carnot cycle.

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Most popular questions from this chapter

A clock which keeps correct time at \(20^{\circ} \mathrm{C}\) has a pendulum rod made of brass. How many seconds will it gain or lose per day when temperature falls to \(0^{\circ} \mathrm{C}\) ? \(\left(\alpha=18 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\) (a) \(155.5 \mathrm{~s}\) (b) \(15.55 \mathrm{~s}\) (c) \(25.55 \mathrm{~s}\) (d) \(18.55 \mathrm{~s}\)

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

In a cyclic process, work done by the system is (a) zero (b) equal to heat given to the system (c) more than the heat given to the system (d) independent of heat given to the system

A brass rod of length \(500 \mathrm{~mm}\) and diameter \(3 \mathrm{~mm}\) is joined to a steel rod of same length and diameter at \(50^{\circ} \mathrm{C}\). If the coefficients of linear expansion of brass and steel are \(2.5 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) and \(1.25 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\), then change in length of the combined rod at \(200^{\circ} \mathrm{C}\) is (a) \(2.4 \mathrm{~mm}\) (b) \(2.8 \mathrm{~mm}\) (c) \(3.2 \mathrm{~mm}\) (d) \(3.6 \mathrm{~mm}\)

An ideal gas is taken through a cyclic thermodynamical process through four steps. The amounts of heat involved in these steps are; \(Q_{1}=5960 \mathrm{~J}, Q_{2}=-5585 \mathrm{~J}\), \(Q_{3}=-2980 \mathrm{~J}, Q_{4}=3645 \mathrm{~J}\) respectively, The corresponding works involvedare; \(W_{1}=2200 \mathrm{~J}, W_{2}=-825 \mathrm{~J}\), \(W_{3}=-1100 \mathrm{~J}\) and \(W_{4}\) respectively. The value of \(W_{4}\) is (a) \(1315 \mathrm{~J}\) (b) \(275 \mathrm{~J}\) (c) \(765 \mathrm{~J}\) (d) \(675 \mathrm{~J}\)

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