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Chapter 15: Problem 51

A Carnot engine operates with an efficiency of 27.0\(\%\) when the temperature of its cold reservoir is 275 \(\mathrm{K}\) . Assuming that the temperature of the hot reservoir remains the same, what must be the temperature of the cold reservoir in order to increase the efficiency to 32.0\(\%\) ?

Short Answer

Expert verified
The cold reservoir temperature must be approximately 256.16 K.

Step by step solution

01

Understanding Carnot Efficiency

The efficiency of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_c}{T_h} \), where \( \eta \) is the efficiency, \( T_c \) is the temperature of the cold reservoir, and \( T_h \) is the temperature of the hot reservoir.
02

Initial Situation Setup

We know the initial efficiency \( \eta_1 = 0.27 \) and \( T_c = 275 \ \mathrm{K} \). Using the formula, we have \( 0.27 = 1 - \frac{275}{T_h} \).
03

Solve for Hot Reservoir Temperature

Rearrange the equation to find \( T_h \): \( \frac{275}{T_h} = 1 - 0.27 \). Therefore, \( T_h = \frac{275}{0.73} \approx 376.71 \ \mathrm{K} \).
04

New Efficiency Equation

For the new efficiency \( \eta_2 = 0.32 \), we use \( 0.32 = 1 - \frac{T_c}{T_h} \). Substitute \( T_h = 376.71 \ \mathrm{K} \).
05

Solve for New Cold Reservoir Temperature

Rearrange the equation: \( \frac{T_c}{376.71} = 1 - 0.32 \). Therefore, \( T_c = 376.71 \times 0.68 \approx 256.16 \ \mathrm{K} \).

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Key Concepts

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

Carnot Engine Efficiency
The Carnot engine is a theoretical construct that offers the highest possible efficiency for a heat engine, operating between two thermal reservoirs. This efficiency is not just an arbitrary number; it follows a strict mathematical relation that governs the conversion of heat into work.
What makes the Carnot engine special is its dependence on the temperatures of its two reservoirs:
  • The efficiency formula is given as \( \eta = 1 - \frac{T_c}{T_h} \), where \( \eta \) is the efficiency.
  • The temperatures \( T_c \) and \( T_h \) are the absolute temperatures of the cold and hot reservoirs, respectively.
Being a theoretical model, Carnot efficiency sets an upper boundary, implying that no real engine can surpass it. Thus, understanding this concept helps us aspire to create more efficient engines by approaching this theoretical limit.
Cold Reservoir Temperature
The cold reservoir's temperature, denoted as \( T_c \), plays a crucial role in determining the efficiency of a Carnot engine. In simpler terms, it affects how much energy can be conserved when heat flows from a hot to a cold area.
Here's how it works:
  • Lowering \( T_c \) increases the efficiency of the engine, as indicated by the formula \( \eta = 1 - \frac{T_c}{T_h} \).
  • Conversely, if the cold reservoir temperature increases, the efficiency decreases.
Understanding this relationship is essential to control and enhance the performance of practical engines. In the exercise, to improve efficiency from 27% to 32%, the \( T_c \) had to be decreased from 275 K to approximately 256.16 K.
Hot Reservoir Temperature
The temperature of the hot reservoir, \( T_h \), is another vital component influencing Carnot engine efficiency. It represents the source temperature from which the engine extracts heat.
Here's the impact of \( T_h \):
  • Higher temperatures in the hot reservoir allow for more significant energy extraction and conversion, leading to improved efficiency.
  • When the hot reservoir remains constant, as in the given problem, any efficiency improvements must come from adjustments to the cold reservoir temperature \( T_c \).
In practice, maximizing \( T_h \) without exceeding material limits can optimize efficiency. For this exercise, \( T_h \) was calculated to be approximately 376.71 K, based on the initial efficiency and cold reservoir conditions.
Thermodynamic Efficiency Formula
The thermodynamic efficiency formula \( \eta = 1 - \frac{T_c}{T_h} \) characterizes the Carnot cycle beautifully. This formula is derived from fundamental principles and illustrates how efficiency is bound by thermal properties.
Here's a breakdown of its application:
  • Efficiency \( \eta \) results from the temperatures of the cold (\( T_c \)) and hot (\( T_h \)) reservoirs.
  • The differences between these temperatures dictate how well energy can be converted into work.
This equation is central to understanding and designing heat engines, enabling scientists and engineers to push the bounds of energy efficiency. By manipulating \( T_c \) or \( T_h \), they can theoretically forecast how modifications will influence overall performance, as demonstrated in the textbook problem with precise calculation steps.

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Most popular questions from this chapter

The pressure of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) doubles during an adiabatic compression. What is the ratio of the final volume to the initial volume?

Multiple-Concept Example 6 deals with the same concepts as this problem does. What is the efficiency of a heat engine that uses an input heat of \(5.6 \times 10^{4} \mathrm{J}\) and rejects \(1.8 \times 10^{4} \mathrm{J}\) of heat?

One-half mole of a monatomic ideal gas expands adiabatically and does 610 \(\mathrm{J}\) of work. By how many kelvins does its temperature change? Specify whether the change is an increase or a decrease.

A Carnot engine has an efficiency of \(0.40 .\) The Kelvin temperature of its hot reservoir is quadrupled, and the Kelvin temperature of its cold reservoir is doubled. What is the efficiency that results from these changes?

A Carnot engine operates with a large hot reservoir and a much smaller cold reservoir. As a result, the temperature of the hot reservoir remains constant while the temperature of the cold reservoir slowly increases. This temperature change decreases the efficiency of the engine to 0.70 from \(0.75 .\) Find the ratio of the final temperature of the cold reservoir to its initial temperature.
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