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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 new cold reservoir temperature should be approximately 256.16 K.

Step by step solution

01

Understanding Efficiency of a Carnot Engine

The efficiency (\( \eta \)) of a Carnot engine is calculated using the formula \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) is the cold reservoir temperature and \( T_h \) is the hot reservoir temperature. We first need to express the given efficiency of 27% and convert it to decimal form: \( \eta = 0.27 \).
02

Calculate Hot Reservoir Temperature

Given \( \eta = 0.27 \) and \( T_c = 275 \mathrm{K} \), we can rearrange the efficiency formula to solve for \( T_h \): \( 0.27 = 1 - \frac{275}{T_h} \). Simplifying, we find \( \frac{275}{T_h} = 0.73 \), which gives \( T_h = \frac{275}{0.73} \approx 376.71 \mathrm{K} \).
03

Set up the New Efficiency Equation

The task is to find the new cold reservoir temperature (\( T_c' \)) for an increased efficiency of 32% (\( \eta' = 0.32 \)). Substituting into the efficiency formula, we get: \( 0.32 = 1 - \frac{T_c'}{376.71} \).
04

Solve for New Cold Reservoir Temperature

Rearrange the equation \( 0.32 = 1 - \frac{T_c'}{376.71} \) to find \( T_c' \): \( \frac{T_c'}{376.71} = 0.68 \). Solving for \( T_c' \), we get \( T_c' = 0.68 \times 376.71 \approx 256.16 \mathrm{K} \).

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Key Concepts

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

Understanding Cold Reservoir Temperature
In a Carnot engine, the cold reservoir temperature, often denoted as \( T_c \), plays a vital role in determining the engine's efficiency. The cold reservoir is the part of the system that absorbs the heat from the engine and is generally at a lower temperature compared to the hot reservoir.
Ideally, the efficiency of a Carnot engine increases as the difference between the hot and cold reservoir temperatures becomes larger. However, the challenge is to manage this temperature difference practically.
The problem at hand involves the cold reservoir temperature starting at \( 275 \text{ K} \). By altering this temperature, you can noticeably impact the efficiency. If you need to increase the efficiency, adjusting \( T_c \) downward within reasonable limits is a strategy worth considering.
Importance of Hot Reservoir Temperature
The hot reservoir temperature, labeled \( T_h \), is another critical component of the Carnot engine. It is the source from which heat enters the engine, driving the cycle.
In a Carnot engine, the temperature of the hot reservoir remains constant in this problem, calculated at approximately \( 376.71 \text{ K} \). The unchanging nature of \( T_h \) signifies its pivotal role as a stable energy source around which other variables can be adjusted.
A higher \( T_h \) results in higher potential efficiency, as the engine can convert more available thermal energy into mechanical work. This principle is rooted in the maximal theoretical efficiency that a Carnot engine can achieve, making \( T_h \) a fundamental factor in calculations.
Calculating Efficiency
Carnot engine efficiency, denoted by \( \eta \), is a measure of how effectively an engine converts heat into work. The formula for this efficiency is \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) represents the cold reservoir temperature and \( T_h \) the hot reservoir temperature.
For the initial efficiency of \( 27\% \), the calculations show how \( \eta \) correlates with the temperatures: \( 0.27 = 1 - \frac{275}{376.71} \). The efficiency increases to \( 32\% \) by adjusting the cold reservoir temperature to approximately \( 256.16 \text{ K} \).
This adjustment emphasizes the inverse relationship between \( T_c \) and \( \eta \), where lowering \( T_c \) effectively boosts the efficiency. Understanding this relationship is essential, enabling the optimization of energy use in idealized cycles such as the Carnot engine.

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Most popular questions from this chapter

A power plant taps steam superheated by geothermal energy to \(505 \mathrm{K}\) (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at \(323 \mathrm{K}\) (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 84000 kilowatts. Determine (a) the maximum efficiency at which this plant can operate and (b) the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

A 52-kg mountain climber, starting from rest, climbs a vertical distance of \(730 \mathrm{m}\). At the top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| /\left|Q_{\mathrm{H}}\right|,\) where \(|W|\) is the magnitude of the work she does and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude of the input heat. Find her efficiency as a heat engine.

The work done by one mole of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) in expanding adiabatically is 825 J. The initial temperature and volume of the gas are \(393 \mathrm{K}\) and \(0.100 \mathrm{m}^{3}\). Obtain (a) the final temperature and (b) the final volume of the gas.

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partition. Both sides contain one mole of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\), with the initial temperature being \(525 \mathrm{K}\) on the left and \(275 \mathrm{K}\) on the right. The partition is then allowed to move slowly (i.e., quasi- statically) to the right, until the pressures on each side of the partition are the same. Find the final temperatures on the (a) left and (b) right.

Engine 1 has an efficiency of 0.18 and requires 5500 J of input heat to perform a certain amount of work. Engine 2 has an efficiency of 0.26 and performs the same amount of work. How much input heat does the second engine require?
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