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Chapter 15: Problem 47

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 \( \eta \) of a Carnot engine is given by the formula: \( \eta = 1 - \frac{T_C}{T_H} \), where \( T_C \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir. Efficiency is typically expressed as a percentage, so convert it to a decimal for calculation.
02

Calculate Current Hot Reservoir Temperature

Given the current efficiency of \( 27\% \), convert this to a decimal: \( \eta = 0.27 \). With a cold reservoir temperature \( T_C = 275 \text{ K} \), the equation becomes \( 0.27 = 1 - \frac{275}{T_H} \). Solve for \( T_H \) to find its current value.
03

Solve For Current \( T_H \)

Rearrange the equation \( 0.27 = 1 - \frac{275}{T_H} \) to find \( T_H \): \( 0.27 = 1 - \frac{275}{T_H} \rightarrow \frac{275}{T_H} = 0.73 \rightarrow T_H = \frac{275}{0.73} \approx 376.71 \text{ K} \).
04

Determine New Cold Reservoir Temperature For 32% Efficiency

Given the new efficiency of \( 32\% \), convert this to a decimal: \( \eta = 0.32 \). The formula becomes \( 0.32 = 1 - \frac{T_C'}{376.71} \). We need to find the new cold temperature \( T_C' \).
05

Solve For New \( T_C' \)

Rearrange the equation \( 0.32 = 1 - \frac{T_C'}{376.71} \) to find \( T_C' \): \( \frac{T_C'}{376.71} = 0.68 \rightarrow T_C' = 0.68 \times 376.71 \approx 256.16 \text{ K} \).
06

Verification

Insert \( T_C' = 256.16 \text{ K} \) back into the efficiency formula to verify: \( \eta = 1 - \frac{256.16}{376.71} \approx 0.32 \), confirming the new temperature provides the desired efficiency.

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Key Concepts

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

Understanding Thermodynamics
Thermodynamics is a branch of physics that studies energy, heat, and their transformations. It examines how energy moves in a system, particularly through mechanisms like work and heat. These principles help us understand engines and refrigerators, including the Carnot engine discussed here.
One of the key laws of thermodynamics is the Second Law, which states that energy spontaneously tends to flow only from being concentrated in one place to becoming diffused and spread out. In simpler terms, heat naturally flows from a hot object to a cold one. This foundational understanding leads us to explore how we can use this flow of energy to do work in engines.
Efficiency Calculation of a Carnot Engine
Efficiency in thermodynamics measures how well an engine converts heat into work. It is given by the formula: \[ \eta = 1 - \frac{T_C}{T_H} \] where
  • \( \eta \) is the efficiency, expressed as a decimal when calculated,
  • \( T_C \) is the temperature of the cold reservoir,
  • \( T_H \) is the temperature of the hot reservoir.
In the Carnot cycle, this formula tells us the maximum possible efficiency that an ideal engine can achieve when operating between two given temperatures.
In the exercise, we calculate how changes in \( T_C \) impact efficiency when \( T_H \) remains constant. If you want to increase efficiency, you need to decrease \( T_C \), or increase \( T_H \), making it essential to understand how each variable affects the result.
Role of Heat Reservoirs in the Carnot Cycle
Heat reservoirs are integral to the Carnot engine's operation. There are two kinds of reservoirs that dictate the flow of heat:
  • **Cold reservoir (\( T_C \))**: where heat is released,
  • **Hot reservoir (\( T_H \))**: where heat is absorbed.
During the Carnot cycle, the engine absorbs heat from the hot reservoir, does work, and then releases some heat into the cold reservoir.
These reservoirs operate under the assumption that their temperatures remain constant, facilitating effective calculations of efficiency. As seen in our problem, adjusting the temperature of these reservoirs directly influences the efficiency of the engine.
Exploring the Carnot Cycle
The Carnot cycle provides an idealized thermal cycle that is reversible and efficient, forming the basis for understanding real-world engines. Comprising four stages, it includes:
  • Isothermal Expansion: the gas absorbs heat from the hot reservoir, expanding and doing work on the surroundings.
  • Adiabatic Expansion: the gas continues to expand without gaining or losing heat, reducing its temperature.
  • Isothermal Compression: the gas releases heat to the cold reservoir, compressing and reducing its volume.
  • Adiabatic Compression: the gas is compressed without heat transfer, increasing its temperature back to the initial state.
This cycle is significant because it defines the maximum efficiency that any engine operating between the same two temperatures could theoretically achieve. Understanding these stages helps in grasping how actual engines work, even though they can never be as efficient as the ideal Carnot engine.

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Most popular questions from this chapter

The inside of a Carnot refrigerator is maintained at a temperature of \(277 \mathrm{~K},\) while the temperature in the kitchen is \(299 \mathrm{~K}\). Using \(2500 \mathrm{~J}\) of work, how much heat can this refrigerator remove from its inside compartment?

Multiple-Concept Example 6 provides a review of the concepts that play roles here. An engine has an efficiency of \(64 \%\) and produces \(5500 \mathrm{~J}\) of work. Determine (a) the input heat and (b) the rejected heat.

Concept Questions A system does \(4.8 \times 10^{4} \mathrm{~J}\) of work, and \(7.6 \times 10^{4} \mathrm{~J}\) of heat flows into the system in the process. (a) Considered by itself, does the work increase or decrease the internal energy of the system? (b) Considered by itself, does the heat increase or decrease the internal energy? (c) Considering the work and heat together, does the internal energy of the system increase, decrease, or remain the same? Explain. Problem Find the change in the internal energy of the system. Make sure that your answer is consistent with your answers to the Concept Questions.

A monatomic ideal gas has an initial temperature of \(405 \mathrm{~K}\). This gas expands and does the same amount of work whether the expansion is adiabatic or isothermal. When the expansion is adiabatic, the final temperature of the gas is \(245 \mathrm{~K}\). What is the ratio of the final to the initial volume when the expansion is isothermal?

Go Concept Questions A system gains a certain amount of energy in the form of heat at constant pressure, and the internal energy of the system increases by an even greater amount. (a) Is any work done and, if so, is it done on or by the system? (b) If there is work, is it positive or negative, according to our convention? (c) Does the volume of the system increase, decrease, or remain the same? Give your reasoning.
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