Problem 24 - University Physics with Modern Physics

Question

20.24. (a) Show that the efficiency e of a Carnot engine and the coefficient of performance \(K\) of a Carnot refrigerator are related by \(K=(1-e) / e\) . The engine and refrigerator operate between the same hot and cold reservoirs. (b) What is \(K\) for the limiting values \(e \rightarrow 1\) and \(e \rightarrow 0 ?\) Explain.

Step-by-Step Solution

Answer

(a) The relationship is proven as \(K = \frac{1-e}{e}\). (b) For \( e \to 1 \), \( K = 0 \); for \( e \to 0 \), \( K = \infty \).

Step 1: Understand the Relation for Efficiency

The efficiency of a Carnot engine, operating between a hot reservoir at temperature \(T_h\) and a cold reservoir at temperature \(T_c\), is defined by the relation: \( e = 1 - \frac{T_c}{T_h} \).

Step 2: Define the Coefficient of Performance for a Refrigerator

The coefficient of performance \(K\) of a Carnot refrigerator operating between the same two reservoirs is defined as: \( K = \frac{T_c}{T_h - T_c} \).

Step 3: Relate Efficiency and Coefficient of Performance

To relate \(e\) and \(K\), express \(K\) in terms of \(e\). Start by rewriting the efficiency equation: \(\frac{T_c}{T_h} = 1 - e\). Substitute this into the expression for \(K\): \(K = \frac{\frac{T_c}{T_h}}{1 - \frac{T_c}{T_h}} = \frac{1-e}{e}\).

Step 4: Evaluate K at Limiting Values of e

When \( e \rightarrow 1 \), \( K = \frac{1-1}{1} = 0 \). This reflects that as the engine becomes perfectly efficient, the refrigerator does no work. When \( e \rightarrow 0 \), \( K = \frac{1-0}{0} = \infty \). This indicates an infinite coefficient of performance, meaning the refrigerator requires no work input.

Key Concepts

Carnot Engine Efficiency

A Carnot engine, named after the French physicist Sadi Carnot, is a theoretical heat engine that operates on an idealized cycle known as the Carnot cycle. The efficiency of a Carnot engine is a measure of how well it converts heat from a hot reservoir into useful work.

This efficiency is given by the formula:

• \( e = 1 - \frac{T_c}{T_h} \)

Here, \( T_h \) is the temperature of the hot reservoir and \( T_c \) is the temperature of the cold reservoir. It's important to note that this expression assumes temperatures are in Kelvin.

Understanding this equation shows us that efficiency depends solely on the temperatures of the reservoirs. As \( T_c \) approaches zero or \( T_h \) approaches infinity, the efficiency \( e \) approaches 1, indicating a perfectly efficient engine. Real-world engines never achieve this ideal efficiency due to deterministic losses in the system."},{

Coefficient of Performance

While an engine converts heat to work, a refrigerator requires work to move heat from a cooler place to a warmer one. The effectiveness of this process is measured by the Coefficient of Performance (COP).

For a Carnot refrigerator, the COP, denoted by \( K \), is expressed as:

• \( K = \frac{T_c}{T_h - T_c} \)

This shows the amount of heat removed from the cold reservoir per unit work input. Notably, higher \( T_c \) or lower \( T_h \) will result in a greater Coefficient of Performance.

The relationship between the efficiency of the Carnot engine and the COP of the refrigerator is derived by substituting the expression for efficiency into the COP formula, resulting in:

• \( K = \frac{1-e}{e} \)

This equation highlights that as engines become more efficient, the refrigerators using the same reservoirs become less effective, and vice versa.

Hot and Cold Reservoirs

In thermodynamics, a reservoir is often considered as a colossal system that can exchange heat without changing its own temperature significantly. The concept of hot and cold reservoirs is critical to understanding the operation of heat engines and refrigerators.

The hot reservoir is the source of heat, typically at a higher temperature. In the Carnot cycle, it is from the hot reservoir that energy is absorbed as heat to perform work in a system like an engine. On the other hand, the cold reservoir absorbs the leftover heat after work is done, at a lower temperature than the hot reservoir.

These reservoirs allow the theoretical description of energy processes without the complex details of actual materials. Understanding these is fundamental for grasping the key principles of energy transfer and efficiency in engines and refrigeration cycles.

Remember, the actual temperatures of these reservoirs directly impact the efficiency and performance as described in the Carnot cycles. This tie greatly affects practical designs where choosing appropriate temperature differences is critical for maximizing performance."}