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A Carnot engine whose high-temperature reservoir is at 620 \(\mathrm{K}\) takes in 550 \(\mathrm{J}\) of heat at this temperature in each cycle and gives up 335 \(\mathrm{J}\) to the low-temperature reservoir. (a) How much mechanical work does the engine perform during each cycle? (b) What is the temperature of the low-temperature reservoir? (c) What is the thermal efficiency of the cycle?

Short Answer

Expert verified
(a) 215 J, (b) 378 K, (c) 39.09% efficiency.

Step by step solution

01

Understand the Given Values

Identify the temperatures and energy quantities given in the problem. The high-temperature reservoir is at \( T_H = 620 \, \text{K} \), the heat taken in by the engine is \( Q_H = 550 \, \text{J} \), and the heat given to the low-temperature reservoir is \( Q_C = 335 \, \text{J} \).
02

Calculate the Work Done by the Engine

Use the first law of thermodynamics, where the work done \( W \) by the engine is the difference in heat input and output: \( W = Q_H - Q_C \). Substitute the values: \( W = 550 - 335 \), which gives \( W = 215 \, \text{J} \).
03

Find the Temperature of the Low-Temperature Reservoir

Use the Carnot efficiency formula: \( \eta = 1 - \frac{T_C}{T_H} \), where \( \eta = \frac{W}{Q_H} \). Calculate \( \eta \) first: \( \eta = \frac{215}{550} \approx 0.3909 \). Rearrange to find \( T_C \): \( T_C = T_H (1 - \eta) \). Substitute the values: \( T_C = 620 (1 - 0.3909) \approx 377.56 \, \text{K} \).
04

Calculate the Thermal Efficiency of the Engine

Thermal efficiency is calculated as \( \eta = \frac{W}{Q_H} \). We have already calculated the work as \( 215 \, \text{J} \). Therefore, \( \eta = \frac{215}{550} \approx 0.3909 \) or \( 39.09\% \).

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. It describes how energy is transferred from one state to another and how it affects matter.

There are four laws of thermodynamics that form the basis of the study:
  • **Zeroth Law:** Helps define temperature equilibrium.
  • **First Law:** Talks about the conservation of energy.
  • **Second Law:** Deals with the direction of energy transfer.
  • **Third Law:** Describes the conditions at absolute zero temperature.
In the context of a Carnot engine, thermodynamics allows us to understand how energy moves from the hot reservoir to the cold one while performing mechanical work. Understanding these concepts helps grasp how heat engines, like the Carnot engine, perform work and convert heat energy efficiently.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. The total energy in any such system is constant.

In mathematical terms, it is described by the equation:\[ \Delta U = Q - W \]where:
  • \( \Delta U \) is the change in internal energy of the system.
  • \( Q \) is the heat added to the system.
  • \( W \) is the work done by the system.
In the Carnot engine exercise, we apply the first law by calculating the work (\( W \)) using:\[ W = Q_H - Q_C \]This equation tells us that the work done by the engine is the difference between the heat absorbed from the hot reservoir (\( Q_H \)) and the heat rejected to the cold reservoir (\( Q_C \)). This fundamental principle helps us understand energy transfer in heat engines.
Carnot Efficiency
Carnot efficiency is a measure of how well an engine can convert heat energy into work compared to the maximum theoretical efficiency.

This efficiency is determined by the temperatures of the hot and cold reservoirs:\[ \eta = 1 - \frac{T_C}{T_H} \]where:
  • \( \eta \) is the efficiency.
  • \( T_C \) is the absolute temperature of the cold reservoir.
  • \( T_H \) is the absolute temperature of the hot reservoir.
The Carnot engine, named after Sadi Carnot, sets the upper limit on the efficiency that any classical thermodynamic cycle can achieve. No real engine can be more efficient than a Carnot engine because real engines have additional losses.In the example provided, we calculated the Carnot efficiency by using the formula and actual values of work and heat, emphasizing the importance of understanding temperature's role in determining efficiency.
Heat Transfer
Heat transfer is the movement of thermal energy from one place or material to another. It is a key concept in understanding how engines and other systems interact with their environment.

There are three main modes of heat transfer:
  • **Conduction:** Transfer of heat through a material without the material itself moving.
  • **Convection:** Transfer of heat through fluid movement.
  • **Radiation:** Transfer of heat in the form of electromagnetic waves without involving particles of matter.
In a Carnot engine, heat transfer occurs when the engine absorbs heat energy from a high-temperature reservoir and expels some of it to a low-temperature reservoir. Understanding this process allows us to calculate how much work the engine can perform and evaluate its efficiency using the principles of thermodynamics. By minimizing unwanted heat transfer, we can maximize the performance and efficiency of heat engines.

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Most popular questions from this chapter

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured-that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

You design an engine that takes in \(1.50 \times 10^{4} \mathrm{J}\) of heat at 650 \(\mathrm{K}\) in each cycle and rejects heat at a temperature of 350 \(\mathrm{K}\). The engine completes 240 cycles in 1 minute. What is the theoretical maximum power output of your engine, in horsepower?

You make tea with 0.250 \(\mathrm{kg}\) of \(85.0^{\circ} \mathrm{C}\) water and let it cool to room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) before drinking it. (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cols, assuming that all the heat lost by the water goes into the air. What is the total entropy change of the system tea \(+\) air?

A freezer has a coefficient of performance of \(2.40 .\) The freezer is to convert 1.80 \(\mathrm{kg}\) of water at \(25.0^{\circ} \mathrm{C}\) to 1.80 \(\mathrm{kg}\) of ice at \(-5.0^{\circ} \mathrm{C}\) in one hour. (a) What amount of heat must be removed from the water at \(25.0^{\circ} \mathrm{C}\) to convert it to ice at \(-5.0^{\circ} \mathrm{C} ?\) (b) How much much wasted heat is delivered to the room in which the freezer sits?

A Carnot engine has an efficiency of 59\(\%\) and performs \(2.5 \times 10^{4} \mathrm{J}\) of work in each cycle. (a) How much heat does the engine extract from its heat source in each cycle? (b) Suppose the engine exhausts heat at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) What is the temperature of its heat source?

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