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20.15. 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 ( \(20.0^{\circ} \mathrm{C} )\) . What is the temperature of its heat source?

Short Answer

Expert verified
(a) Heat extracted, \(Q_H = 4.24 \times 10^4 \mathrm{J}\). (b) Heat source temperature, \(T_H \approx 715 \mathrm{K}\).

Step by step solution

01

Identify the Efficiency Formula

The efficiency (\(\eta\)) of a Carnot engine is given by the formula \(\eta = \frac{W}{Q_H}\), where \(W\) is the work done and \(Q_H\) is the heat extracted from the hot reservoir. We are told that \(\eta = 59\% = 0.59\) and \(W = 2.5 \times 10^4\, \mathrm{J}\).
02

Calculate Heat Extracted

Rearrange the efficiency formula to solve for \(Q_H\). \(Q_H = \frac{W}{\eta}\). Substitute the known values: \(Q_H = \frac{2.5 \times 10^4}{0.59}\). Calculate \(Q_H\).
03

Efficiency and Temperature Relation

The efficiency of a Carnot engine is also given by \(\eta = 1 - \frac{T_C}{T_H}\), where \(T_C\) is the temperature of the cold reservoir and \(T_H\) is the temperature of the hot reservoir, both in Kelvin. Room temperature \(T_C\) is given as \(20.0^{\circ} \mathrm{C}\), which is \(273.15 + 20.0 = 293.15 \mathrm{K}\).
04

Determine Temperature of Heat Source

Rearrange the efficiency-temperature relation to solve for \(T_H\): \(T_H = \frac{T_C}{1 - \eta}\). Substitute the values: \(T_H = \frac{293.15}{1 - 0.59}\). Calculate \(T_H\).

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Key Concepts

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

Efficiency of a Carnot Engine
Efficiency is a measure of how well an engine converts heat energy into work. In the specific case of the Carnot engine, its efficiency is defined as the ratio of the work output to the heat input from the hot reservoir. The formula used is: \[ \eta = \frac{W}{Q_H} \]where \( \eta \) is the efficiency, \( W \) is the work done, and \( Q_H \) is the heat drawn from the heat source. In this exercise, the Carnot engine's efficiency is given as 59%, meaning it can convert 59% of the heat energy from the heat source into useful work.
This leaves 41% of the energy that must be expelled as waste heat. Understanding this efficiency allows us to better understand the constraints and limitations of heat engines.
Heat Source in Thermodynamics
The heat source is a critical component of a Carnot engine. It provides the thermal energy required for the engine to perform work. In this exercise, calculating the heat extracted from the heat source gives additional insight into how much energy is required for the engine operation.Using the formula: \[ Q_H = \frac{W}{\eta} \]we can determine the energy input from the heat source given the efficiency and work output. The heat source's quality (e.g., temperature) significantly influences the engine's capability to convert heat into work efficiently.
Role of Temperature in Engine Efficiency
Temperature plays a vital role in determining the efficiency of a Carnot engine. The efficiency is dependent on the temperatures of both the hot and cold reservoirs. The relationship is given by the formula:\[ \eta = 1 - \frac{T_C}{T_H} \]where \( T_C \) is the cold reservoir temperature and \( T_H \) is the heat source's temperature. These temperatures are measured in Kelvin to facilitate accurate calculations. In this scenario, the temperature of the cold reservoir is room temperature (20°C), which is equivalent to 293.15 K.
This formula helps us understand that the higher the heat source temperature compared to the cold reservoir, the higher the efficiency. Engineers often strive to increase the heat source temperature to achieve greater efficiency.
Understanding Thermodynamics and the Carnot Engine
Thermodynamics is the science of energy transfer, and when applied to engines, it primarily deals with how heat energy is converted to work and vice versa. The Carnot engine is an idealized model that provides the maximum possible efficiency for a heat engine operating between two thermal reservoirs. This is rooted in two main laws:
  • The First Law of Thermodynamics, which is essentially a version of the law of energy conservation adapted for thermodynamic systems.
  • The Second Law of Thermodynamics, which introduces the concept of irreversibility and provides the foundation for the principle that no engine can be more efficient than a Carnot engine operating between the same two thermal reservoirs.
Understanding these principles helps explain why real-world engines can never quite reach the efficiency of a Carnot engine, as they are always subject to inefficiencies due to factors like friction and heat losses.

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Most popular questions from this chapter

20.9. A refrigerator has a coefficient of performance of \(2.10 .\) In cach cycle it absorbs \(3.40 \times 10^{4} \mathrm{J}\) of heat from the cold reservoir. (a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat is discarded to the high-temperature reservoir?

A Gasoline Engine. Agasoline engine takes in \(1.61 \times 10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g}\) . (a) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

20.25. A sophomore with nothing better to do adds heat to 0.350 \(\mathrm{kg}\) of ice at \(0.0^{\circ} \mathrm{C}\) until it is all melted. (a) What is the change in entropy of the water? (b) The source of heat is a very massive body at a temperature of \(25.0^{\circ} \mathrm{C}\) . What is the change in entropy of this body? (c) What is the total change in entropy of the water and the heat source?

20.11. A window air-conditioner unit absorbs \(9.80 \times 10^{4} \mathrm{J}\) of heat per minute from the room being cooled and in the same time period deposits \(1.44 \times 10^{5} \mathrm{J}\) of heat into the outside air (a) What is the power consumption of the unit in watts? (b) What is the energy efficiency rating of the unit?

20.13. A Camot 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?

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