/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A Carnot engine has an efficienc... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 15

A Carnot engine has an efficiency of 66% and performs 2.5 \(\times\) 10\(^4\) 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\(^\circ\)C). What is the temperature of its heat source?

Short Answer

Expert verified
(a) Heat extracted is approximately 3.79 脳 10鈦� J. (b) Temperature of the heat source is about 862.21 K.

Step by step solution

01

Understanding Efficiency

The efficiency of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) and \( T_h \) are the temperatures of the cold and hot reservoirs, respectively. Given the efficiency \( \eta = 0.66 \), which means 66%, we'll use this information later to find the temperature of the heat source.
02

Calculating Heat Energy Input

Efficiency \( \eta \) is also defined as the ratio of work done \( W \) to the heat absorbed from the heat source \( Q_h \). The equation is \( \eta = \frac{W}{Q_h} \). Given \( W = 2.5 \times 10^4 \) J and \( \eta = 0.66 \), we solve for \( Q_h \) using \( 0.66 = \frac{2.5 \times 10^4}{Q_h} \). Rearranging gives \( Q_h = \frac{2.5 \times 10^4}{0.66} \approx 3.79 \times 10^4 \) J.
03

Understanding Temperatures in Kelvin

Convert the exhaust temperature from Celsius to Kelvin since thermodynamic calculations require the temperature in Kelvin. The conversion is given by \( T(K) = T(^{\circ}C) + 273.15 \). For a room temperature of 20.0掳C, we have \( T_c = 20 + 273.15 = 293.15 \) K.
04

Finding the Temperature of the Heat Source

Using the Carnot efficiency equation \( \eta = 1 - \frac{T_c}{T_h} \), where \( \eta = 0.66 \) and \( T_c = 293.15 \) K, we can find \( T_h \). Solving \( 0.66 = 1 - \frac{293.15}{T_h} \) yields \( \frac{293.15}{T_h} = 0.34 \), and rearranging gives \( T_h = \frac{293.15}{0.34} \approx 862.21 \) K.

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Key Concepts

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

Efficiency of Heat Engines
In the world of thermodynamics, the efficiency of a heat engine is a crucial aspect to understand. A heat engine extracts heat from a source, performs work, and exhausts some of the heat to a sink. The efficiency (\(\eta\)) measures how well an engine converts the absorbed heat (\(Q_h\)) into work (\(W\)). It can be calculated using the formula:
\(\eta = \frac{W}{Q_h}\).
For a Carnot engine, which is a theoretical perfect engine, the efficiency is also given by the expression:
\(\eta = 1 - \frac{T_c}{T_h}\).
Here, \(T_c\) is the temperature of the cold reservoir and \(T_h\) is the temperature of the hot source. All these temperatures must be in Kelvin.
A Carnot engine represents an ideal because, in reality, 100% efficiency is unattainable due to the second law of thermodynamics. This means that no engine can convert all the heat it absorbs into work without losing some to the surroundings.
Thermodynamic Temperature Conversion
When dealing with thermodynamic processes, it's essential to use the Kelvin scale for temperatures. The Kelvin scale begins at absolute zero, the coldest possible temperature, which helps eliminate negative values that can complicate calculations.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, converting the room temperature of 20.0掳C, we do:
  • \(T(K) = T(^{\circ}C) + 273.15\)
  • \(T(K) = 20 + 273.15 = 293.15\) K
In thermodynamic calculations involving engines, using Kelvin ensures consistency in calculating formulas like the Carnot efficiency. This temperature conversion is important because avoiding negative values allows for smoother and more precise calculations when using scientific equations.
Work and Heat Energy Relationship
In the study of heat engines, understanding how work and heat energy are interconnected is indispensable. The energy absorbed as heat from the heat source (\(Q_h\)) and the energy lost to the cold sink influence the amount of work (\(W\)) the engine can perform.
An important relationship here is: \(W = Q_h - Q_c\), where \(Q_c\) is the heat expelled to the cold sink. This equation demonstrates that the work done by the engine is the difference between the heat absorbed from the source and the heat lost to the sink.
Moreover, we explored the efficiency calculation \(\eta = \frac{W}{Q_h}\), concluding that efficiency helps determine how much of the absorbed heat is converted to useful work. By improving efficiency, an engine can perform more work for the same amount of heat input, but natural limitations prevent reaching the theoretical 100% efficiency.

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Most popular questions from this chapter

An average sleeping person metabolizes at a rate of about 80 \(W\) by digesting food or burning fat. Typically, 20% of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal internal temperature of the body (where the metabolism takes place) is 37\(^\circ\)C, and the skin is typically 7 C\(^\circ\) cooler. By how much does the person's entropy change per second due to this heat transfer?

If the proposed plant is built and produces 10 \(MW\) but the rate at which waste heat is exhausted to the cold water is 165 \(MW\), what is the plant's actual efficiency? (a) 5.7%; (b) 6.1%; (c) 6.5%; (d) 16.5%.

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 kg of ice. During this time, how much work \(W\) is performed by the engine?

A gasoline engine takes in 1.61 \(\times\) 10\(^4\) J of heat and delivers 3700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 \(\times\) 10\(^4\) J/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?

A gasoline engine has a power output of 180 \(kW\)(about 241 hp). Its thermal efficiency is 28.0%. (a) How much heat must be supplied to the engine per second? (b) How much heat is discarded by the engine per second?
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