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Chapter 20: Problem 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 \(\left(20.0^{\circ} \mathrm{C}\right) .\) What is the temperature of its heat source?

Short Answer

Expert verified
(a) Heat extracted is about 4.24脳10鈦� J per cycle. (b) Heat source temperature is approximately 715 K.

Step by step solution

01

Define Carnot Efficiency

The efficiency \( \eta \) of a Carnot engine is defined 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. The efficiency for this engine is given as 59\(\%\), which is equivalent to 0.59 in decimal form. The efficiency can also be calculated using: \[ \eta = \frac{W}{Q_H} \]where \( W \) is the work done per cycle and \( Q_H \) is the heat absorbed from the hot reservoir.
02

Calculate Heat Absorbed from Hot Reservoir (Q_H)

Given that the engine's efficiency \( \eta = 0.59 \) and the work done per cycle \( W = 2.5 \times 10^4 \) J, use:\[ \eta = \frac{W}{Q_H} \]Rearranging for \( Q_H \) gives:\[ Q_H = \frac{W}{\eta} \] Substitute the known values:\[ Q_H = \frac{2.5 \times 10^4 \text{ J}}{0.59} \approx 4.24 \times 10^4 \text{ J} \]Thus, the heat extracted from the heat source is approximately \( 4.24 \times 10^4 \) J per cycle.
03

Convert Room Temperature to Kelvin

The exhaust or cold reservoir temperature \( T_C \) is given as \( 20.0^\circ \text{C} \). To convert to Kelvin, use:\[ T_C (\text{K}) = 20.0 + 273.15 = 293.15 \text{ K} \]
04

Calculate Hot Reservoir Temperature (T_H)

Using the efficiency equation:\[ \eta = 1 - \frac{T_C}{T_H} \]Rearrange to solve for \( T_H \):\[ T_H = \frac{T_C}{1 - \eta} \]Substitute \( T_C = 293.15 \text{ K} \) and \( \eta = 0.59 \):\[ T_H = \frac{293.15}{1 - 0.59} \approx 715.00 \text{ K} \]Thus, the temperature of the heat source is approximately \( 715.00 \text{ K} \).

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Key Concepts

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

Thermodynamic Efficiency
Carnot engines serve as a model of ideal efficiency. Understanding thermodynamic efficiency can help gauge how well an engine converts heat into work. The efficiency \( \eta \) of a Carnot engine is derived from the temperatures of the two thermal reservoirs involved. The formula is expressed as:
  • \( \eta = 1 - \frac{T_C}{T_H} \)
Here, the symbol \( T_C \) represents the temperature of the cold reservoir, while \( T_H \) refers to the hot reservoir. This formula underscores that the efficiency depends on the difference between the two temperatures.
If the efficiency is also defined in terms of work and heat absorbed, another formula used is:
  • \( \eta = \frac{W}{Q_H} \)
In this case, \( W \) denotes the work performed by the engine, and \( Q_H \) is the heat taken from the hot reservoir. By rearranging this equation, you can determine how much heat is required to perform a certain amount of work at a given efficiency. For instance, in our exercise, if the engine's efficiency is 59\(\%\), this translates to a decimal of 0.59.
Heat Transfer
In a Carnot engine, heat transfer involves moving thermal energy from a high-temperature source to perform work, then releasing remaining energy to a low-temperature sink. Heat absorption is marked as \( Q_H \), signifying the energy the engine draws from the hot reservoir. When calculating this value, knowing the efficiency of the engine and the work done is crucial. Applying the efficiency formula gives:
  • \( Q_H = \frac{W}{\eta} \)
Knowing that \( W \) is the work the engine performs, you can determine how much heat is absorbed using the formula.
In the example, with work done being \( 2.5 \times 10^4 \text{ J} \) and efficiency at 59\(\%\), the heat absorbed is \( Q_H = \frac{2.5 \times 10^4}{0.59} \), which is approximately \( 4.24 \times 10^4 \text{ J} \). This reveals the amount of energy that the heat engine draws from the source for each cycle.
Temperature Conversion
Temperature conversion plays a critical role in understanding heat engines, especially when dealing with temperature scales like Celsius and Kelvin. To compute thermodynamic efficiency or calculate reservoir temperatures, it's important to convert Celsius to Kelvin, as absolute temperature is necessary in these formulas.
For example, if the cold reservoir, or exhaust temperature, is given as \( 20.0^{\circ} \text{C} \), converting this to Kelvin requires adding 273.15:
  • \( T_C = 20.0 + 273.15 = 293.15 \text{ K} \)
Converting to Kelvin ensures temperature values operate effectively within the efficiency equations.
Similarly, to find the temperature of the heat source or hot reservoir (\( T_H \)), knowing \( T_C \) and efficiency \( \eta \) allows for using the relation:
  • \( T_H = \frac{T_C}{1 - \eta} \)
In our scenario, with \( T_C = 293.15 \text{ K} \) and \( \eta = 0.59 \), the source temperature comes out as roughly \( 715.00 \text{ K} \. \) Understanding such conversions helps grasp how the interrelation of temperatures influences engine performance.

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Most popular questions from this chapter

A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(\mathrm{K}\) and 300 \(\mathrm{K}\) . (a) If the engine receives 6.45 \(\mathrm{kJ}\) of heat energy from the reservoir at 520 \(\mathrm{K}\) in each cycle, how many joules per cycle does it discard to the reservoir at 300 \(\mathrm{K}\) ? (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

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?

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 \(10.0\) -L gas tank containing 3.20 moles of ideal He gas at \(20.0^{\circ} \mathrm{C}\) is placed inside a completely evacuated, insulated bell jar of volume 35.0 L. A small hole in the tank allows the He to leak out into the jar until the gas reaches a final equilibrium state with no more leakage. (a) What is the change in entropy of this system due to the leaking of the gas? (b) Is the process reversible or irreversible? How do you know?

An object of mass \(m_{1},\) specific heat \(c_{1},\) and temperature \(T_{1}\) is placed in contact with a second object of mass \(m_{2},\) specific heat \(c_{2},\) and temperature \(T_{2}>T_{1} .\) As a result, the temperature of the first object increases to \(T\) and the temperature of the second object decreases to \(T^{\prime}\) . (a) Show that the entropy increase of the system is $$\Delta S=m_{1} c_{1} \ln \frac{T}{T_{1}}+m_{2} c_{2} \ln \frac{T^{\prime}}{T_{2}}$$ and show that energy conservation requires that $$m_{1} c_{1}\left(T-T_{1}\right)=m_{2} c_{2}\left(T_{2}-T^{\prime}\right)$$ (b) Show that the entropy change \(\Delta S,\) considered as a function of \(T,\) is a maximum if \(T=T^{\prime},\) which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of entropy as a measure of disorder.
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