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Chapter 20: Problem 37

20.37. You design a Carnot engine that operates between temperatures of 500 \(\mathrm{K}\) and 400 \(\mathrm{K}\) and produces 2000 \(\mathrm{J}\) of work in each cycle. (a) Calculate your engine's efficiency. (b) Calculate the amount of heat discarded during the isothermal compression at 400 \(\mathrm{K}\) . (c) Sketch the 500 \(\mathrm{K}\) and 400 \(\mathrm{K}\) isotherms on apV-diagram (no calculations); then sketch the Carnot cycle followed by your engine. (d) On the same diagram, sketch the 300 \(\mathrm{K}\) isotherm; then sketch, in a different color if possible, the Carnot cycle starting at the same point on the 500 \(\mathrm{K}\) isotherm but operating in a cycle between the 500 \(\mathrm{K}\) and 300 \(\mathrm{K}\) isotherms. (e) Compare the areas inside the loops (the net work done) for the two cycles. Notice that the same amount of heat is extracted from the hot reservoir in both cases. Can you explain why less heat is "wasted" during the 300 \(\mathrm{K}\) isothermal compression than during the 400 \(\mathrm{K}\) compression?

Short Answer

Expert verified
Efficiency is 20%, 8000 J heat discarded, 300 K cycle does more work and wastes less heat.

Step by step solution

01

Understand the Carnot engine efficiency formula

The efficiency \( \eta \) of a Carnot engine is given 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, both in Kelvin.
02

Calculate the Carnot engine efficiency

For this Carnot engine, \( T_h = 500 \) K and \( T_c = 400 \) K. Substitute these values into the efficiency formula: \[ \eta = 1 - \frac{400}{500} = 0.2. \] Thus, the efficiency is 20%.
03

Use work and efficiency to find heat input

The work output \( W \) is related to the heat input \( Q_h \) by the equation \( \eta = \frac{W}{Q_h} \). Since \( W = 2000 \) J, and \( \eta = 0.2 \), \( \frac{2000}{Q_h} = 0.2 \). Solving for \( Q_h \), we get \( Q_h = 10000 \) J.
04

Determine heat discarded

The heat discarded \( Q_c \) can be found using the relation \( Q_h = W + Q_c \). Thus, \( 10000 = 2000 + Q_c \), giving \( Q_c = 8000 \) J.
05

Sketch p-V diagram with given temperatures

On a p-V diagram, plot the isotherms for 500 K and 400 K. Sketch the Carnot cycle as a rectangle with isothermal expansion at 500 K, adiabatic expansion, isothermal compression at 400 K, and adiabatic compression closing the cycle.
06

Sketch additional isotherm and Carnot cycle

On the same diagram, add the 300 K isotherm. Sketch another Carnot cycle starting at the same point on the 500 K isotherm, following the same process with isothermal processes at 500 K and 300 K.
07

Compare work done in different cycles

The area inside the loop on the p-V diagram represents the net work done by the cycle. Since the 300 K cycle extends further from the x-axis, it encapsulates a larger area, indicating more net work done. Less heat is discarded at a lower 300 K temperature compared to 400 K because less energy is removed, allowing more work output.

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Key Concepts

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

Thermodynamics
Thermodynamics studies the relationships between heat, work, temperature, and energy. Within this realm, the Carnot engine is a theoretical model used to understand how an engine operates between two heat reservoirs.
  • A heat reservoir is a system that can absorb or release heat without undergoing a change in temperature.
  • The Carnot engine uses two reservoirs: a hot reservoir at a higher temperature, and a cold reservoir at a lower temperature.
Understanding these basic concepts is crucial because the Carnot engine allows us to explore the maximum possible efficiency an engine can achieve under given conditions. It sets an ideal benchmark for real-world engines, which always have lower efficiencies due to irreversible processes and practical limitations. In our exercise, the Carnot engine operates between 500 K as the hot reservoir and 400 K as the cold reservoir, resulting in a cycle that ideally extracts and utilises energy through perfect thermodynamic processes.
Efficiency Calculation
The efficiency of a heat engine is a measure of how well it converts heat energy into work. For a Carnot engine, this efficiency is calculated using a simple formula: \[ \eta = 1 - \frac{T_c}{T_h} \] where \( \eta \) denotes efficiency, \( T_c \) the temperature of the cold reservoir, and \( T_h \) the temperature of the hot reservoir.
  • In our exercise, \( T_h = 500 ext{ K} \) and \( T_c = 400 ext{ K} \), so the efficiency is: \( 1 - \frac{400}{500} = 0.2 \, (20\%) \).
This means that only 20% of the heat energy absorbed from the hot reservoir is converted into useful work, while the rest is wasted as heat. This calculation highlights the theoretical limits of efficiency, emphasizing the importance of higher temperature differences between the reservoirs for higher engine efficiency.
p-V Diagrams
The p-V diagram is a graphical representation of the pressure-volume relationship within a thermodynamic cycle. These diagrams help visualize the steps and changes in a Carnot cycle.
  • The isothermal processes occur at constant temperatures (500 K for expansion and 400 K for compression).
  • The adiabatic processes involve no heat transfer and connect the isothermal processes on the diagram.
On a p-V diagram, the Carnot cycle appears as a loop with the shape of a rectangle: one side represents the isothermal expansion, the opposite side isothermal compression, and the other two sides adiabatic expansions and compressions. By adding the 300 K isotherm, you can compare how varying the cold reservoir temperature affects the cycle's work output, further emphasizing visual understanding of efficiency and energy transformations in heat engines.
Heat Transfer
Heat transfer is a crucial part of the Carnot cycle, as it underpins the engine's ability to do work. In the Carnot cycle, heat is transferred in two main ways:
  • Isothermal expansion at the hot temperature (500 K), where the engine absorbs heat \( Q_h \).
  • Isothermal compression at the cold temperature (400 K), where the engine releases heat \( Q_c \).
In the given exercise, the heat absorbed from the hot reservoir is 10,000 J, while 8,000 J is discarded to the cold reservoir. The remaining energy completes the work cycle, producing 2,000 J of work. The nature of heat transfer in a Carnot engine allows for these calculations, demonstrating the profound impact of thermodynamic laws on energy efficiency and conservation. Notably, adjusting the temperature of the cold reservoir to 300 K reduces 'wasted' heat, highlighting the sensitivity of engine performance to reservoir temperatures.

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Most popular questions from this chapter

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?

20.36 . A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized Intermational Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3}\) . Calculate the entropy change of the air during the expansion.

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?

20.18. A Carnot device extracts 5.00 \(\mathrm{kJ}\) of heat from a body at \(-10.0^{\circ} \mathrm{C} .\) How much work is done if the device exhausts heat into the environment at \((a) 25.0^{\circ} \mathrm{C} ;(\mathrm{b}) 0.0^{\circ} \mathrm{C} ;(\mathrm{c})-25.0^{\circ} \mathrm{C} ;\) In each case, is the device acting as an engine or as a refrigerator?

20.50. A stirling-cycle Engine. the Otto cycle, except that the compression and expansion of the gas are done at constant temperature, not adiabatically as in the Otto cycle. The Stirling cycle is used in external combustion engines (in fact, burning fuel is not necessary; any way of producing a temperature difference will do -solar, geothermal, ocean temperature gradient, etc. \(.\) which means that the gas inside the cylinder is not used in the combustion process. Heat is supplied by burning fuel steadily outside the cylinder, instead of explosively inside the cylinder as in the Otto cycle. For this reason Stirling-cycle engines are quieter than Otto-cycle engines, since there are no intake and exhaust valves (a major source of engine noise). While small Stirling engines are used for a variety of purposes, Stiring engines for automobiles have not been successful because they are larger, heavier, and more expensive than conventional automobile engines. In the cycle, the working fluid goes through the following sequence of steps (Fig. 20.30\()\) : (i) Compressed isothermally at temperature \(T_{1}\) from the initial state \(a\) to state \(b\) , with a compression ratio \(r .\) (ii) Heated at constant volume to state \(c\) at temperature \(T_{2}\) . (iii) Expanded isothermally at \(T_{2}\) to state \(d\) . (iv) Cooled at constant volume back to the initial state \(a\) . Assume that the working fluid is \(n\) moles of an ideal gas (for which \(C_{V}\) is independent of temperature). (a) Calculate \(Q, W,\) and \(\Delta U\) for each of the processes \(a \rightarrow b, b \rightarrow c, c \rightarrow d,\) and \(d \rightarrow a\) . (b) In the Stirling cycle, the heat transfers in the processes \(b \rightarrow c\) and \(d \rightarrow a\) do not involve external heat sources but rather use regeneration: The same substance that transfers heat to the gas inside the cylinder in the process \(b \rightarrow c\) also absorbs heat back from the gas in the process \(d \rightarrow a\) . Hence the heat transfers \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) do not play a role in determining the efficiency of the engine. Explain this last statement by comparing the expressions for \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) calculated in part (a). (c) Calculate the efficiency of a Stirling-cycle engine in terms of the temperatures \(T_{1}\) and \(T_{2}\) . How does this compare to the efficiency of a Carnot-cycle engine operating between these same two temperatures? (Historically, the Stirling cycle was devised before the Carnot cycle.) Does this result violate the second law of thermodynamics? Explain. Unfortunately, actual Stirling-cycle engines cannot achieve this efficiency due to problems with the heat-transfer processes and pressure losses in the engine.
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