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20.22. 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 \(\mathrm{kg}\) of ice. During this time, how much work \(W\) is performed by the engine?

Short Answer

Expert verified
Work performed cannot be determined without temperatures of reservoirs.

Step by step solution

01

Identify the Given Information

The heat engine operates between a hot reservoir of boiling water and a cold reservoir of ice. The mass of ice melted is given as 0.0400 kg, and the time of operation is 5 minutes.
02

Determine the Latent Heat of Fusion

The latent heat of fusion of ice is a known value, approximately 334,000 J/kg. This is the amount of energy required to melt 1 kg of ice without changing its temperature.
03

Calculate Heat Rejected (Q_c)

The energy required to melt the ice, which is the heat rejected by the engine, can be calculated using the formula \( Q_c = mL \), where \( m = 0.0400 \) kg is the mass of ice, and \( L = 334,000 \) J/kg is the latent heat of fusion. \[ Q_c = 0.0400 \times 334,000 = 13,360 \text{ J} \]
04

Recall Carnot Efficiency and Relationship to Work

Carnot efficiency for a reversible heat engine is given by \( \eta = 1 - \frac{T_c}{T_h} \). The work done by the engine can be calculated as \( W = Q_h - Q_c \), where \( Q_h \) is the heat absorbed from the hot reservoir.
05

Use Efficiency to Find Q_h and W

In an ideal Carnot engine, \( Q_c = Q_h \times \frac{T_c}{T_h} \). Rearrange to find \( Q_h \), and use this to compute work \( W = Q_h - Q_c \). Since only \( Q_c \) is given, more information is needed, such as the temperatures of the reservoirs.

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Key Concepts

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

Latent Heat of Fusion
The latent heat of fusion is a crucial concept when understanding phase changes, specifically from solid to liquid. In simple terms, it denotes the amount of heat needed to convert a solid into a liquid at constant temperature. For ice, the latent heat of fusion is approximately 334,000 J/kg. This value implies that to melt 1 kilogram of ice without raising its temperature, 334,000 joules of energy are required. In our problem, with 0.0400 kg of ice being melted by the engine, we use the heat of fusion to calculate the energy transfer involved.
  • For 0.0400 kg, the energy needed can be determined by multiplying the mass with the latent heat of fusion, leading to 13,360 J.
  • This energy represents the amount rejected by the Carnot engine.
Latent heat is a key component in thermodynamics, especially in studying how energy moves during phase changes, without altering temperature.
Heat Rejected
In the context of a thermodynamic cycle, heat rejected is the energy discharged or released by an engine to the surrounding environment. For the Carnot heat engine, the ice acts as the cold reservoir where this heat is rejected. By melting the ice, the engine transfers energy, effectively rejecting it. This is quantitatively determined using the latent heat of fusion formula.
  • In our example, with a mass of 0.0400 kg of ice, the heat rejected is 13,360 J.
This process highlights the functioning of the heat engine in maintaining its cycle. It captures the unavoidable process of losing some energy to the cold reservoir, pivotal for any real or theoretical engine. Remember, while heat rejected illustrates loss from the engine's perspective, it follows the principle of energy conservation in physical processes.
Carnot Efficiency
Carnot Efficiency is a theoretical metric determining how effectively a heat engine converts absorbed heat into useful work. It's a defining formula given by \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) and \( T_h \) are the absolute temperatures of the cold and hot reservoirs, respectively. This equation reveals that efficiency inversely correlates with the ratio of these temperatures.
  • A lower \( T_c \) or a higher \( T_h \) increases efficiency.
  • However, reaching 100% efficiency is impossible due to the second law of thermodynamics.
This Carnot cycle serves as the benchmark for real-world engines, which can never exceed this ideal. Understanding efficiency is vital when analyzing how much work an engine can optimally perform given its heat intake and loss with the universal statistics defined by Carnot.
Work Performed
In thermodynamics, the work performed by a heat engine is the net energy output after it has completed a cycle. In a Carnot engine, we can compute the work performed by analyzing the difference between the heat absorbed (\( Q_h \)) and heat rejected (\( Q_c \)). The relationship is defined as \( W = Q_h - Q_c \).
  • To solve our problem, we first need the temperatures of both reservoirs to derive \( Q_h \) using the Carnot efficiency.
  • Subsequent calculations would then simplify to finding \( W \) after subtracting \( Q_c \) from the obtained \( Q_h \).
The significance of work in this context emphasizes the use of thermodynamic processes to harness usable energy. It's a practical application showing energy transformation, critical in any mechanical system.

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Most popular questions from this chapter

20.5. A certain nuclear-power plant has a mechanical-power output (used to drive an electric generator) of 330 \(\mathrm{MW}\) . Its rate of heat input from the nuclear reactor is 1300 \(\mathrm{MW}\) . (a) What is the thermal efficiency of the system? (b) At what rate is heat discarded by the system?

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.

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?

20.29. 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 isothernal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air. What is the total entropy change of the system tea \(+\) air?

20.56. The maximum power that can be extracted by a wind turbine from an air stream is approximately $$ P=k d^{2} v^{3} $$ where \(d\) is the blade diameter \(v\) is the wind speed, and the constant \(k=0.5 \mathrm{W} \cdot \mathrm{s}^{3} / \mathrm{m}^{5} .\) (a) Explain the dependence of \(P\) on \(d\) and on \(v\) by considering a cylinder of air that passes over the turbine blades in time \(t(\text { Fig. } 20.31)\) . This cylinder has diameter \(d .\) length \(L=v t\) and density \(\rho .\) (b) The Mod-SB wind turbine at Kahaku on the Hawaiian island of Oahu has a blade diameter of 97 \(\mathrm{m}\) (slightly longer than a football field sits atop a \(58-\mathrm{m}\) tower. It can produce 3.2 \(\mathrm{MW}\) of electric power. Assuming 25\(\%\) efficiency, what wind speed is required to produce this amount of power? Give your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) . (c) Commercial wind turbines are commonly located in or downwind of mountain passes. Why?

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