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(a) A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00 MJ into the engine. How much heat transfer occurs to the environment? (b) What is unreasonable about the engine? (c) Which premise is unreasonable?

Short Answer

Expert verified
Heat transfer to the environment is 1.00 MJ. The engine is unreasonable because its efficiency (80%) is greater than the maximum possible efficiency given by the Carnot limit (41.5%). The premise that an engine could exceed the Carnot efficiency is unreasonable.

Step by step solution

01

Convert temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin by adding 273.15 to each Celsius temperature. The temperature of the hot reservoir, in Kelvin, is 450 + 273.15 = 723.15 K. The temperature of the cold reservoir, in Kelvin, is 150 + 273.15 = 423.15 K.
02

Calculate heat transfer to the environment

The heat transfer to the environment can be found by subtracting the work done by the engine from the heat input to the engine. Using the conservation of energy, the heat transferred to the cold reservoir (environment) is the heat input minus the work output: Q_cold = Q_hot - W = 5.00 MJ - 4.00 MJ = 1.00 MJ.
03

Assess the efficiency of the engine

Calculate the efficiency of the engine using the formula efficiency = 1 - (Q_cold / Q_hot). Using the calculated values, efficiency = 1 - (1.00 MJ / 5.00 MJ) = 0.80 or 80%.
04

Compare efficiency with the Carnot efficiency

The maximum efficiency for any heat engine operating between two temperatures is given by the Carnot efficiency, which is efficiency_carnot = 1 - (T_cold / T_hot). Calculating the Carnot efficiency gives: efficiency_carnot = 1 - (423.15 K / 723.15 K) = 1 - 0.585 = 0.415 or 41.5%.
05

Determine the reasonableness of the engine's operation

By comparing the engine's efficiency (80%) with the Carnot efficiency (41.5%), we can see that the actual efficiency exceeds the maximum possible (Carnot) efficiency for an engine operating between these two temperatures, which is not possible according to the second law of thermodynamics.
06

Identify the unreasonable premise

Based on the second law of thermodynamics and the Carnot efficiency, the premise that this engine could have an efficiency of 80% is unreasonable. The maximum efficiency must be less than or equal to the Carnot efficiency.

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Key Concepts

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

Carnot Efficiency
Understanding Carnot efficiency is crucial when analyzing the performance of heat engines. In layman's terms, the Carnot efficiency is the ideal or the highest possible efficiency that a heat engine operating between two temperatures can achieve. This concept honors the French physicist Sadi Carnot, who first described it.

The formula for Carnot efficiency is given by: \[\text{efficiency}_{\text{carnot}} = 1 - \left(\frac{T_{\text{cold}}}{T_{\text{hot}}}\right)\] where \(T_{\text{hot}}\) is the temperature of the heat source and \(T_{\text{cold}}\) is the temperature of the heat sink, both in Kelvin.

In the textbook problem, the Carnot efficiency is calculated to be 41.5%. This means that no real engine operating between these temperatures can have an efficiency greater than 41.5%. Any claim to the contrary would indicate an impossible scenario given our current understanding of thermodynamics, as it would suggest a violation of the second law of thermodynamics.
Heat Transfer Calculation
Calculating heat transfer is a fundamental part of thermodynamics that involves quantifying the energy exchange due to temperature differences. For a heat engine, this means determining how much energy is absorbed from the hot source and how much is expelled to the cold sink.

In the context of the exercise, the heat absorbed by the engine (the heat input) is given as 5.00 MJ. The work output, or the energy converted into work, is 4.00 MJ. Using the law of conservation of energy, the heat expelled to the environment (the heat output) can be found by subtracting the work from the heat input:\[Q_{\text{cold}} = Q_{\text{hot}} - W\]
This equation states that the heat transfer to the cold reservoir, \(Q_{\text{cold}}\), is simply the initial heat input, \(Q_{\text{hot}}\), minus the work produced by the engine, \(W\). Thus, the heat transfer to the environment is calculated as 1.00 MJ.

This step is essential not just to determine engine performance but also to account for energy dispersion in any thermodynamic system.
Second Law of Thermodynamics
The second law of thermodynamics is a fundamental principle that has profound implications on the efficiency of heat engines. In essence, it states that heat cannot spontaneously transfer from a colder body to a hotter one, and that the total entropy, or disorder, in an isolated system can never decrease over time.

One key consequence of the second law is that it sets a limit on the maximum efficiency of heat engines. As shown in the problem, the calculated efficiency of the hypothetical engine is 80%, which exceeds the calculated Carnot efficiency. This defies the second law, which asserts that it's impossible for any engine to be more efficient than a Carnot engine operating between the same two temperatures.

This law not only dictates the fundamental limits of energy conversion but also challenges and inspires engineers to design engines that approach but do not surpass the efficiency dictated by Carnot's theoretical model.

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Most popular questions from this chapter

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