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Chapter 6: Problem 14

We can think of wind in the atmosphere as a giant heat engine \(^{67}\) operating between the \(288 \mathrm{~K}\) surface and the top of the troposphere \(^{68}\) at \(230 \mathrm{~K}\). What is the maximum efficiency this heat engine could achieve in converting solar heating into airflow?

Short Answer

Expert verified
Answer: The maximum efficiency of the heat engine is approximately 20.14%.

Step by step solution

01

Identify the given information

We are given the two temperatures between which the heat engine operates: - \(T_H = 288 \mathrm{~K}\) (higher temperature - surface) - \(T_C = 230 \mathrm{~K}\) (lower temperature - top of the troposphere)
02

Write down the formula for the Carnot efficiency

The formula for the efficiency of a heat engine based on the Carnot cycle is: \(\eta = 1 - \frac{T_C}{T_H}\)
03

Substitute the given temperatures into the formula

We substitute the given temperatures into the formula for the Carnot efficiency: \(\eta = 1 - \frac{230 \mathrm{~K}}{288 \mathrm{~K}}\)
04

Calculate the efficiency

We perform the calculation to find the efficiency: \(\eta = 1 - \frac{230}{288} = 1 - 0.7986 \approx 0.2014\)
05

Express the efficiency as a percentage

To express the efficiency as a percentage, we multiply the result by 100: \(\eta = 0.2014 \times 100\% = 20.14\%\) The maximum efficiency this heat engine could achieve in converting solar heating into airflow is approximately 20.14%.

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Most popular questions from this chapter

What would the maximum thermodynamic efficiency be of some heat engine operating between your skin temperature and the ambient environment \(20^{\circ} \mathrm{C}\) cooler than your skin?

In a house achieving a heat loss rate of \(200 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) equipped only with two \(1,500 \mathrm{~W}\) space heaters, what is the coldest it can get outside if the house is to maintain an internal temperature of \(20^{\circ} \mathrm{C} ?\)

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Let's say you come home from a winter vacation to find your house at \(5^{\circ} \mathrm{C}\) and you want to heat it to \(20^{\circ} \mathrm{C}\). Let's say the house contains: \(500 \mathrm{~kg}\) of air; \(^{62} 1,000 \mathrm{~kg}\) of furniture, books, and other possessions; plus walls and ceiling and floor that amount to \(6,000 \mathrm{~kg}\) of effective \(^{63}\) mass. Using the catch-all specific heat capacity for all of this stuff, how much energy will it take, and how long to heat it up at a rate of \(10 \mathrm{~kW}\) ? Express in useful, intuitive units, and feel free to round, since it's an estimate, anyway.
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