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Chapter 19: Problem 21

At what cold-reservoir temperature (in \(^{\circ} \mathrm{C}\) ) would a Carnot engine with a hot-reservoir temperature of \(427^{\circ} \mathrm{C}\) have an efficiency of \(60 \% ?\)

Short Answer

Expert verified
The cold-reservoir temperature at which a Carnot engine with a hot-reservoir temperature of \(427^{\circ} \mathrm{C}\) would have an efficiency of \(60 \% \) is approximately \(6.91^{\circ} \mathrm{C}\).

Step by step solution

01

Convert the Hot-reservoir Temperature to Kelvin

The hot-reservoir temperature is given as \(427^{\circ} \mathrm{C}\). We need to convert this temperature into Kelvin. The conversion equation is \(T_{hK} = T_{hC} + 273.15\), where \(T_{hK}\) is the hot-reservoir temperature in Kelvin, and \(T_{hC}\) is the hot-reservoir temperature in Celsius. After substitution, we get \(T_{hK} = 427 + 273.15 = 700.15 \mathrm{K}\).
02

Substitute Efficiency in the Form of a Fraction

The efficiency of the engine is given as \(60 \%\). Convert this percentage to a decimal before using it in the efficiency formula. We get \(e = 0.60\) or \(60 \% = \frac{60}{100} = 0.60\).
03

Re-arrange the Efficiency Formula and Compute Cold Reservoir Temperature

Now we substitute \(e\) and \(T_{hK}\) into the efficiency equation \(e = 1 - \frac{T_c}{T_h}\) and calculate \(T_c\). This can be rearranged into \(T_c = T_h*(1-e)\). After substitution, we get \(T_c = 700.15 * (1 - 0.60) = 280.06 \mathrm{K}\).
04

Convert the Cold-reservoir Temperature to Celsius

Now we convert the cold-reservoir temperature from Kelvin to Celsius using the conversion formula \(T_{cC} = T_{cK} - 273.15\). Substituting the values, we get \(T_{cC} = 280.06 - 273.15 = 6.91^{\circ} \mathrm{C}\).

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the context of a Carnot engine, which is an idealized heat engine, thermodynamics plays a crucial role. It governs the laws that dictate how energy is converted from heat into work and how efficiency is limited by these laws.

At the heart of a Carnot engine's functionality is the concept of reversible processes and the idea that no engine can be more efficient than a Carnot engine operating between two given temperatures. Understanding thermodynamics and specifically the second law, which introduces the concept of entropy, is essential to comprehend why the efficiency of real engines is always less than that of an ideal Carnot engine.
Temperature Conversion
Temperature conversion is a vital step when working with thermal systems like Carnot engines because different equations may require temperature values in specific units. Most scientific work requires temperatures in the Kelvin scale, which is an absolute temperature scale.

To convert Celsius to Kelvin, the formula used is: \( T_K = T_C + 273.15 \). The reverse conversion, from Kelvin to Celsius, is just as simple: \( T_C = T_K - 273.15 \). These conversions are necessary for calculating thermal efficiency, as seen with the Carnot engine exercise. Accurate conversions ensure that the laws of thermodynamics are applied correctly.
Kelvin Scale
The Kelvin scale is the base unit of temperature in the International System of Units (SI) and is widely used in scientific applications, including thermodynamics. It is an absolute temperature scale with zero Kelvin equivalent to absolute zero - the lowest theoretical temperature where particles cease to have any thermal motion.

One unique feature of the Kelvin scale is that it uses the same increment size as the Celsius scale; thus, a difference of one Kelvin is the same as a difference of one degree Celsius. The Kelvin scale is crucial when discussing a Carnot engine's efficiency because it requires absolute temperatures to properly calculate the efficiency using the efficiency equation.
Efficiency Equation
The efficiency equation for a Carnot engine is pivotal to understanding how efficient the engine can possibly be. The equation is a simple representation of the idea that the efficiency of an engine is the ratio of the work output to the heat input. Specifically, for a Carnot engine, the efficiency (e) is given by the formula: \( e = 1 - \frac{T_c}{T_h} \), where \( T_c \) is the absolute temperature of the cold reservoir and \( T_h \) is the absolute temperature of the hot reservoir, both in Kelvin.

Understanding this equation allows one to calculate the potential maximum efficiency of a heat engine given two operating temperatures. In teaching this concept, it's critical to reinforce the fact that absolute temperatures must be used (hence the Kelvin scale) and that the temperatures need to be in the same temperature scale to ensure the calculated efficiency is accurate.

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Most popular questions from this chapter

a. A large nuclear power plant has a power output of \(1000 \mathrm{MW}\). In other words, it generates electric energy at the rate \(1000 \mathrm{MJ} / \mathrm{s}\). How much energy does this power plant supply in one day? b. The occans are vast. How much energy could be extracted from \(1 \mathrm{km}^{3}\) of water if its temperature were decreased by \(1^{\circ} \mathrm{C} 2\) For simplicity, assume fresh water. c. A friend of yours who is an inventor comes to you with an idea. He has done the calculations that you just did in parts a and \(b\), and he's concluded that a few cubic kilometers of ocean water could meet most of the energy needs of the United States. This is an insignificant fraction of the U.S. coastal waters. In addition, the oceans are constantly being reheated by the sun, so energy obtained from the occan is essentially solar energy. He has sketched out some design plans- -highly secret, of course, because they're not patented - and now he needs some investors to provide money for a prototype. A working prototype will lead to a patent. As an initial investor, you'll receive a fraction of all future royalties. Time is of the essence because a rival inventor is working on the same idea. He nceds \(\$ 10,000\) from you right away. You could make millions if it works out. Will you invest? If so, explain why. If not, why not? Either way, your explanation should be based on scientific principles. Sketches and diagrams are a reasonable part of an explanation.

The coefficient of performance of a refrigerator is \(5.0 .\) The compressor uses \(10 \mathrm{J}\) of energy per cycle. a. How much heat cnergy is exhausted per cycle? b. If the hot-reservoir temperature is \(27^{\circ} \mathrm{C},\) what is the lowest possible temperature in \(^{\circ} \mathrm{C}\) of the cold reservoir?

A Carnot engine operating between energy reservoirs at temperatures \(300 \mathrm{K}\) and \(500 \mathrm{K}\) produces a power output of \(1000 \mathrm{W}\) What are (a) the thermal efficiency of this engine, (b) the rate of heat input, in \(\mathrm{W}\), and (c) the rate of heat output, in \(\mathrm{W} ?\)

\(50 \mathrm{J}\) of work are done per cycle on a refrigerator with a coefficient of performance of \(4.0 .\) How much heat is (a) extracted from the cold reservoir and (b) exhausted to the hot reservoir per cycle?

Prove that the coefficient of performance of a Carnot refrigerator is \(K_{\text {Grast }}=T_{\mathrm{C}} /\left(T_{\mathrm{H}}-T_{\mathrm{C}}\right).\)
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