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Chapter 20: Problem 29

Consider a Carnot engine that works between thermal reservoirs with temperatures of \(1000.0 \mathrm{~K}\) and \(300.0 \mathrm{~K}\). The average power of the engine is \(1.00 \mathrm{~kJ}\) per cycle. a) What is the efficiency of this engine? b) How much energy is extracted from the warmer reservoir per cycle? c) How much energy is delivered to the cooler reservoir?

Short Answer

Expert verified
Answer: The engine's efficiency is 70%, the amount of energy extracted from the warmer reservoir per cycle is about 1428.57 J, and the amount of energy delivered to the cooler reservoir per cycle is about 428.57 J.

Step by step solution

01

Determine the efficiency of the engine

We will use the formula for Carnot engine efficiency: \(\eta = 1 - \frac{T_c}{T_h}\), where \(\eta\) is the efficiency, \(T_c\) is the temperature of the colder reservoir, and \(T_h\) is the temperature of the hotter reservoir. Given \(T_h = 1000.0 \mathrm{~K}\) and \(T_c = 300.0 \mathrm{~K}\), we can calculate the efficiency: \(\eta = 1 - \frac{300.0}{1000.0} = 1 - 0.3 = 0.7\) The engine's efficiency is 70%.
02

Calculate the energy extracted from the warmer reservoir per cycle

We know the power, which represents the useful work done per cycle, is equal to \(1.00 \mathrm{~kJ}\) or \(1000.0 \mathrm{~J}\). Efficiency is defined as the ratio of useful work done to the energy extracted from the hotter reservoir we will call \(Q_h\). Therefore, we can set up the equation: \(\eta = \frac{W}{Q_h}\) Now, we can solve for \(Q_h\): \(Q_h = \frac{W}{\eta} = \frac{1000.0 \mathrm{~J}}{0.7} \approx 1428.57 \mathrm{~J}\) The amount of energy extracted from the warmer reservoir per cycle is about 1428.57 J.
03

Calculate the energy delivered to the cooler reservoir per cycle

Energy conservation implies that the energy extracted from the warmer reservoir is equal to the sum of the useful work done by the engine and the energy delivered to the cooler reservoir, which we will call \(Q_c\). Therefore, we can write the equation: \(Q_h = W + Q_c\) Now, we can solve for \(Q_c\): \(Q_c = Q_h - W = 1428.57 \mathrm{~J} - 1000.0 \mathrm{~J} \approx 428.57 \mathrm{~J}\) The amount of energy delivered to the cooler reservoir per cycle is about 428.57 J.

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Most popular questions from this chapter

A heat pump has a coefficient of performance of 5.0. If the heat pump absorbs 40.0 cal of heat from the cold outdoors in each cycle, what is the amount of heat expelled to the warm indoors?

Suppose a person metabolizes \(2000 .\) kcal/day. a) With a core body temperature of \(37.0^{\circ} \mathrm{C}\) and an ambient temperature of \(20.0^{\circ} \mathrm{C}\), what is the maximum (Carnot) efficiency with which the person can perform work? b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings? c) With a skin area of \(1.50 \mathrm{~m}^{2}\), a skin temperature of \(27.0^{\circ} \mathrm{C}\) and an effective emissivity of \(e=0.600,\) at what net rate does this person radiate heat to the \(20.0^{\circ} \mathrm{C}\) surroundings? d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs At body temperature, the latent heat of vaporization of water is \(575 \mathrm{cal} / \mathrm{g}\). At what rate, in grams per hour, does this person lose water? e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of \(37.0^{\circ} \mathrm{C}\).

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