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Ch 20: The Second Law of Thermodynamics

Chapter 20, Problem 20

A Carnot engine is operated between two heat reservoirs at temperatures of 520 K and 300 K. (c) What is the thermal efficiency of the engine?

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Hey everyone welcome back in this problem. We have an ideal and reversible heat engine. Okay. That absorbs 7000 joules of heat and operates between two thermal pools. Okay. The high temperature source is 420°C and the sink is at 300°C. We are asked to find the efficiency of the engine. Okay. Okay. So let's recall that the efficiency E. Is equal to the work divided by Q. H. The quantity of heat. Mhm. We can also write this as one minus the absolute value of Q. C. Divided by Q. H. Okay. The ratio of heat quantities. Alright, let's fill out the information we know and see where we can go from there. Okay, so we're told that the heat engine absorbs 7000 joules of heat. Okay. And so are the absolute value of Q. H. Okay, absorbing heat is going to be 7000 jules. Okay. We're also told the temperature the high temperature is 420°C. Okay. The absolute value of th or just th is equal to 420 degrees Celsius Which is equal to 693.15 Calvin adding 273.15. We're told that the temperature at the sink. So the low temperature or the temperature of the cold reservoir T. C. Is equal to 300°C which is equal to 573.15 Calvin. Alright, so looking at our efficiency equation, we have Q. H. But we don't have QC. Okay now let's recall. We can relate the ratio of the quantity of heat cues to the ratio of the temperatures. So we can write the absolute value of QC divided by Q H. Is equal to the absolute value of the ratio of temperatures. TC over th Okay. All right now, first instinct might be to find Q. C. Okay, we don't know Q. C. We know Q. H. So go ahead and find Q. C. Plug it into the equation. But what you'll see is that in our efficiency equation when finding E. We only use the ratio QC over Q. H. Okay, so we don't even need to find Q. C. On its own. We just need to find the ratio of the two. Okay, so on the right hand side we're going to get the temperature and the cold reservoir 573.15 Calvin, divided by the temperature in the hot reservoir. 693.15 Calvin. Okay, the unit of Calvin will cancel be left dimension liss. We get that the ratio QC two qh Is equal to 0. 68. Okay, Alright, now we have this ratio we can go ahead and find E. Okay, so now e the efficiency that we are trying to find is equal to 1 -0. 268. And we can see that the possible answers have been rounded to the nearest percent. So this is going to be 0.17, Which is equal to 17%. Okay. I guess this is approximately equal just based on a rounding. Okay. And we get an efficiency of 17% for this engine. Okay. The answer is going to be B. E. is equal to 17%. Thanks everyone for watching. See you in the next video.
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Textbook Question
A Carnot engine is operated between two heat reservoirs at temperatures of 520 K and 300 K. (a) If the engine receives 6.45 kJ of heat energy from the reservoir at 520 K in each cycle, how many joules per cycle does it discard to the reservoir at 300 K?
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Textbook Question
A Carnot engine is operated between two heat reservoirs at temperatures of 520 K and 300 K. (b) How much mechanical work is performed by the engine during each cycle?
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Textbook Question
A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it?
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Textbook Question
A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (c) What is the coefficient of performance of the refrigerator?
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Textbook Question
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 kg of ice. During this time, how much work W is performed by the engine?
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