A Carnot engine whose hot-reservoir temperature is 400℃ has a thermal efficiency of 40%. By how many degrees should the temperature of the cold reservoir be decreased to raise the engine's efficiency to 60%?

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Hello, fellow physicists today, we're going to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, consider a heat engine that operates on a cargo cycle. The engine's thermal efficiency is 35% when the temperature of the hot reservoir is 450 degrees Celsius to increase the efficiency to 65% determine the necessary reduction in the temperature of the cold reservoir. So that is our end goal. OK. So we're given some multiple choice answers. They're all in the same units of degrees Celsius. So let's read them off to see what our final answer might be. A is 470 B is 253 C is and D is 352. So first off, let us recall the equation for the efficiency of a Carno engine. So a ac we're gonna denote it as A to C for the Carno engine. So the equation for the efficiency of a Carno engine is equal to one minus the temperature of the cold reservoir divided by the temperature of the hot reservoir. So now we need to plug in our known variables and rearrange to isolate. And so for the temperature of the cold reservoir, so we're given the thermal efficiency as 35%. So that's gonna represent A to C but we need to turn the percent into a decimal. So it's gonna be 0.35 and you just take 35 divided by 100 is equal to one minus the temperature of the cold reservoir divided by the temperature of the hot reservoir, which we need to convert the temperature of the hot reservoir from 450 degrees C to Kelvin. So 450 degrees Celsius, all you gotta do is take that and add 273 which is equal to 723 Kelvin. So 723 Kelvin. So now we need to rearrange this equation to solve for the temperature of the cold reservoir. So when we do a little bit of algebra, do some rearranging here, we get 0. multiplied by 723. Kelvin, which is equal to when we plug into a calculator, 469.95 which is approximately Kelvin. But we now we need to increase the efficiency to 65%. So now A to C is equal or A to C I should say prime is equal to 0.65 because 65 divided by 100 0.65. So now the temperature of the cold reservoir prime is equal to the temperature of the hot reservoir multiplied by one minus a ac prime. So now we need to plug in our known variables to solve or temperature of the cool reservoir prime is equal to 450 plus 273. So we already converted it. So now we take our temperature for the hot reservoir which was 723. Kelvin multiplied by one minus A to C prime, which was 0.65 which when we plug into a calculator, we should get 253 Calvin, but we need to convert 253 Kelvin to degrees Celsius. So 253 Kelvin. So all we have to do is just take and minus 273 which is equal to negative 20 degrees Celsius. So now we must find the difference in the temperature of the cool reservoirs. So define the temperature difference. We take 450 degrees C and we, and we subtract negative two negative 20 degrees Celsius because we take the temperature of the hot reservoir minus the temperature of the cold reservoir. So when we do that, we get that the temperature is 470 degrees Celsius, which is our final answer. So therefore, or thus to increase the efficiency for 35% to 65% the temperature must be decreased by 470 degrees Celsius. Ok. So that means looking at our multiple choice answers, the correct answer is the letter A 470 degrees Celsius. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

A Carnot engine whose hot-reservoir temperature is 400℃ has a thermal efficiency of 40%. By how many degrees should the temperature of the cold reservoir be decreased to raise the engine's efficiency to 60%?

Verified Solution

Key Concepts

Carnot Efficiency

Thermal Reservoirs

Temperature Conversion