Carnot engine efficiency and exhaust problem

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The discussion revolves around calculating the intake temperature of a Carnot heat engine with an exhaust temperature of 121°C and an efficiency of 13.4%. The initial calculation incorrectly used Celsius instead of Kelvin, leading to an incorrect result. The correct approach involves converting the temperatures to Kelvin for accurate calculations. Once the temperatures are converted, the efficiency formula can be applied correctly. This highlights the importance of using the appropriate temperature scale in thermodynamic equations.
lzh
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Homework Statement


The exhaust temperature of a Carnot heat engine is 121◦C. What is the intake temperature if the efficiency of the engine is 13.4 %? Answer in units of ◦C.


Homework Equations


e=1-(Tc/Th)


The Attempt at a Solution


This seems like just a plug and chug problem:
0.134=1-(121/Th)
Th=139.723 ◦C
However, this answer is wrong-according to my Quest web homework. Am I missing something here? Thanks
 
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Hi lzh,

lzh said:

Homework Statement


The exhaust temperature of a Carnot heat engine is 121◦C. What is the intake temperature if the efficiency of the engine is 13.4 %? Answer in units of ◦C.


Homework Equations


e=1-(Tc/Th)


The Attempt at a Solution


This seems like just a plug and chug problem:
0.134=1-(121/Th)
Th=139.723 ◦C
However, this answer is wrong-according to my Quest web homework. Am I missing something here? Thanks

The Celsius scale is not a thermodynamic temperature scale. You need to use a temperature scale such as Kelvin here.
 
Oh! I see! Thanks! I got it now!
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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