Problem 3008. Carnot cycle
Carnot engine is considered the most efficient heat engine.When operated between two temperatures T1 and T2, the efficiency is given by: efficiency=1-T1/T2 where T1<=T2 and both are in kelvin. Given the temperatures in degree celsius, find the efficiency.
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I think their is a problem with the test case 3.
I guess you're right
T1 and T2 do not necessarily imply x(1) and x(2).the higher temp. is always in den.
I have used Max(x) and Min(x) ... then also the test case does not satisfy...
The problem is the isequal() without any tolerance...
The conversion between Kelvin and Celsius is 273.15, not 273.
Read this before you try this problem: the author is using 273 for the conversion between Celsius and Kelvin, the two temperatures are not necessarily ordered, and the shown formula is correct although it is not in its most computationally realiable form. Good luck.
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