Problem 3008. Carnot cycle

Created by Ritwik Rudra

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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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Last Solution submitted on Nov 13, 2024

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Philemon Chavrier on 12 Feb 2015

I think their is a problem with the test case 3.

Romain Dupouy on 12 Feb 2015

I guess you're right

Ritwik Rudra on 12 Feb 2015

T1 and T2 do not necessarily imply x(1) and x(2).the higher temp. is always in den.

Anupam Agarwal on 13 Feb 2015

I have used Max(x) and Min(x) ... then also the test case does not satisfy...

Sylvain on 13 Feb 2015

The problem is the isequal() without any tolerance...

James on 13 Feb 2015

The conversion between Kelvin and Celsius is 273.15, not 273.

Rafael S.T. Vieira on 28 Sep 2020

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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Problem 3008. Carnot cycle

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Problem 3008. Carnot cycle

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