Math, asked by SnehaGupta6543, 1 year ago

A carnot engine with sinks temperature at 17 degree has 50% efficiency. By how much should its source temperature be changed to increase its efficiency to 60%

Answers

Answered by jaswanth200518
5

THE ABOVE ONE IS THE ANSWER

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Answered by prachikalantri
0

Answer: T=17+273

T=290k

\eta=1-\frac{T}{T_0}

0.5=1-\frac{290}{T_0}

[\frac{290}{T_0}=\frac{1}{2}  ]\\T_0=580k

\eta=1-\frac{T}{T_1} \\0.6=1-\frac{1}{T_1}

T_1=\frac{T}{0.4} =\frac{290}{0.4} =725k

T_1-T_0=145k

#SPJ3

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