Kushagra Madhukar
Last Activity: 4 Years ago
Dear student,
Please find the answer to your question.
Efficiency of carnot engine (n1) = 40% = 0.4;
Initial intake temperature (T1) = 500 K
and new efficiency (n2) = 50% = 0.5
Now, Efficiency, n = 1 – T2/T1
or, T2/T1 = 1 – n
In first case, T2/500 = 1 – 0.4 = 0.6
=> T2 = 0.6 x 500 = 300 K
In second case, 300/T1 = 1 – 0.5 = 0.5
=> T1 = 300/0.5 = 600 K
Thanks and regards,
Kushagra