Question

What is the temperature of source in Carnot cycle of 10% efficiency when heat exhausts at 270 K?

• A. 400 K
• B. 500 K
• C. 300 K
• D. 600 K

Answer 1

The correct option is C 300 K

$\eta =1-\frac{T_2}{T_1}$
Therefore $\frac{T_2}{T_1}=1-\eta =1-\frac{10}{100}=\frac{90}{100}$

or $T_1=\frac{100T_2}{90}$
$=\frac{100}{90}\times 270=300K$