Question

A carnot engine works as a refrigerator in between 250 K and 300 K. If it acquires 750 calories from heat source at low temperature, then what is the heat generated at higher temperature. (in calories) ?

Answer 1

$\begin{array}{c}n=\frac{T_1-T_2}{T}=\frac{{\theta }_1-{\theta }_2}{{\theta }_1}\\ =\frac{300-250}{300}=\frac{\theta -750}{d}\\ \Rightarrow \frac{50}{300}=1-\frac{750}{\theta }\\ \Rightarrow \frac{750}{\theta }=1-\frac{1}{6}=\frac{5}{6}\\ \therefore \theta =\frac{750\times 6}{5}\\ =900\end{array}$