Initially :
\(\frac{1}{4}\)=1−\(\frac{300}{T_H}\)
⇒ TH = 400 K
Finally, : Efficiency becomes \(\frac{1}{2}\)
∴\(\frac{1}{2}\)=1−\(\frac{300}{T_H}\)
∴TH=600 K
⇒ The temperature of the source increases by 200°C.
The correct option is (B) : Increases by 200°C.
In 1st case, Carnot engine operates between temperatures 300 K and 100 K. In 2nd case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in 2nd case) will be:
From the second law of thermodynamics, two important results are derived where the conclusions are taken together to constitute Carnot’s theorem. It may be stated in the following forms.