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.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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.