The correct answer is (C) : \(266.7\, K\)
Efficiency , \(η=1-\frac{T_L}{T_H}\)
Given : η = 50%
\(\therefore\frac{1}{2}=1-\frac{T_L}{T_H}\)
when η increases by 30% then,
\(\frac{1}{2}(1.3)=1-(\frac{T_L-40}{T_H})\)
\(⇒\frac{1}{2}(1.3)=\frac{1}{2}+\frac{40}{T_H}\)
\(\therefore T_H=266.7\ K\)
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:
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
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.