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:
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) 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.