A Carnot engine has efficiency of $50 \%$ .If the temperature of sink is reduced by $40^{\circ} C$, its efficiency increases by $30 \%$ .The temperature of the source will be :
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, \(\frac{1}{2}(1.3)=1-(\frac{T_L-40}{T_H})\) \(⇒\frac{1}{2}(1.3)=\frac{1}{2}+\frac{40}{T_H}\) \(\therefore T_H=266.7\ K\)
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.
‘No engine can be more efficient than a perfectly reversible engine working between the same two temperatures’.
‘The efficiency of all reversible engines, working between the same two temperatures is the same, whatever the working substances’.
Carnot’s Ideal Heat Engine
A cylinder having perfectly non-conducting walls, a perfectly conducting base and is provided with a perfectly non-conducting piston which moves without friction in the cylinder. The cylinder contains one mole of perfect gas as the working substance.
Source: A reservoir maintained at a constant temperature T1 from which the engine can draw heat by perfect conduction. It has an infinite thermal capacity and any amount of heat can be drawn from it at constant temperature T1.
Heat insulating stand. A perfectly mom-conducting platform acts as a stand for adiabatic processes.
Sink: A reservoir maintained at a constant lower temperature T2 (T2 < T1) to which the heat engine can reject any amount of heat. The thermal capacity of sink is infinite so that its temperature remains constant at T2, no matter how much heat is given to it.
