A Carnot engine with efficiency eta operates | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Taking the differential,we get $d\eta = \frac{T_2}{T_1^2} dT_1 - \frac{

Vedclass · Accepted Answer

-$2$

Vedclass · Answer

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Taking the differential,we get $d\eta = \frac{T_2}{T_1^2} dT_1 - \frac{1}{T_1} dT_2$.For a small change in $T_1$ (keeping $T_2$ constant),$\Delta \eta_1 = \frac{T_2}{T_1^2} \Delta T_1 = \frac{T_2}{T_1} \left( \frac{\Delta T_1}{T_1} ight) = (1 - \eta) \frac{\Delta T_1}{T_1}$.Given $\frac{\Delta T_1}{T_1} = 0.4 \%$,so $\Delta \eta_1 = (1 - \eta) 	imes 0.004$.For a small change in $T_2$ (keeping $T_1$ constant),$\Delta \eta_2 = -\frac{1}{T_1} \Delta T_2 = -\frac{T_2}{T_1} \left( \frac{\Delta T_2}{T_2} ight) = -(1 - \eta) \frac{\Delta T_2}{T_2}$.Given $\frac{\Delta T_2}{T_2} = 0.2 \%$,so $\Delta \eta_2 = -(1 - \eta) 	imes 0.002$.Taking the ratio,$\frac{\Delta \eta_1}{\Delta \eta_2} = \frac{(1 - \eta) 	imes 0.004}{-(1 - \eta) 	imes 0.002} = -2$.

Similar Questions

$A$ reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $62^\circ C$,the efficiency of the engine is doubled. The temperatures of the source and sink are:

For which of the following options will the efficiency of a Carnot engine be the highest?

$A$ Carnot's engine operates with a source at $127^{\circ} C$ and a sink at $27^{\circ} C$. If the source supplies $40 \ kJ$ of heat energy,the work done by the engine is (in $kJ$)

The efficiency of an ideal heat engine working between the freezing point and boiling point of water is ........ $\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module