Consider a reversible engine of efficiency fr | vedclass

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

Question

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperatur

Vedclass · Accepted Answer

$372 \,K$ and $310 \,K$

Vedclass · Answer

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.Given $\eta_1 = \frac{1}{6}$, we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$, which implies $\frac{T_2}{T_1} = \frac{5}{6}$ or $T_2 = \frac{5}{6} T_1$ (Eq. $1$).When the sink temperature is reduced by $62^{\circ} C$, the new sink temperature is $T_2' = T_2 - 62$. The new efficiency is $\eta_2 = 2 	imes \eta_1 = 2 	imes \frac{1}{6} = \frac{1}{3}$.Thus, $\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$.Substituting $T_2 = \frac{5}{6} T_1$ into the equation: $\frac{1}{3} = 1 - \frac{\frac{5}{6} T_1 - 62}{T_1} = 1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{6} + \frac{62}{T_1}$.Rearranging gives $\frac{1}{3} - \frac{1}{6} = \frac{62}{T_1}$, so $\frac{1}{6} = \frac{62}{T_1}$, which means $T_1 = 372 \,K$.Using Eq. $1$, $T_2 = \frac{5}{6} 	imes 372 = 310 \,K$.

Consider a reversible engine of efficiency $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{\circ} C$, its efficiency gets doubled. The temperature of the source and sink respectively are

Similar Questions

$A$ reversible engine and an irreversible engine are working between the same temperatures. The efficiency of the ...........

$A$ reversible engine converts $1/6$ of its input heat into work. When the temperature of the sink is reduced by $62^{\circ}C$,the efficiency of the engine is doubled. Find the temperatures of the source and the sink.

In a Carnot engine,when ${T_2} = {0^o}C$ and ${T_1} = {200^o}C$,its efficiency is ${\eta _1}$. When ${T_1} = {0^o}C$ and ${T_2} = -{200^o}C$,its efficiency is ${\eta _2}$. What is the ratio ${\eta _1}/{\eta _2}$?

$A$ Carnot engine whose heat sink is at $27\,^{\circ} C$ has an efficiency of $25 \%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency?

In practice,all heat engines have an efficiency less than that of a Carnot engine because

Vedclass Test Series

Exam Paper Generator

Online Exam Module