For a Carnot engine,the efficiency is given b | vedclass

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

Question

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Initially,$\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6} \qu

Vedclass · Accepted Answer

$37^{\circ}C$ and $99^{\circ}C$

Vedclass · Answer

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Initially,$\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6} \quad \dots(i)$When the sink temperature $T_2$ is reduced by $62^{\circ}C$ (which is equivalent to $62 K$ in absolute scale),the new sink temperature is $T_2' = T_2 - 62$.The new efficiency is $\eta' = 1 - \frac{T_2 - 62}{T_1} = 2\eta$.Substituting $\eta = 1/6$,we get $\eta' = 2 	imes (1/6) = 1/3$.So,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3} \quad \dots(ii)$.From $(i)$,$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$.Substitute $T_2$ into $(ii)$: $\frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{2}{3} \implies \frac{5}{6} - \frac{62}{T_1} = \frac{2}{3}$.$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6} \implies T_1 = 62 	imes 6 = 372 K$.Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ}C$.Now,$T_2 = \frac{5}{6} 	imes 372 = 310 K$.Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ}C$.Thus,the sink temperature is $37^{\circ}C$ and the source temperature is $99^{\circ}C$.

For a Carnot engine,the efficiency is given by $W/Q_1 = 1/6$. If the temperature of the sink is reduced by $62^{\circ}C$,the efficiency doubles. The initial temperatures of the sink and the source are,respectively:

Similar Questions

In a Carnot engine,as the gas gives heat energy to the sink,the temperature of the sink

$A$ Carnot engine takes $3 \times 10^6$ calories of heat from a reservoir at $627^{\circ} C$ and gives it to a sink at $27^{\circ} C$. The work done by the engine is

The efficiency of a Carnot engine is $\eta$ when its hot and cold reservoirs are maintained at temperatures $T_1$ and $T_2$, respectively. To increase the efficiency to $1.5 \eta$, the increase in temperature $(\Delta T)$ of the hot reservoir, while keeping the cold reservoir constant at $T_2$, is

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

If the temperature of the sink is at absolute zero,then the efficiency of the Carnot engine will be ........ $\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module