A reversible Carnot heat engine converts frac | vedclass

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

Question

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{W}{Q} = 1 - \frac{T_2}{T_1}$,where $W$ is work done,$Q$ is input heat,$T_1$ is

Vedclass · Accepted Answer

$600 \ K, 450 \ K$

Vedclass · Answer

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{W}{Q} = 1 - \frac{T_2}{T_1}$,where $W$ is work done,$Q$ is input heat,$T_1$ is the source temperature,and $T_2$ is the sink temperature.Initially,$\eta = \frac{1}{4}$. Therefore,$1 - \frac{T_2}{T_1} = \frac{1}{4} \implies \frac{T_2}{T_1} = \frac{3}{4} \implies T_2 = \frac{3}{4}T_1$ (Equation $i$).When the sink temperature is reduced by $50 \ K$,the new sink temperature is $T_2' = T_2 - 50$. The new efficiency is $\eta' = 33 \frac{1}{3} \% = \frac{1}{3}$.Thus,$1 - \frac{T_2 - 50}{T_1} = \frac{1}{3} \implies 1 - \frac{T_2}{T_1} + \frac{50}{T_1} = \frac{1}{3}$ (Equation $ii$).Substituting $\frac{T_2}{T_1} = \frac{3}{4}$ from Equation $i$ into Equation $ii$:$1 - \frac{3}{4} + \frac{50}{T_1} = \frac{1}{3} \implies \frac{1}{4} + \frac{50}{T_1} = \frac{1}{3}$.$\frac{50}{T_1} = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}$.$T_1 = 50 	imes 12 = 600 \ K$.Now,$T_2 = \frac{3}{4} 	imes 600 = 450 \ K$.Therefore,the initial temperatures are $600 \ K$ and $450 \ K$.

$A$ reversible Carnot heat engine converts $\frac{1}{4}$ of its input heat into work. When the temperature of the sink is reduced by $50 \ K$,its efficiency becomes $33 \frac{1}{3} \%$. The initial temperatures of the source and the sink respectively are

Similar Questions

$A$ Carnot engine absorbs heat from a reservoir maintained at temperature $1000 \,K$. The engine rejects heat to a reservoir whose temperature is $T$. If the magnitude of absorbed heat is $400 \,J$ and work performed is $300 \,J$, then the value of $T$ is (in $\,K$)

The source temperature of a heat engine is $727^{\circ}C$ and the sink temperature is $227^{\circ}C$. What is the maximum possible efficiency of this engine?

The efficiency of a Carnot engine is $25 \%$,when the temperature of the sink is $300 \ K$. The increase in the temperature of the source required for the efficiency to become $50 \%$ is (in $K$)

An ideal Carnot engine,whose efficiency is $40 \%$ receives heat at $500 \; K$. If its efficiency is $50 \%$ then the intake temperature for the same exhaust temperature is ......... $K$.

An ideal gas heat engine operates in a Carnot cycle between $227\,^{\circ}C$ and $127\,^{\circ}C$. It absorbs $6.0 \times 10^4\,cal$ at the higher temperature. The amount of heat converted into work is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module