For which combination of working temperatures | vedclass

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

Question

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

Vedclass · Accepted Answer

$40 K, 20 K$

Vedclass · Answer

(D) The efficiency of a Carnot engine is given by the formula $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.To maximize the efficiency $\eta$,the ratio $\frac{T_2}{T_1}$ must be minimized.Let us calculate the ratio for each option:$(a)$ $\frac{60}{80} = 0.75$$(b)$ $\frac{80}{100} = 0.80$$(c)$ $\frac{40}{60} \approx 0.667$$(d)$ $\frac{20}{40} = 0.50$Comparing these values,the ratio is minimum for option $(d)$. Therefore,the efficiency is highest for the combination $40 K$ and $20 K$.

For which combination of working temperatures is the efficiency of a Carnot engine the highest?

Similar Questions

$A$ Carnot engine operates between $300 \ K$ and $600 \ K$. If the work done per cycle is $800 \ J$,then the heat supplied per cycle is ....... $J/cycle$.

$A$ Carnot engine works between $27^{\circ} C$ and $127^{\circ} C$. Heat supplied by the source is $500 \ J$. The heat ejected to the sink is (in $J$):

$A$ heat engine has an efficiency of $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{\circ}C$,its efficiency gets doubled. The temperature of the source is $.....^{\circ}C$.

Show the four steps of a Carnot engine in a $P-V$ graph,write the equation for each step,and obtain the work done by the system. Also,derive the efficiency of a Carnot engine.

$A$ Carnot engine absorbs $3 \times 10^{6} \text{ cal}$ of heat from a reservoir at $627^{\circ}C$ and rejects it at $27^{\circ}C$. The work done by the engine is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module