The efficiency of a Carnot engine is 50% and | vedclass

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

Question

(C) The efficiency $\eta$ 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 temp

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(C) The efficiency $\eta$ 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 $\eta_1 = 50\% = 0.5$ and $T_2 = 500\, K$.$0.5 = 1 - \frac{500}{T_1} \Rightarrow \frac{500}{T_1} = 0.5 \Rightarrow T_1 = 1000\, K$.Now,the source temperature $T_1$ is kept constant at $1000\, K$ and the efficiency is to be increased to $\eta_2 = 60\% = 0.6$.$0.6 = 1 - \frac{T_2'}{1000}$.$\frac{T_2'}{1000} = 1 - 0.6 = 0.4$.$T_2' = 0.4 	imes 1000 = 400\, K$.Thus,the required temperature of the sink is $400\, K$.

The efficiency of a Carnot engine is $50\%$ and the temperature of the sink is $500\, K$. If the temperature of the source is kept constant and its efficiency is to be raised to $60\%$,then the required temperature of the sink will be........ $K$.

Similar Questions

An inductance of $1 \ H$ is connected in series with an $AC$ source of $220 \ V$ and $50 \ Hz$. The inductive reactance (in ohm) is (in $pi$)

$\begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 \\ & \left|\left(f_{x x} f_{y y}-f_{x y}^2\right)\right|_{(0,0)} \end{aligned}$

If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to

The area (in sq units) of the region bounded by $x=-1, x=2, y=x^2+1$ and $y=2x-2$ is

Which one of the following is common to multicellular fungi,filamentous algae,and the protonema of mosses?

Vedclass Test Series

Exam Paper Generator

Online Exam Module