Question

A carnot engine whose efficiency is 40% takes in heat from a source maintained at a temperature of 500 K. it is designed to have an engine of efficiency 60% then the intake temperature for the same exhaust(sink) temperature must be?
1) 600 K
2) The efficiency of the Carnot engine cannot be made larger than 50%
3) 1200 K
4) 750 K
​

Answer 1

ANSWER:

• Option 4) 750 K is correct.

GIVEN:

• A carnot engine whose efficiency is 40% takes in heat from a source maintained at a temperature of 500 K.

• It is designed to have an engine of efficiency 60% at same exhaust temperature.

TO FIND:

• The intake temperature for the same exhaust(sink) temperature.

EXPLANATION:

For Carnot's engine,

$\boxed{\bold{\large{\gray{\eta = 1-\dfrac{T_s}{T_S}}}}}$

$\sf \eta \leadsto Efficiency$

$\sf T_s \leadsto Exhaust\ temperature(sink)$

$\sf T_S\leadsto Intake\ temperature(source)$

A carnot engine whose efficiency is 40% takes in heat from a source maintained at a temperature of 500 K.

$\sf \eta = \dfrac{40}{100}$

$\sf T_S = 500\ K$

$\sf \dfrac{40}{100} = 1-\dfrac{T_s}{500}$

$\sf \dfrac{40}{100} = \dfrac{500 - T_s}{500}$

$\sf 40 = \dfrac{500 - T_s}{5}$

$\sf 200 - 500 = - T_s$

$\sf - 300 = - T_s$

$\sf T_s = 300\ K$

It is designed to have an engine of efficiency 60% at same exhaust temperature.

$\sf \eta = \dfrac{60}{100}$

$\sf T_s = 300\ K$

$\sf \dfrac{60}{100} = 1-\dfrac{300}{T_S}$

$\sf \dfrac{6}{10} = \dfrac{T_S - 300}{T_S}$

$\sf 6T_S = 10T_S - 3000$

$\sf 6T_S - 10T_S = - 3000$

$\sf - 4T_S = - 3000$

$\sf T_S = \dfrac{3000}{4}$

$\sf T_S = 750 \ K$

The intake temperature for the same exhaust(sink) temperature = 750 K.

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━

Given:-

$\bf{n_{1} \: = \: \frac{40}{100} \: = \: 0.4}$

$\bf{T_{1} \: = \: 500 K}$

$\bf{Let \: T_{2} \: be \: a. }$

$\bf{n_{2} \: = \: \frac{60}{100} \: = \: 0.6}$

$\bf{Let \: T_{3} \: be \: b}$

$\bf{Let \: T_{4} \: be \: a. }$

━━━━━━━━━━━━━━━━━━━━

$\large{\mathrm{\gray{\underline{To \: Find :-}}}}$

$\bf{{\implies}T_{3}}$

$\bf{{\implies}T_{2} \: and \: T_{4}}$

━━━━━━━━━━━━━━━━━━━━

$\large{\mathrm{\gray{\underline{Solution :-}}}}$

$\bf{\bold{Firstly,\: we \: will \: find \: T_{2} \: and \: T_{4}}}$

Using Formula :-

$\bf{\boxed{\boxed{\pink{n_{1} \: =\: \frac{T_{1}}{T_{2}}}}}}$

[Put values]

$\bold{⇒ 0.4 = 1 - a/500}$

Taking LCM

$\bold{⇒ 0.4 = 500 - a / 500}$

$\bold{⇒ 0.4 * 500 = 500 - a}$

$\bold{⇒ 200 = 500 - a}$

$\bold{⇒200 - 500 = -a}$

$\bold{⇒ - 300 = -a}$

$\bold{⇒ a = 300 K}$

$\bf{\bold{ Now, \: we \: will \: find \: T_{3}}}$

Using Formula :-

$\bf{\boxed{\boxed{\red{n_{2} \: =\: \frac{T_{4}}{T_{3}}}}}}$

[Put values]

$\bold{⇒ 0.6 = 1 - 300/b}$

$\bold{⇒ b = 750 K}$

$\implies{\underline{\boxed{\sf{\purple{ b\: = \:750\:k}}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━