Tag: conversion of heat into work: heat engine and it's efficiency

Questions Related to conversion of heat into work: heat engine and it's efficiency

A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of $227^0C$; the heat is rejected at $27^0C$, the cycle is reversible, then what amount of heat is rejected?

physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

  1. 24kJ/s

  2. 223kJ/s

  3. 150kJ/s

  4. none of the above

Show answer
Correct Option: C
Explanation:

The temperature in kelvin scales are $T _1=273+27=300K,T _2=273+237=500K$

$T _1$is temperature of sink And$T _2$ is temperature of source hence by efficiency we get

$\eta=1-\dfrac{T _1}{T _2}=1-\dfrac{Q _1}{Q _2}$


$Q _1=\dfrac{T _1}{T _2}*Q _2=\dfrac{300}{500}*250=150kW$

An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at $27^0C$, then what is the source temperature (in $^0C$)?

physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

  1. 477

  2. 346

  3. 564

  4. none of the above

Show answer
Correct Option: A
Explanation:
$\eta =\dfrac { 60 }{ 100 } $

${ T } _{ 2 }={ 27 }^{ o }C$
${ T } _{ 2 }={ 300 }^{ o }K$
${ T } _{ 1 }=?$

$\eta =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

Here, ${ T } _{ 1 }=Source\quad temperature$
${ T } _{ 2 }=sink\quad temperature$
$\eta =Efficiency$
$\eta =100-rejecion$

$\therefore \dfrac { 60 }{ 100 } =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

$\therefore 0.6=\dfrac { { T } _{ 1 }-300 }{ { T } _{ 1 } } $

$\therefore { T } _{ 1 }\left( 1-0.6 \right) =300$

${ T } _{ 1 }=\dfrac { 300 }{ 0.4 } =\dfrac { 3000 }{ 4 } ={ 750 }^{ o }K$

$\therefore { T } _{ 1 }={ \left( 750-273 \right)  }^{ o }C$
$\therefore { T } _{ 1 }={ 477 }^{ o }C$

One reversible heat engine operates between 1600 K and $T _2$ K, and another reversible heat engine operates between $T _2$ K and 400 K. If both the engines have the same heat input and output, then the temperature $T _2$ must be equal to:

physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

  1. 600

  2. 800

  3. 650

  4. 675

Show answer
Correct Option: B
Explanation:

${ T } _{ 1 }=1600K$


${ T } _{ 2 }={ T } _{ 2 }K$


${ T } _{ 1 }^{ 1 }={ T } _{ 2 }K$

${ T } _{ 2 }^{ 1 }=400K$

Same input and same output.Then the efficiency is same.
$\eta =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

$\therefore \dfrac { 1600-{ T } _{ 2 } }{ 1600 } =\dfrac { { T } _{ 2 }-400 }{ { T } _{ 2 } } $

$=1600{ T } _{ 2 }-{ T } _{ 2 }^{ 2 }=1600{ T } _{ 2 }-640000$

$\therefore { T } _{ 2 }^{ 2 }=640000$
$\therefore { T } _{ 2 }=800K$

The freezer in a refrigerator is located at the top section so that;

physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

  1. The entire chamber of the refrigerator is cooled quickly due to convection

  2. The motor is not heated

  3. The heat gained from the environment is high

  4. The heat gained from the environment is low

Show answer
Correct Option: A
Explanation:

The freezer in a refrigerator is located at the top section so that the entire chamber of the refrigerator is cooled quickly due to convection.  
The correct option is A.

If the door of a refrigerator is kept open, the room in which the refrigerator is kept 

physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

  1. gets cooled

  2. gets heated

  3. neither gets cooled nor gets heated

  4. gets cooled or heated depending on the initial temperature of the room

Show answer
Correct Option: C