Home
Class 12
PHYSICS
A reversible engine converts one-sixth o...

A reversible engine converts one-sixth of heat input into work. When the temperature of sink is reduced by `62^(@)C,` its efficiency is doubled. Find the temperature of the source and the sink,

A

` 99^@C , 37 ^@ C `

B

` 80^@C, 37 ^@ C `

C

`95^@C, 37^@ C `

D

`90^@C, 37^@C `

Text Solution

Verified by Experts

The correct Answer is:
A
` eta _ 1 = 1 - ( T _ L ) / (T _ H ) = ( W ) / ( Q _ 1 ) = ( 1 )/( 6 ) `
or ` 5T _ B - 6 T _ L = 0" " `…(i)
` eta _ 2 = 1 - ( T _ L - 62 ) /( T _ H ) = 2 eta _ 1 = ( 1 ) / ( 3 ) ` (Given )
` rArr1 - ( 1 ) /( 3 ) = ( T _ L - 62 ) /( T _ H ) `
or ` 2 T_H - 3 T _ L = - 186 " " `...(ii)
Solving (i) and (ii) we get
` therefore T _ u = 372 K = 99^@ C `
` T _ L = ( 5 ) /( 6 ) T _ H = ( 5 ) /( 6) xx372 K = 310 K = 37^@ C `
Promotional Banner

Similar Questions

Explore conceptually related problems

An engine has an efficiency of 1/6 . When the temperature of sink is reduced by 62^(@)C , its efficiency is doubled. Temperature of the source is

An engine has an efficiency of 1/6 . When the temperature of sink is reduced by 62^(@)C , its efficiency is doubled. Temperature of the source is

An engine has an efficiency of 0.25 when temperature of sink is reduced by 58^(@)C , If its efficiency is doubled, then the temperature of the source is

A Carnot engine has an efficiency of 1//6 . When the temperature of the sink is reduced by 62^(@)C , its efficiency is doubled. The temperature of the source and the sink are, respectively.

A Carnot engine has an efficiency of 20% . When temperature of sink is reduced by 80°C its efficiency is doubled. The temperature of source is

A Carnot reversible engine converts 1//6 of heat input into work. When the temperature of the sink is redused by 62 K, the efficiency of Carnot’s cycle becomes 1//3 . The sum of temperature (in kelvin) of the source and sink will be

An engine takes heat from a reservior and converts its 1//6 part into work. By decreasing temperature of sink by 62^(@)C , its efficiency becomes double. The temperatures of source and sink must be

A cannot engine has efficiency (1)/(6) . If temperature of sink is decreased by 62^(@)C then its efficiency becomes (1)/(3) then the temperature of source and sink:

A Carnot engine efficiency is equal to 1/7 . If the temperature of the sink is reduced by 65 K , the efficiency becomes 1/4 . The temperature of the source and the sink in the first case are respectively

The efficiency of a cannot's engine is 20%. When the temperature of the source is increased by 25^(@)C, then its efficiency is found to increase to 25%. Calculate the temperature of source and sink.