The adiabatic compression ratio in a carnot reversible cycle is 9, when the temperature iof the source is `227^(@)C`. Calculate the temperature of the sink. Given `gamma=1.5`

Answer

`-106.33^(@)C`

Explanation

Here,`(V_(2))/(V_(1))=9, T_(1)=227+273= 500K`,
`T_(2)=?, gamma=1.5`
For an adiabatic change,
`T_(2)V_(2)^((gamma-1))=T_(1)V_(1)^((gamma-1))`
or `T_(2)=T_(1)((V_(1))/(V_(2)))^((gamma-1))`
`=500(1/9)^((1.5-1))=(500)/3= 166.67`
`=166.67-273= -106.33^(@)C`