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    				| 1. | 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» Correct Answer - `-106.33^(@)C` 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` | |