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.

The efficiency of a cannot's engine is 20%. When the temperature of th

1.	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.
Answer» Solution :Temperature of source `= T_(1)` Temperature of sink `= T_(2)` EFFICIENCY `= eta = 20%=0.2` `eta = 1 - (T_(2))/(T_(1)), (T_(2))/(T_(1)) = 1 - 0.2 = 0.8 i.e., T_(2) = 0.8 T _(1)` When the temperature of the source is INCREASED by `25^(@)C,` the new efficiency is 25%. `T_(2)` is same. `0.25 = 1 - (T_(1))/(T_(1)+_25) , i.e, (T_(2))/(T_(1) + 25) = 1 - 0.25 = 0.75` `T_(2) = 0.75 (T_(1) + 25) = 0.74 T_(1) + 0.75 XX 25` But ` T_(2) = 0.8 T_(1)` `therefore 0.8 T _(1) = 0.75 T_(1) + 18. 75 ` `0.5 T_(1) = 18.75 ,T_(1) = (18.75)/(0.05) = 375K` `T_(2) = 0.8 T_(1) = 0.8 xx 375= 300 K`