1.

A Carnot's engine is working between 7^(@)C and 287^(@)C. It is desired to increase its efficiency to 70%. By how much should the temperature of source be increase ?

Answer»

`273^(@)C`
`173^(@)C`
`373.3^(@)C`
`-273.3^(@)C`

SOLUTION :Let `T_(1)` be temperature of source when
`ETA =(70)/(100) cdot" then "(7)/(10)=(T_(1)-280)/(T_(1))" or "T_(1)=933.3 K.`
INCREASE in temperature `=933.3-560=373.3^(@)C`.
`therefore` Correct choice is (c ).


Discussion

No Comment Found

Related InterviewSolutions