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 ?

Options

A. `273^(@)C`
B. `173^(@)C`
C. `373.3^(@)C`
D. `-273.3^(@)C`

Answer

C

Explanation

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 ).