In a Carnot engine, the temperature of the source and sink are `500K and 375K` respectively. If the engine consumes `600xx10^(3)` cals./cycle, find
(i) the efficiency of engine (ii) work done/cycle
(iii) heat rejected per cycle.

Answer

`25% ; 6.3xx10^(5)J ; 4.5xx10^(5)cal.`

Explanation

Here, `T_(1)= 500K`,
`T_(2)= 375 K`,
`Q_(1)= 600xx10^(3)cals`.
`eta=? W=?,Q_(2)=?`
`eta=1 -(T_(2))/(T_(1))=1 -(375)/(500)= 1 -3/4=1/4= 25%`
As `(Q_(2))/(Q_(1))=(T_(2))/(T_(1))`
`:. Q_(2)=Q_(1)(T_(2))/(T_(1))=6xx10^(5)xx(375)/(500)`
`= 4.5xx10^(5)cals`.
`W= Q_(1)-Q_(2)=6xx10^(5)-4.5xx10^(5)`
`=1.5xx10^(5)cals`.
`=1.5xx10^(5)xx4.2J`
`6.3xx10^(5)J`