Calculate the external workdone by the system in Kcal, when 40 Kcal of heat is supplied to the system and internal energy rises by 8400J.

Explanation

`dQ=dU+dW`
`dU=8400J=(8400)/(4200)K Cal=2 K Cal`
`therefore 40 K Cal =2 K Cal+` external work done
The external work done `=40-2=38` K Cal