1.

If heat given to a system is 6 kcal and work done is 6kj, then calculate the change in internal energy.

Answer»

Solution :`DELTA Q = Delta U + Delta W`
`:. Delta U = Delta Q - Delta W = (6 xx 1000 xx 4.2) - (6 xx 1000)`
`= 6000(4.2-1) = 6000 xx 3.2 = 19.2 xx 1000 = 19.2 KJ`


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