Question

If 150 calories of heat are added to a system and a work of 200 joule is done on the system then calculate change in internal energy of the system :- (1) 430 joule (2) 830 joule (3) –830 joule (4) –430 joule

Answer 1

Answer:

(2) 830 J

Explanation:

ΔQ =+150 cal = 150 × 4.184 =627.6 J  (∵ heat is added to the system it is +ve)

ΔW = +200 J (∵ work is done on the system it is +ve)

ΔU =ΔQ + ΔW

=627.6 + 200

= 827.6 J ≅ 830 J