Question

The amount of heat given to a system is 40 J and the amount of work done on the system is 20 J, then the change in internal energy of the system is

(1) – 20 J

(2) + 20 J

(3) – 60 J

(4) + 60 J

Answer 1

(4) + 60 J

According to first law of thermodynamics

dQ = dU + dW 

Given dQ = 40 J

dW = -20 J (As work done on the system is negative by sign convention).

⇒ dU = dQ - dW = 40 - (-20)

= 60 J