Question

In a process, a system does 140 J of work on the surroundings and only 40 J of heat is added to the system hence change in internal energy is :

• A. 180 J
• B. -180 J
• C. 100 J
• D. -100 J

Answer 1

The correct option is D -100 J

Solution:- (D) $-100J$

As the work is done by the system and heat is added to the system, thus the work done will be negative and the value of $q$ will be positive.

Therefore,

$W=-140J$

$q=40J$

As we know that,

$\Delta E=q+W$

$\therefore \Delta E=40+(-140)=-100J$

Hence the change in internal energy will be $-100J$.