Question

A system undergoes a process in which system releases 200 J heat and work done by the surroundings is 300 J. Change in internal energy of the system is

• A. −100 J
• B. 100 J
• C. −500 J
• D. 500 J

Answer 1

The correct option is B

100 J

$\Delta U=q+w$
Since system releases heat, q will be negative. It is also given that work done by the surroundings is 300 J which is the same as the work done on the system. Hence, w has a +ve sign. Keeping this in mind, we get:

= $-200+300=100J$