Question

The change in internal energy in a reaction when $2$ KJ of heat is released by the system and $6$ KJ of work is done on the system will be
(A) $ + 4$ KJ
(B) $4$ KJ
(C) $+ 3$ KJ
(D) $ - 8$ KJ

Answer 1

Hint: According to the first law of thermodynamics, the change in internal energy is the sum of work done and heat. Heat and work are the two modes of energy transfer. These two are path dependent quantities. Hence they are called path functions whereas the sum of two is a state function i.e. internal energy. It only depends on the initial and final position and not on the path followed.

Complete Step by Step Solution:
The sign conventions followed in thermodynamics are as follows. The work done on the system is positive whereas the work done by the system is negative. The heat released by the system is negative whereas the heat given to the system is positive. The sign conventions are based on the fact that the process is positive if it increases the energy of the system, for example providing heat and work done on the system. Whereas processes such as work done by system or heat released lowers the internal energy thus the sign conventions are negative.

According to given data, Q (heat) = -$2$ KJ
W (work done) = $6$ KJ
Hence, \[U{\text{ }} = \;W{\text{ }} + {\text{ }}Q\]
Where U is change in internal energy of system
W is work done
Q is the heat of system
\[U = 2 - 6\]
$U = + 4$
Hence, the correct answer is $ + $$4$ KJ.

Note: Positive energy signifies that the final internal energy is greater than initial whereas the negative energy signifies that initial energy is greater than final energy. Work and heat can only be added or subtracted when the two have the same units.