A system releases 7.9kj of heat while 6.5kj work is done on it. Calculate the total change in internal energy during these operations in kJ.
14.4kJ
The first law of thermodynamics which states that the total change in internal energy (ΔU) of a system is the sum of the total heat (Q) released by the system to its surroundings and the total work done (W) on the system by its surrounding, can be expressed mathematically as follows:
ΔU = Q + W -------------------(i)
From the question;
Q = 7.9kJ
W = 6.5kJ
Substitute these values into equation(i) as follows;
ΔU = 7.9kJ + 6.5kJ
ΔU = 14.4kJ
Therefore, the change in internal energy is 14.4kJ
