Question
The internal energy change in a system that has absorbed 2 Kcal of heat and done 500 J of work is
-
8900 J
-
6400 J
-
5400 J
-
7900 J
Solution
D.
7900 J
Heat given to a system (ΔQ) is equal to the sum of the increase in the internal energy (Δu) and the work done (ΔW) by the system against the surrounding and 1 cal = 4.2. J
According to first law of thermodynamics
ΔU = Q-W
= 2 x 4.2 x 1000-500
= 8400-500
=7900 J
