The internal energy change in a system that has absorbed 12 Kcals of heat and done 23000 J of work is :
Answer & explanation
Correct answer: option 3
Q = \(\Delta\) U + W
\(\Delta\) U = Q - W
= 12 x 4.2 x 1000 - 23000
= 27400 J
The internal energy change in a system that has absorbed 12 Kcals of heat and done 23000 J of work is :
Correct answer: option 3
Q = \(\Delta\) U + W
\(\Delta\) U = Q - W
= 12 x 4.2 x 1000 - 23000
= 27400 J