Question

A system undergoes a two-step process: Step 1: The system absorbs 70 J of heat while 31 J of work is done on it. Step 2: The system absorbs 31 J of heat while performing 70 J of work. What is the change in internal energy for the entire process?

Answer 1

To solve this problem we simply need to apply the concepts of energy conservation.

Let us break down the problem by the steps suggested by the statement:

For the first step we have to

$Q=70J$

$W=31J$

Where Q is heat absorbed and W the work done

$U = Q+W$

$U = 70 + 31 = 101 J$

For the second step we have to

$Q = 31 J$

$W =-70 J$

$U = 70-31 = -39 J$

$\Delta U_T = U_1+U_2$

$\Delta U_T = 101 - 39 = 62 J$

Therefore the change in internal energy for entire process is 62J