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lawyer [7]
4 years ago
9

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 sys

tem absorbs 31 J of heat while performing 70 J of work. What is the change in internal energy for the entire process?
Physics
1 answer:
Marizza181 [45]4 years ago
4 0

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

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