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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?

Respuesta :

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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

[tex]Q=70J[/tex]

[tex]W=31J[/tex]

Where Q is heat absorbed and W the work done

[tex]U = Q+W[/tex]

[tex]U = 70 + 31 = 101 J[/tex]

For the second step we have to

[tex]Q = 31 J[/tex]

[tex]W =-70 J[/tex]

[tex]U = 70-31 = -39 J[/tex]

[tex]\Delta U_T = U_1+U_2[/tex]

[tex]\Delta U_T  = 101 - 39 = 62 J[/tex]

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

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