Question

Suppose a system returns to its original overall, internal energy ((∆U) after the following changes. In step 1, 25 J of work is done on the system as it releases 37 J of energy as heat. If, in step 2, the system performs 12 J of work, how much heat is transferred?

Answer 1

Answer:

Heat transfer in step 2 = 47.75 J

Explanation:

Internal energy = heat + work done

U = Q + W

In a cyclic process the total internal energy change of the system = 0.

In the process there are two steps. The total heat exchange in the process is the sum of heat exchanges in the two processes.

We have to find the heat exchange in step 2.

In step 1,

W = 1.25 J          Q = -37 J

$U_{1}$ = -37 + 1.25 = -35.75 J

In step 2, the internal energy change will be negative of that in step 1.

U = 35.75 J

W = -12 J

U = Q + W

35.75 = Q -12

Q = 47.75 J

Heat transfer in step 2 = 47.75 J