If one lightbulb burns out, but the other bulbs remain lit, the circuit must be a
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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
= -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