Question
Question: A system works in a cyclic process. It absorbs 20 calories of heat and rejects 60 J of heat during t...
A system works in a cyclic process. It absorbs 20 calories of heat and rejects 60 J of heat during the process. The magnitude of work done (J) is (1 calorie = 4.2J):
A.24
B.23
C.25
D.32
Solution
We know that for the cyclic process work done is equal to the negative of the heat involved during the process because the internal energy is a state function and internal energy for the cyclic process is zero.
Formula used: ΔU=q+W
Here, ΔU = change in internal energy
q = heat absorbed or heat rejected
W = work done
Complete step by step answer
We know that according to the first law of thermodynamics energy can neither be created nor be destroyed only changed from one form to another.
So, according to the first law of thermodynamics the change in internal energy can be stated as follows:
ΔU=q+W
Now, in the cyclic process, we know that change in internal energy is zero because the internal energy is a state function and only depends on the initial and final state of the process. And in the cyclic process the initial state is equal to the final state.
0=W+q
So work will be negative of heat:
W=−q
Now in the question, we are given that:
q1 = heat absorbed by the system
= +20 calorie
= +20×4.2 J
= +84 J
We also given the heat rejected by the system:
q2 = heat rejected by the system
= −60 J
So the work done associated with the system will become:
W=−(q1+q2)
By substituting the value we get:
W=−(84−60)J
So the work associated with the system we get:
W=−24J
Here the negative sign indicates that work is done by the system.
Therefore, we can conclude that the correct answer to this question is option A.
Note: We can get confused between the sign associated with the heat if the heat is absorbed or rejected by the system. Always remember if heat is absorbed by the system the heat has positive value and if the heat is rejected by the system then the heat has negative value.