Suppose a system returns to its original internal energy after the following changes.In step 1,25 J of work is done on the system as it releases 37 J of heat energy.If,in step 2,the system performs 12 J of work,what is the value of q?

Question

Quizplus · Accepted Answer

According to the first law of thermodynamics, the change in the internal energy of a system is equal to the work done on the system plus the heat added to the system:

ΔU = q + w

where ΔU is the change in internal energy, q is the heat added to the system, and w is the work done on the system.

In step 1, the system has 25 J of work done on it as it releases 37 J of heat energy. Therefore,

ΔU = q + w = -12 J

The negative sign indicates that the internal energy of the system decreased.

In step 2, the system performs 12 J of work. Since the system returns to its original internal energy, the change in internal energy is zero:

ΔU = 0 = q + w

Substituting in the value of w, we get:

q = -w = -12 J

However, we want to know the value of q in joules, not in negative joules or in work units. 
Therefore, q = +24 J.

Suppose a System Returns to Its Original Internal Energy After