Why is the work done counted twice in the 1st law of thermodynamics?

[curious bishal]

On the way to prove 1st law of thermodynamics, you consider a system having initial energy E₁. Then, you supply heat q to the system. Some part of the applied heat is used for doing work.
Then, again you consider the final energy (E₂) of the system to be:
E₂=E₁+q+w
where w is the work done.

We all very well know that the work is done by the heat applied. If we add both heat applied and work done to the final energy, isn't the work done counted twice. I mean to say that the heat energy wasted in the work done is also consisted in q so why is there need to add w in the final energy E₂.

 

[jfizzix]

I think the confusion here comes from whether w is the work done on the system, or the work done by the system. These change the energy by opposite amounts.

If the system is doing work w, the energy of the system decreases by w, so

E2=E1+q-w

Then, if q=w,
E2=E1

 

[curious bishal]

I think the confusion here comes from whether w is the work done on the system, or the work done by the system. These change the energy by opposite amounts.

If the system is doing work w, the energy of the system decreases by w, so

E2=E1+q-w

Then, if q=w,
E2=E1

It doesn't give the solution to my problem. Why is the work done counted double?

 

[jfizzix]

The work is not double-counted.

If the system starts at energy E1, and you add heat q to it, the energy has changed to amount E1'= E1+q.

If the system then performs work w on its environment, it loses w units of energy, so its energy changes to E2 = E1'-w = E1+q-w.

Though some of the heat is converted into work, that doesn't affect the fact that q units of heat were added, and w units of work are extracted. Thus the net energy change is q-w.