How to apply the First Law of Thermodynamics to this problem?

[vcsharp2003]

Thats the best idea I can come up with.

I think trying to force yourself to use the First Law of Thermodynamics for such a problem is unnecessary. It can simply be solved by using the law of conservation of energy that we learn in the chapter of Work, Power and Energy. The spring mass system is losing some spring potential energy and that energy lost is converted to thermal energy. That's all there is to this problem.

 

[kuruman]

But, why would W = 0 beats me.

The system of spring, 400-g mass and water is isolated. No external work or heat crosses the system's boundary. So W = 0. If you think the work is not zero, what exactly is it that does work on the system?

 

[vcsharp2003]

If you think the work is not zero, what exactly is it that does work on the system?

That seems correct. There is no external work happening since the work done by the water on the mass is not external work, but like internal work. Right? I'm not sure if the work done against water by the mass is to be included in thermodynamic work W.

 

[kuruman]

That seems correct. There is no external work happening since the work done by the water on the mass is not external work, but like internal work. Right?

Right. One part of the system may do mechanical work on another part, but the other part does equal and opposite work on the first part. This is guaranteed by Newton's 3rd law.

 

[vcsharp2003]

Right. One part of the system may do mechanical work on another part, but the other part does equal and opposite work on the first part. This is guaranteed by Newton's 3rd law.

That's a beautiful explanation. I think I am getting how to apply thermodynamics to this situation.

 

[kuruman]

When the mass oscillates up and down until it stops and the water plus 400-g mass have their temperature raised, we say that "heat is added to them". What does that really mean? It means that the water molecules and the molecules in the mass have, on average, more kinetic energy than before which appears as an increase in temperature.

Say you have a gas in a piston. If you compress the gas adiabatically (no heat added), you do mechanical work W on the gas and the temperature rises meaning that, on average, the gas molecules move faster than before. If you then lock the piston in place and put it over a flame adding heat Q at constant volume, the gas molecules will move, on average even faster resulting in an even higher temperature. What have you really done during this last step? The much faster moving molecules in the flame collided with the walls of the piston molecules making them move faster which in turn made the gas molecules colliding with the piston walls move faster than before. So this addition of "heat" is really transfer of kinetic energy from the outside into the system. However, the complicated microscopic processes of kinetic energy transfer between a prodigious number of molecules is significantly facilitated if you lump the total energy transferred as "heat entering the system."

Then you have a generalized form of the work energy theorem. $ΔK = W$ becomes $ΔK = Q + W$ to account for the intractable amount of kinetic energy that goes into the system when "heat" is added. Remembering that the average kinetic energy of an atom in an ideal monatomic gas at temperature $T$ is $\bar K=\frac{3}{2}kT$, the total kinetic energy content of $N$ atoms is $K=\frac{3}{2}NkT.$ Thus, in an ideal monatomic gas the total kinetic energy is the same as the internal energy $U.$ We use a different symbol because in polyatomic gases energy coming in from the outside does not only go into translational kinetic energy but also into rotations and vibrations of the molecules.

 

[nasu]

Right. One part of the system may do mechanical work on another part, but the other part does equal and opposite work on the first part. This is guaranteed by Newton's 3rd law.

Newton's 3rd law guarantees the equality of forces but not of the amounts of work.

 

[Chestermiller]

For this particular problem, the general form of the 1st law of thermodynamics presented by the OP in post #1 should include a term involving the change in the stored elastic energy of the spring, and should read: $$\Delta U+\Delta KE+\Delta PE+\Delta \left(\frac{1}{2}kx^2\right)=Q-W$$So, in this problem where $\Delta KE$ and $\Delta PE$ are both zero, we have $$\Delta U+\Delta \left(\frac{1}{2}kx^2\right)=Q-W=0$$