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The first law says that the change in internal energy is equal to the work done on the system (W) minus the work done by the system (Q). However, can $Q$ be any kind of work, such as mechanical work? For example, does a string heat up if it is attached to a block and a tension is applied to move the block? And what about the work done on the system by gravity when an object falls? Is there any internal energy change when that happens?

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$Q$ is the amounts of heat supplied to the system, and $W$ is the work done by the system on its surroundings. If $W<0$ then the work is done upon the system by its surroundings. Similarly with $Q$. – Ana May 13 '13 at 15:44
up vote 6 down vote accepted

The work in the first law is exactly the usual work $W=\int Fdx\rightarrow\int PdV$. For point particles, this is enough to completely specify the behavior of the system using Newton's first law, or energy methods. However, for macroscopic objects, the motion of the internal components (in thermodynamics these would be particles) have some additional degrees of freedom. Statistically, we can use the temperature to tell us things about "how much" energy the system has on it's own, so changes in temperature can tell us how much it's energy changes.

So the first law $\Delta U=Q-W$ tells us that we have to take care of heat transfer ($Q$) as well as the work that is done on the system $W$. Of course, these two quantities are not completely decoupled as I have described, but this is my intuition for how this works.

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Q is heat and W is work. Q is considered positive when it is added to the system and work is considered positive when work is done by the system. Also remember that Q and W are energies in transit I.e energies associated when system moves from one state of equilibrium to another state of equilibrium. One cannot say that "Q" amount of heat is present in the system or W amount of work is present in the system for certain conditions. We can say that the body gained Q amount of heat during the process and did W amount of work during the process.

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Consider a piston cylinder arrangement. Pressure * Area equals force and this force moves by a distance ds (consider piston moving upward by a distance ds) then work done = Fdx = PAreads =Pdv. Volume of cylinder equals Area *height

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