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$2$ kg of water is heated by stirring. If this process raises the temperature of water from $15^{\circ}$ C to $25^{\circ}$ C, how much work, in joules, was done to the water by the stirring?

I calculated $W = Q = mc\Delta T = (2000)(4.18)(10) = 83,600 J$.

Is this right? I don't know if my original claim that $W = Q$ is right.

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    – ACuriousMind ♦
    Jan 19, 2019 at 11:00

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The first law is simply $\Delta U = Q - W$. For an incompressible substance (e.g., water), you have $\Delta U = mc \Delta T$. Now, the problem says nothing about heat transfer into the water, only that its temperature raises. I'd say this is is a clear hint that $Q=0$, and the work is done on the water adiabatically.

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Drew's correct answer directly addresses the question about the work done by stirring the fluid. There is an additional point in your question that I feel would be important to clarify.

First principle defines $\Delta U$ as the sum of the incoming heat and work (there is freedom about the convention for signs, but, at the and of the day, work which is performed on the system increases its internal energy). So, internal energy may increase with different combination of incoming work and/or heat. Unfortunately, in everyday language we do not have a special word for indicating an increase of internal energy and we use the verb "to heat" even if the "heating" is not due to a difference of temperature but to work done by friction forces.

Therefore, stirring a fluid means that some work has been done and, when the system relaxes to equilibrium, it has entirely converted into internal energy.

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