# Basic First Law of Thermodynamics Question [closed]

$$2$$ 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.

• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. Jan 19, 2019 at 11:00

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.
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.