How can there be an entropy change in this system?
Suppose if I have a system consisting of liquid water, $1\, \mathrm{kg}$ at $290\,\mathrm{K}$, I stir it, and do say, $10\, \mathrm{J}$ of work on it, I can calculate the temperature change of the system given that:
$$U = cT \quad\mbox{ and }\quad S = c \ln \Omega$$ for $c$ constant.
From the fundamental equation of thermodynamics: $$dU = dQ + dW = 0 + dW = 0 + 10 = 10\,\mathrm{J}$$ Hence: $$dT = \frac{dU}{cM} = \frac{1}{410}\,\mathrm{K}$$
But how can there be a change of entropy in the universe when $dQ = 0$. I understand that we can calculate it using the formula for $S$ given, but I don't understanding how the fundamental equation allows this?
$$dQ = 0$$ and $$dS = T^{-1}\,dQ$$
Hence, it may be concluded that: $$dS = 0$$
Can someone tell me where my understanding is lacking, because obviously the entropy change is not zero in this case?