As we learned last week, heat capacity is amount of energy required to raise the temperature of an object by a small amount. $$C \sim \frac{đ Q}{\partial T}$$ $$đ Q = C dT \text{ At constant what?}$$ If we hold the volume constant, then we can see from the first law that $$dU = đQ - pdV$$ since $dV=0$ for a constant-volume process, $\newcommand\myderiv[3]{\left(\frac{\partial #1}{\partial #2}\right)_{#3}}$ $$C_V = \myderiv{U}{T}{V}$$ But we didn't measure $C_V$ on Monday, since we didn't hold the volume of the water constant. Instead we measured $C_p$, but what is that? To distinguish between different sorts of heat capacities, we need to specify the sort of path used. So, for instance, we could write $$đQ=Tds$$ $$đQ=C_αdT+?dα$$ $$TdS=C_αdT+?dα$$ $$dS= \frac{C_\alpha}{T} dT + \frac{?}{T}d\alpha$$ $$C_\alpha = T \myderiv{S}{T}{\alpha}$$
This may look like an overly-tricky derivative, so let's go through the first law and check that we got it right in a few cases. I'll do the $C_V$ case. We already know that $$dU=đQ-pdV$$ $$C_V= \myderiv{U}{T}{V}$$ $$= \myderiv{U}{S}{V} \myderiv{S}{T}{V}$$ $$= T \myderiv{S}{T}{V}$$ where the second step just uses the ordinary chain rule.