Molar specific heat of water = molar mass of water
$ \times $ specific heat of water
$ =\frac{18g}{mol}\times \frac{1\,cal}{{{g}^{o}}C} $
$ =18\frac{cal}{mo{{l}^{o}}C} $
$ =9R $ [where, $ R=2\text{ }cal\text{ }mo{{l}^{-{{2}_{o}}}}\text{ }{{C}^{-1}} $ ]