# How to calculate the enthalpy change of water?

Calculate the $\Delta H$ of the reaction where $2.6\:\mathrm{g}$ of water, $C_\mathrm{s}= 4.184 \mathrm{\frac{J}{g\:K}}$ is heated, raising the temperature increases from $298\:\mathrm{K}$ to $303\:\mathrm{K}$.

I used the following equation: $\Delta H=mC_\mathrm{s}\Delta T$

$m = 2.6~\mathrm{g}$, $C_\mathrm{s} = 4.184~\mathrm{J/g}$, $\Delta T= 5$

What I got: $(2.6)(4.184)(5)= 54.392~\mathrm{J/g}$

Did I solve this correctly?

Yes, but your answer should be just in joules not $\mathrm{J/g}$, and your units are off.
$$(2.6~\mathrm{\color{\red}{g}})\left(4.184~\mathrm{\frac{J}{\color{\red}{g}\ K}}\right)(5~\mathrm{K})= 50~\mathrm{J}$$