# Are enthalpy and entropy affected by temperature?

Dec 13, 2015

Yes, they are. Here is how:

$\Delta \overline{S} = {\int}_{{T}_{1}}^{{T}_{2}} \frac{{\overline{C}}_{P}}{T} \mathrm{dT}$

$\Delta \overline{H} = {\int}_{{T}_{1}}^{{T}_{2}} {\overline{C}}_{P} \mathrm{dT}$

where ${\overline{C}}_{P}$ is the molar heat capacity in a constant pressure system, and $\Delta S$ and $\Delta H$ are the change in entropy and enthalpy, respectively.

So, let's say you had water, whose chemistry molar heat capacity is defined as $\text{75.375 J/mol"*"K}$. Then, we have, from $273$ to $\text{298 K}$:

$\Delta \overline{S} = {\int}_{\text{273 K")^("298 K}} \frac{{\overline{C}}_{P}}{T} \mathrm{dT}$

$= {\overline{C}}_{P} | \left[\ln | {T}_{2} | - \ln | {T}_{1} |\right] {|}_{\text{273 K")^("298 K}}$

$= 75.375 \cdot \left(\ln \left(298\right) - \ln \left(273\right)\right)$

$\textcolor{b l u e}{\text{= 6.604 J/mol"*"K}}$

$\Delta \overline{H} = {\int}_{\text{273 K")^("298 K}} {\overline{C}}_{P} \mathrm{dT}$

$= {\overline{C}}_{P} \left({T}_{2} - {T}_{1}\right)$

$= 75.375 \left(298 - 273\right)$

$\textcolor{b l u e}{\text{= 1884.38 J/mol}}$