# Is ΔH=mCΔT and q=mCΔT the same thing? If so, which is more scientific?

May 18, 2017

$q = m c \Delta T$ is correct, but only at constant pressure, but it can be shown that $\Delta H = m c \Delta T$ is also correct.

But $q \ne \Delta H$ unless we are at constant system pressure. So no, they are definitely not quite the same thing. Neither is more "scientific", whatever that means...

$q$ is heat flow in any condition. $\Delta H$ is the change in enthalpy, and is the heat flow only at constant system pressure.

By definition,

${\left(\frac{\partial H}{\partial T}\right)}_{P} = {C}_{P} \equiv c$

That is, the specific heat capacity you use is really for constant pressure. So then,

${\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dH} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} \mathrm{dT}$

and in a small-enough temperature range,

$\Delta H = {C}_{P} \Delta T$

And therefore, whereas ${C}_{P}$ is typically used in $\text{J/mol"cdot"K}$, we can redefine this using $c$, which has units of $\text{J/g"^@ "C}$, and thus

$\textcolor{b l u e}{\Delta H = m c \Delta T}$

That is "more" correct. However, this only applies at constant pressure anyway, so really, $q = \Delta H$ and neither is actually wrong.