Triple point temperature and freezing point Why does the triple point temperature have very similar values to the freezing point, in most substances?
 A: From the Clausius Clapeyron equation,
$$ \frac{dP}{dT}_{i \rightarrow j} = \frac{\ell_{i \rightarrow j}}{T(v^{j}-v^{i})} $$
Where $ \ell_{i \rightarrow j} $ is the latent heat of going from phase $i$ to phase $j$, $\, v$ is the specific volume, and all other variables have their common meaning. 
Taking for example water, this equation gives the change in temperature as the pressure is increased from the low value of the triple point ($P_{tp}=610 \, Pa$) to atmospheric pressure, where the normal melting point of water is defined ($P=1.01 \times 10^{5} \, Pa$).
Notice that the specific volume is the reciprocal of the density. The density for liquid water is $1000 \, kg\,m^{-3}$, and the density of ice is $916 \, kg\,m^{-3}$. The latent heat is $3.34 \times 10^{5} J\,kg^{-1}$.
Now, assuming the change of temperature is small, which is the case,
$$ \Delta T = \frac{T(v^{j}-v^{i})}{\ell_{i \rightarrow j}} \Delta P  $$
$$ = \frac{273 (1.00 \times 10^{-3} - 1.09 \times 10^{-3})(1.01\times10^{5})}{3.34\times 10^{5}} = -0.0074\, K$$
It helps to give you an idea of why the change in temperature is so small.
(The previous example was borrowed from Ashley Carter's Classical and statistical thermodyanmics). 
