I am supposed to derive an expression for the specific heat capacity $c_p$ for air with a water vapour mass mixing ratio of $\mu$. The hint is: (Use the fact that $c_p= (\frac{\partial H}{\partial T})_p$, where H is the enthalpy per unit mass.
I do not want a complete solution, but I have absolutely no idea where to start. What is the relation between enthalpy and mixing ratio, enthalpy is just internal energy and isobaric work.
Enthalpy of a mixture of gases is determined by this formula: $$h=\Sigma (mf)_ih_i\;[\mathrm {kJ/kg}]$$ $h$ is the enthalpy of unit mass of the mixture,
$(mf)_i$ is mass fraction of gas $i$, $(mf)_i=\large{\frac{m_i}m}$ ($m_i$ is the mass of gas $i$ and $m$ is the mass of mixture)
$h_i$ is the enthalpy of unit mass of gas $i$ that can be determined by using $\textrm{one}^1$ of Dalton's law of additive pressures or Amagat's law of additive volumes and temperature of the mixture.
In this case, if we assume that $\large{\frac{m_v}{m_a}}=\mu$; as $m=m_v+m_a$, then we have $$(mf)_v=\large{\frac{\mu}{\mu+1}}$$ and $$(mf)_a=\large{\frac 1{\mu+1}}$$ Thus, $$\large{h}=\large{\frac{\mu}{\mu+1}}h_v+\large{\frac 1{\mu+1}}h_a$$
$^{\mathrm 1}$ For mixture of ideal gases, both of these laws end to same result, but for mixture of real gases, choosing between these two laws depends on the problem.
