= Solution
Write $\beta$ for the <thermal expansion coefficient>, $\nu$ for <kinematic viscosity>, $\kappa$ for <thermal diffusivity>, and $k_T=\rho c_p\kappa$ for <thermal conductivity>. For a temperature difference $\Delta>0$ across depth $h$,
$$
\mathrm{Ra}=\frac{g\beta\Delta h^3}{\nu\kappa},\qquad \mathrm{Nu}=\frac{qh}{k_T\Delta}.
$$
The <four-thirds convective heat-transfer law> posits $\mathrm{Nu}=C\mathrm{Ra}^{1/3}$ at high <Rayleigh number>, and therefore
$$
\boxed{q=K\Delta^{4/3},\qquad K=Ck_T\left(\frac{g\beta}{\nu\kappa}\right)^{1/3}}.
$$
The layer depth cancels. A useful physical justification is a mixed interior with thin conductive <boundary layers> of thickness $\delta$, each maintained near a fixed critical local <Rayleigh number>: $g\beta\Delta_b\delta^3/(\nu\kappa)\sim\mathrm{Ra}_{\rm crit}$ and $q\sim k_T\Delta_b/\delta$. Eliminating $\delta$ gives the exponent $4/3$. Whether $\Delta$ denotes the whole-layer contrast or the drop across one <boundary layer> changes the coefficient, not the exponent.
Assume <Boussinesq approximation>, constant material properties, a laterally extensive statistically turbulent <Rayleigh-Bénard convection> layer, negligible side-wall <heat transfer>, no rotation or imposed shear, approximately uniform bulk temperature, and the same specified <Prandtl number> regime when choosing $C$. The closure is empirical and regime dependent; high <Rayleigh number> alone does not make it exact or extend it to the impulsive conductive initial transient.
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