Questioning solution of lake ice melting exercise My thermodynamics teacher solved one exercise but I'm not convinced of his solution. The exercise asks this:

A lake is covered by an ice sheet of thickness $D$ given. Ice latent fusion heat $\lambda$, ice density $\rho$ and thermal conductivity coefficient $\kappa$ are also given. The external air temperature remains constant at $T_e=0° C$, the ice sheet melts in a certain time $\Delta t$. The problem requires to find the temperature of the lake water $T_l$. The water temperature and the air temperature are kept constant in space and time as boundary conditions, they are temperature sources.

This is my teacher's solution:
$\delta Q$ required to melt a $dx$ thick quantity of ice is: $$\delta Q= \lambda \rho A dx$$ where A is the area of the lake.
The heat pass through the ice by conduction so in steady state holds: $$\delta Q = - \kappa A \frac{T_l-T_e}{x} dt$$
Equating the two $\delta Q$ expressions and integrating:
$$\int_D^0 xdx =\int_0^{\Delta t} \kappa \frac{T_l-T_e}{\lambda \rho}dt$$
So plugging in the numerical values:
$$T_l=T_c+\frac{D^2\lambda\rho}{2\kappa\Delta t}=3.44°C$$
What doesn't convince me of this solution is that the heat, instead of passing through the ice sheet to the air, contributes to melt the nearer to the water portion of the ice sheet and stops there, no part of the heat reaches the air. So how can we talk about heat conduction though the entire ice sheet to the air ad use the above formulas? And moreover if there was a steady state heat conduction through the ice sheet this would require the existence of gradient of temperature in the entire ice sheet from $0$ to $3.44°C$, but this absurd because this is over the ice fusion temperature. And in the end if the heat from the water just melts the nearer ice how can the melting speed be influenced by the thickness of the above ice?
 A: The answer given is inconsistent with the set up for the problem statement. It also misses one term in the heat flow.
Imagine a lake with water at an unknown temperature $T_l$ ($^o$C) covered by ice with thickness $z_o$ (m) at temperature $T_i$. The ice is covered by air at $T_a$. In this setup, $T_l > T_i = T_a$.
Heat flows from the lake water $\dot{q}_h$ (W) into the ice. The expression for convection heat flow is
$$ \dot{q}_h = h\ A\ (T_l - T_i) $$
where $h$ is the convection coefficient (W/m$^2$ K) for the water against the ice and $A$ is the lake area (m$^2$). Alternatively, the heat flow can be modeled as a constant $\dot{q}_c$ occurring directly at the interface between the lake water and the ice.
The heat flow from the lake water into the ice continues in three possible ways. First, ice melts according to
$$\dot{q}_m = \rho_i \ A\ \Delta_{fus}\tilde{H}\ \frac{dz}{dt} $$
where $\rho_i$ is the ice density (kg/m$^3$), $\Delta_{fus}\tilde{H}$ is the specific fusion enthalpy (J/kg), $dz$ is the change in thickness (m), and $dt$ is the change in time (s).
Secondly, the melted water is heated from $T_i$ to $T_l$ as
$$\dot{q}_T =  \rho_w\ \tilde{C}_p\ A\ (T_l - T_i)\ \frac{dz}{dt}$$
where $\rho_w$ is the water density and $\tilde{C}_p$ is the specific heat capacity of the water (J/kg K).
Finally, we may suggest that some portion of $\dot{q}_h$ or $\dot{q}_c$ could be conducted through the ice to the air as
$$\dot{q}_k = k\ A \frac{T_l - T_a}{z} $$
where $k$ is the thermal conductivity of the ice (W/m K). A problem arises with this statement. Steady state conduction requires or establishes a linear temperature gradient across the object, but we have stated that the ice is at a constant temperature throughout. One approach is to allow the ice to support a hypothetical linear gradient. Another approach is to state that conduction through the ice is zero because the ice remains at a constant temperature throughout. Finally, this problem could be resolved in principle when the external air temperature is allowed to be even a fraction of a degree below the melting temperature of ice because we would then be able to infer a linear temperature gradient across the ice itself.
Whichever approach is taken for conduction, the two or three heat terms (melting, heating the melted water, and conduction) are equated to the heat flow from the lake water to the ice. With all three terms restricted in a final dimensionless equation, the end result is a first order differential equation of the form below.
$$0 = A\ \frac{dZ}{d\tau} + \frac{B}{Z} + C$$
where $A, B, C$ are dimensionless constants (accounting for melting, conduction, and convection/constant heat flow from lake respectively), $Z = z/z_o$, and $\tau = t/t_c$ with $t_c$ as a characteristic time constant. When conduction through the ice is neglected, the equation has the form below instead.
$$0 = A\ \frac{dZ}{d\tau} + C$$
In summary, the solution does not involve equating the latent heat to the conducted heat as through the heat flow for these two events are in series. Latent heat is not conducted (it is lost to melting the ice), and conducted heat does not melt the ice (it is transported through the ice). Melting and conduction happen in parallel. The ratio that goes to melting versus conduction is determined essentially by the ratio $\dot{q}_m/\dot{q}_k$. Alternatively conduction should be ignored.
Finally, at a minimum to obtain an exact solution for the water temperature, one needs a value for either the convection coefficient $h$ for the lake water or the constant heat flux $\dot{q}_c/A$ from the lake water to the ice.
A: I believe your instructors solution is fine except that the thermal conductivity should be that of the water rather than ice. Also, the following assumptions need to be true:

*

*All heat transfer is via conduction. This is true if $T_{lake} <= 4°C$ when warmer water begins to sink instead of rise so that convection is minimized.

*The thermal conductivity of the water is constant between $0$°C and $4$°C. This is not true but the change is small enough to ignore for this type of calculation.

*The lake is still and remains still during the melt (no forced convection).

*The lake, meltwater, and ice remain in full contact during the melt

*There is no temperature change due to the phase transition from solid to liquid

*$T_{melt/ice} = 0 $ during the melting process (T at the phase boundary is zero)

*The temperature at the surface of the lake is $T_{lake}$ at t = $0°C$.  Not true but it simplifies the calculation.

*The melt water does not diffuse, it retains its surface area and grows upward from the lake/ice boundary at t = $0$ like a thin disk of slightly less dense fluid atop a denser fluid. This is true if $T_{lake} <= 4°C$ when colder water begins to rise instead of sink and if the ice is not too thick so the ice melts before diffusion and convection occur.

Analogous to electrical circuits, the heat flow equation can be written:
$$ T_{lake} - T_{air} = \frac {dQ}{dt}(R_{melt} - R_{ice})$$ where $R_{melt}$ and $R_{ice}$ are the thermal resistance of the water and ice, respectively.
Looking at the temperature difference between ice and the air, it's clear that there will be no heat flow through the ice during the melting process due to assumption 6:
$$\frac{dQ_{ice}}{dt} = -k_{ice}A \frac {T_{melt/ice}-T_{ice/air}}{D}$$ since
$T_{melt/ice}-T_{ice/air} = 0$
This means that all of the temperature drop occurs across the meltwater:
$$ T_{lake} - T_{air} = \frac {dQ}{dt}R_{melt}$$ Now we can pick up from where your professor began:
$$ \frac{dQ}{dt} = - k_{melt} A \frac{T_{lake}-T_{air}}{x}$$
$$\int_D^0 xdx =\int_0^{t} k_{melt} \frac{T_{lake}-T_{air}}{\lambda \rho}dt$$
$$ T_{lake}=T_{air}+\frac{D^2\lambda\rho}{2k_{melt}t}$$
This is the same answer as listed in the post except that k is for water rather than ice. How accurate is this analysis? I think as a simple quick and dirty calculation that's accurate to within 20% and that can be done on the back of an envelope its fine.
