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?

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    $\begingroup$ For a start I would have expected there to be a temperature gradient in the water, from 0 Celsius in the water touching the ice, down to 4 Celsius (approx) at the bottom of the lake. And that is ignoring the likelihood of currents in the water. I am also unconvinced by your teacher's analysis of the situation. $\endgroup$
    – Peter
    Commented Jun 11, 2022 at 5:05
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    $\begingroup$ @Peter Yes, but also if the water temperature is kept constant as a boundary condition, like my teacher said, the analysis doesn't work for me. I will add this to the post to be more clear. $\endgroup$
    – Mattia
    Commented Jun 11, 2022 at 9:41
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    $\begingroup$ Please, use an informative title: "thermodynamics exercise" can be anything (consider editing the question accordingly). $\endgroup$
    – Quillo
    Commented Jun 11, 2022 at 9:47
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    $\begingroup$ @Quillo ok I edited the title $\endgroup$
    – Mattia
    Commented Jun 11, 2022 at 9:50
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    $\begingroup$ I agree. This problem statement makes no sense. $\endgroup$ Commented Jun 11, 2022 at 10:04

2 Answers 2


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.

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    $\begingroup$ Between 0° and 4° C water has a negative coefficient of expansion that lowers the density of water so that the cooler melt water doesn't sink near the surface. So at these temperatures, convective heat transfer at the surface doesnt occur. I believe the problem was given a deliberate answer of under 4° so that convection can be ignored. $\endgroup$ Commented Jun 12, 2022 at 4:09
  • $\begingroup$ Thanks for the answer! I have a pair of questions, the first: if the ice temp is 0°C how can there be heat conduction through the ice to the air? Wouldn't any quantity of heat that reaches the ice just melt some quantity of ice instead of passing through it? The second: why you say that some heat transferred to the ice is used to heat the fusion water? Like @SteveSaban said I think that if the fusion water doesn't sink, it remains between the ice and the lower hotter water influencing heat transfer; but if the water sinks it should be heated without interfering with the heat transfer $\endgroup$
    – Mattia
    Commented Jun 12, 2022 at 10:40
  • $\begingroup$ If the melted water does not heat up from 0 oC and remains "stuck" at ice surface, we end up with a sandwich type system with (warm) lake water, 0 oC water, 0 oC ice, and 0 oC air. No ice will melt once we have this structure. The melted ice water has to be heated above 0 oC so that heat can flow from it to the ice. Otherwise, good question about conduction. I've expanded to explain the options. @SteveSaban I've also modified to present a case to use constant heat flow from the lake water as an alternative. $\endgroup$ Commented Jun 12, 2022 at 13:38
  • $\begingroup$ I'll be curious to see the difference between the final numerical answers of the two solutions. It seems that the solution given in the post is meant to be quick and easy approach that gives an approximate answer. If your professors answer agrees with Jeffrey's within a few percent, it's a useful approach. $\endgroup$ Commented Jun 12, 2022 at 16:21
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    $\begingroup$ The heat flow (flux) from the top of the water to the bottom of the ice is unknown. It is either by convection or by direct conduction. The better statement of reality is likely that no heat flows from the melting ice in contact with the (higher temperature) water to go through the ice to the air because the solid ice does not sustain a temperature gradient. $\endgroup$ Commented Jun 13, 2022 at 2:26

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:

  1. 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.
  2. 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.
  3. The lake is still and remains still during the melt (no forced convection).
  4. The lake, meltwater, and ice remain in full contact during the melt
  5. There is no temperature change due to the phase transition from solid to liquid
  6. $T_{melt/ice} = 0 $ during the melting process (T at the phase boundary is zero)
  7. The temperature at the surface of the lake is $T_{lake}$ at t = $0°C$. Not true but it simplifies the calculation.
  8. 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.

  • $\begingroup$ But in the meltwater there is a linear temperature gradient to sustain steady state conduction, and new meltwater has to be heated to the correct temperature given by that gradient, so some heat goes doesn't go to the ice. And I don't know how much is realistic to assume that heat flow between ice and a temperature source depends only on what there is between but not on the physical properties of the ice (and the source). $\endgroup$
    – Mattia
    Commented Jun 14, 2022 at 10:57
  • $\begingroup$ Anyway this could be an interesting approach once accounted for the new meltwater heat and if for the most part of the process there a significant quantity of meltwater so that the heat conduction depend mostly on its thickness and not on other parameters. So here one doesn't need to know before the heat flux to solve the problem. $\endgroup$
    – Mattia
    Commented Jun 14, 2022 at 11:03
  • $\begingroup$ It's how the melting of the ice is modeled that makes this a weird problem to visualize. The latent heat of melting is the amount of energy required to convert a mass of ice at 0° to the same mass of water at 0°. If there is no temperature change between the two phases, how does heat flow to melt the ice? In this model, it flows by allowing the ice/melt boundary between the two move upward and allowing the lake/melt boundary to stay fixed in position and at lake temperature. $\endgroup$ Commented Jun 15, 2022 at 20:00
  • $\begingroup$ Yes but but to allow heat flow there must be a temperature gradient in the meltwater, namely between lake/melt boundary and ice/melt boundary. New meltwater must be heated accordingly to this gradient. $\endgroup$
    – Mattia
    Commented Jun 15, 2022 at 21:15
  • $\begingroup$ It will be after the melt process is complete. Prior to that $\Delta T_{lake/melt } $will be $T_{Lake} -T{melt/ice}$. If it is assumed that the time it takes for the phase change to occur << the time it takes to heat the lake/melt boundary (which I believe to be true ), then your concerns are negligible until the ice is melted provide the ice isn't too thick to start with. A more appropriate problem is a disk of ice in a beaker on a constant T surface with the air, ice, and other parameters as in the original. Then this becomes a very good solution. $\endgroup$ Commented Jun 15, 2022 at 21:37

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