If you have an ice block in the International Space Station would it melt more slowly than on Earth at the same temperature and pressure. The reason I ask is that ice can melt due to pressure. The ice block on Earth has a certain weight which presses down on the lower layers of ice within itself sort of like how my body puts pressure on my feet when I stand. If this causes the ice block to melt more quickly on Earth then would the same ice block take longer to melt in zero gravity?

  • $\begingroup$ Could you please be more specific what the experiment should look like? Are the temperature and pressure in the space station the same as on earth or are there differences? Your assumptions are not clear here. $\endgroup$
    – Merlin1896
    Aug 21, 2015 at 16:42
  • $\begingroup$ Assume that the only difference is gravity. $\endgroup$
    – Alex
    Aug 21, 2015 at 16:44

2 Answers 2


Well, the temperature profile of an object on the macroscopic scale is governed by,

$${{dT} \over {dt}}=\kappa \cdot \nabla^2 T $$

Where T is the temperature, t is time, and $\kappa$ is the thermal diffusivity. It's the constant that matters here. The larger the constant, the faster the temperature changes. The equation itself is called the Heat Equation.

$$\kappa={{\lambda} \over {\rho \cdot c_p}}$$

So the thermal diffusivity is a function of the thermal conductivity, density, and heat capacity.

So basically, we need to lay out ground rules. First, these "constants" may be functions of temperature. Secondly, we're interested in a case with no gravity.

Now, on Earth, a melting block of ice will drip water as it melts. This water will collect at the bottom of the ice block and then move away from it.

However, in space, there's no force to "whisk" the newly liberated water away from the ice. Instead, the water will stay attached to the surface due to cohesion. Imagine a shell of water surrounding the ice block suspended in mid air.

This actually has an impact on the melting rate. The density of ice hovers around $\rho_i=917 {{kg} \over {m^3}}$. The density of water hovers at around $\rho_w=1000 {{kg} \over {m^3}}$. I'll assume the ambient temperature of the ISS is room temperature, $25 \ C^o$, so we don't have appreciable changes in density. The percent change in density is $8.3$%. For the other variables transitioning ice to water, we have thermal conductivity going from $2.22$, ice, to $0.6$, water, and the heat capacity going from $2.05$, ice, to $4.193$, water. Overall the thermal diffusivity sees a $88$ % decrease! This means that the rate of temperature change at the surface of the water collecting at the block will go down $88$% after phase transition begins. Overall, compared to earth, you'd expect the time needed heat the surface of the melting block to go up considerably. However, something very interesting happens at this phase transition. The single PDE becomes two! By the way, all of these values for the constants come up with a quick search engine search.

You see, now there is an air-water boundary and a water-ice boundary. In fact, phase transition splits the single PDE into a system of PDEs. To simplify things, we'll ignore the air-water PDE. That is the PDE that governs water temperature. Now, for the water-ice PDE, the one that governs ice melting, nothing changes, the ice keeps melting, now it just does so submerged in water. The thing that is interesting though, is that now there is an "conductive layer" between the air and the ice. This is because the thermal conductivity of water is about $20$ times greater than the conductivity of air. This means that the water layer will conduct heat from the air to the ice faster than if there was no water layer at all! However, this is assuming the temperature of the air and water are roughly equal. That isn't the case here. Using the right hand side of the Heat Equation, we see that it's the laplacian of the temperature that gives rise to changes in temperature. At the ice-water layer the laplacian is incredibly small because the water has just melted. We can approximate $\nabla^2$ to,

$$\nabla^2 T \sim {{T_o+T_i} \over {dx}}$$

Where $T_o$ is the outer surface temperature, and $T_i$ is the inner surface temperature. We'll approximate $dx \sim 0.001 m$. The value $dx$ represents a cold layer that forms a barrier between material layers. Because we're considering zero gravity, this "cold layer" doesn't sink. For the ice-air boundary, $T_o=25$ and $T_i=1$, we get $\nabla^2 T \sim 2.4 \cdot 10^7$. For the water-ice layer we have $T_o=1$ and $T_i=0$, we get $\nabla^2 T \sim 10^6$. This shows that the formation of the water layer changes the rate of temperature change at the surface by a factor of 25. This combined with the $88$ % decrease and balanced with the increased conductivity of water will slow the melting to about a tenth of the original rate.

So we'd expect the time needed to melt the ice to up 10 fold, right? Nope, it's going to melt only about 5 to 10 times slower. The reason is because even on Earth this protective layer of water is formed, it's just muted compared to the one formed in space. I'd compare this phenomena to clothing. We wear jackets in cold weather because it forms a layer between us and the cold.

  • $\begingroup$ I really do like your answer, but I have a few suggestions: You are dealing with three different kappas/lambdas for the different interfaces: ice-air, ice-water, water-air. It might be good, to give them indices in order to be absolutely clear, which interfaces are meant. Also, one additional question: Why don't you have to consider the interface water-air more explicitly? If the heat transport there was much slower, then this would have a huge impact on the whole process, right? $\endgroup$
    – Merlin1896
    Aug 21, 2015 at 21:42
  • $\begingroup$ I agree with @Merlin1896 that one needs to take into account how quickly heat energy can traverse the water-air interface, and that there might be a bottleneck there. On the other hand, there could be some kind of "thermal impedance matching" going on as well that results in a more efficient conduction of heat to the ice. (It's been a while since I've had to think about heat flow.) $\endgroup$ Aug 21, 2015 at 21:47
  • $\begingroup$ @MichaelSeifert seems as though I underestimated the layer effect. I fixed up my answer. $\endgroup$
    – Zach466920
    Aug 21, 2015 at 22:37
  • $\begingroup$ @Merlin1896 seems as though I did mess up, just not in the way you thought. I miscalculated the laplacian due to the water-ice layer. Because they're basically the same temperature, there isn't any heat flow to counteract the water-air heat transfer slow down. I added this into my answer. $\endgroup$
    – Zach466920
    Aug 21, 2015 at 22:41
  • $\begingroup$ @Zach466920 In zero gravity there will also be a layer of cold air around the block with no convection to carry it away (downwards). Still air is a very good insulator. $\endgroup$ Aug 21, 2015 at 23:46

Yes space is a vacuum which means no pressure and air and no gravity. As you said ice melts quickly in high pressure so in space ice will take a longer time to melt if it's not so hot there.

  • 1
    $\begingroup$ The question was about "an ice block in the International Space Station", not in the vacuum of space. $\endgroup$ Aug 21, 2015 at 16:53

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