Proving height of water as ice melts in it is constant

Question

edit: assuming there are no evaporative losses and the ice shrinks as a cube, it should stay constant...?

I'm trying to prove mathematically that the height of water in a closed container will stay constant as a cube of ice in it melts.

Here's a digram of the problem (show $H=H_0$):

I have used the following information in my proof (a cube of ice in water)

My working so far:

At time $t_0$

• Initial volume of water$=(AH_0-yx^2 )⇒y=ρ_i/ρ_w ⋅x⇒(AH_0-x^3⋅ρ_i/ρ_w ) $
• Initial mass of water$=V⋅ρ=(AH_0-x^3⋅ρ_i/ρ_w )⋅ρ_w$
• Initial mass of ice$=x^3⋅ρ_i$

At time $t_f$

• Volume of water$=AH-αy(αx)^2=AH-(αx)^3⋅ρ_i/ρ_w $
• Mass of water$=(AH-(αx)^3⋅ρ_i/ρ_w )⋅ρ_w$
• Mass of ice$=(αx)^3⋅ρ_i$
• Change in mass of ice$=x^3⋅ρ_i-α^3 x^3⋅ρ_i=x^3 (1-α)^3⋅ρ_i$

Doing a mass balance: mass of water at $t_f$ = mass of water at $t_0$ + change in mass of ice

$(AH-(αx)^3⋅ρ_i/ρ_w ) ρ_w=(AH_0-x^3⋅ρ_i/ρ_w ) ρ_w+x^3 (1-α)^3 ρ_i$ $AHρ_w-(αx)^3 ρ_i=AH_0 ρ_w-x^3 ρ_i+x^3 (1-α)^3 ρ_i$

I've tried cancelling terms after this but I'm still left with some that I can't get rid of. Where am I going wrong?

Answer 1

Your error is here:

• Change in mass of ice$=x^3⋅ρ_i-α^3 x^3⋅ρ_i=x^3 (1-α)^3⋅ρ_i$

It is not true: $x^3⋅ρ_i-α^3 x^3⋅ρ_i=x^3 (1-α^3)⋅ρ_i$

Answer 2

By Archimedes' law, the volume of water displaced by the floating ice has the same weight as the ice. That is exactly the volume the melted ice occupies, thus the level doesn't change (as long as the temperature doesn't change, there is no evaporation, ...).

Note that the form of the ice cube and of the container (as long as the ice floats freely) is completely irrelevant.

Answer 3

In order for the ice float, we know that the net upward force on the ice is equal to the weight of the ice. We also know that the force of buoyancy equals the weight of the water displaced. Therefore, the weight of the ice is equal to the weight of water displacing the ice. Knowing this, we see that

$\forall_{wd} \rho_{water} g = \forall_{ice} \rho_{ice} g$, and it follows that $\forall_{wd} = \frac{\forall_{ice} \rho_{ice}}{\rho_{water}} = \frac{m_{ice}}{\rho_{water}}$.

Since this is a closed system, we see that

$m_{water} + m_{ice} = m_{tot}$,

$m_{water} = m_{tot}-m_{ice}$,

and $m_{tot}$ is constant.

Let's represent the melting mass of ice as a function of time where $m_{ice}(t_{0}) = m_{0}$ and $m_{ice}(t_{\infty}) = 0$.

We see that height of the water at time $t$ is $h(t) = \frac{\forall_{water}(t)+\forall_{wd}(t)}{Area_{container}}$.

Substituting for the volume of water and displaced ice, we see that

$h(t) = \frac{\frac{m_{water}(t)}{\rho_{water}}+\frac{m_{ice}(t)}{\rho_{water}}}{Area_{container}} = \frac{\frac{m_{tot}-m_{ice}(t)}{\rho_{water}}+\frac{m_{ice}(t)}{\rho_{water}}}{Area_{container}} = \frac{{m_{tot}-m_{ice}(t)+m_{ice}(t)}}{\rho_{water}\times Area_{container}} = \frac{m_{tot}}{\rho_{water} Area_{container}}$.

We can conclude that the height is not a function of time and is constant... I actually didn't expect that. Please check me.