Its freezing out here!!

Level pending

The time taken to freeze water present in an open container with insulated walls is of the form $c_1 \times \rho_w^a \times L^b \times h^c \times k^d \times T^e$ where $c_1$ is a constant, $\rho_w$ is the density of water, $L$ is the latent heat of fusion of water, $h$ is the height of the water column, $k$ is the thermal conductivity of ice and $-T^{\circ}C$ is the temperature of the surroundings.Assume that the initial temperature of water is $0^{\circ}C$ .

Find $4c_1+a+b+c+d+e$

The answer is 4.

1 solution

Pratik Shastri
Feb 12, 2014

Let us assume that $x$ depth of ice is already formed and a layer of thickness $dx$ of ice is forming,

Let $A$ be the area, $V$ be the volume of the container and $M$ be the mass of water present.

So, $dm=\frac{M}{V} \times dx \times A$

$dm=\rho_w \times A \times dx$

Now, the heat change $dQ=dm \times L$

$=\rho_w \times A \times dx \times L$

And now, using $\frac{dQ}{dt}=\frac{kA\triangle T}{l}$ ,

$\frac{dQ}{dt}=\frac{kAT}{x}$

$\frac{\rho_w AdxL}{dt}=\frac{kAT}{x}$

$\rho_w ALxdx=kATdt$

Now, integrating $x$ from $0$ to $h$ and $time$ from $0$ to $t$ ,

$\displaystyle\int_{0}^{h}\rho_w ALxdx=\displaystyle\int_{0}^{t} kATdt$

We finally get $Time=\frac{\rho_w Lh^2}{2kT}$

So, $4c_1+a+b+c+d+e=\boxed{4}$

Is there another way to do this problem without calculus manipulation?

Marcus Lara - 7 years, 2 months ago

I don't think so..

Pratik Shastri - 7 years ago

Its freezing out here!!

The answer is 4.

1 solution

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