MHB How long does it take for a snowball to melt completely?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Assume that a snowball melts so that its volume decreases at a rate proportional to its surface area. If it takes three hours for the snowball to decrease to half its original volume, how much longer will it take for the snowball to melt completely?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Aryth, BAdhi, lfdahl, MarkFL, Opalg, and Pranav. You can find lfdahl's solution below.

[sp]\[\frac{\mathrm{d} V(r)}{\mathrm{d} t} = -kA(r),\; \; \; k > 0 \\\\ \frac{4}{3}\pi \frac{\mathrm{d}r^3 }{\mathrm{d} t}= -k4\pi r^2 \Rightarrow r' = -k \Rightarrow r(t) = r_o-kt \;\;\;\; (1)\]

Where $r(0)=r_o$ is the initial radius of the snowball. At $t=3$ half of the initial volume has melted away - under the assumption, that the melting is an isotropic process, i.e. can be expressed in terms of the reduction of $r(t)$:

\[ V(3)= \frac{4}{3}\pi r^3(3)=\frac{1}{2}\frac{4}{3}\pi r_o^3 \Rightarrow r(3)= \frac{1}{\sqrt[3]{2}}r_o\]

Solving for $k$ using $(1)$:

\[r(3)= \frac{1}{\sqrt[3]{2}}r_o = r_o-k\cdot 3 \Rightarrow k=\frac{r_o}{3}\left ( 1-\frac{1}{\sqrt[3]{2}} \right )\]

All of the snowball has melted away at time: $t=\frac{r_o}{k}$ so the time from half to zero volume must be:

\[t_{\frac{1}{2}} = \frac{r_0}{k}-3 \Rightarrow t_{\frac{1}{2}} = \frac{3}{\sqrt[3]{2}-1}\approx 11.54 \; \; hrs\][/sp]