Sometime this past winter, I was in a snowball fight... and I had two left when we had to call it quits. As it happened, they were both spheres and one had exactly twice the diameter as the other.
I left the two on the ground, when we quit... and the weather started to get warmer. The snowballs started to melt. The melting only occurred at their surface, so the speed at which the balls melted was proportional to only the surface of the (remainder of the) snowballs.
How much (volume) was left of the small snowball when half the volume of the larger had melted?
If anyone can point out an error in my woking i'd appreciate it!
let initial radius of snowball 1 = R
let initial radius of snowball 2 = 2R
S1(t), S2(t) = surface area of snowballs 1 & 2 respectively
V1(t), V2(t) = volume
r1(t), r2(t) = radius  ie r1(0)=R & r2(0) = 2R
time, t = T when snowball 2 is half it's initial volume
melting over time is proportional to surface area = constant, k

V2(0) = 4/3.pi.(2R)^3
V2(T) = 0.5 V2(0) = 16/3.pi.R^3
=> r2(T) = 4^(1/3).R

S2(0) = 4.pi.(2R)^2
S2(T) = 4.pi.[4^(1/3).R]^2
k = {S2(T)S2(0)} / S2(0) = {164^(5/3)} / 16 = 0.37

S1(0) = 4.pi.R^2
S1(T) = 0.37 x 4.pi.R^2 = 4.pi.r1(T)^2
=> r1(T) = (0.37^0.5).R
Now V1(0) = 4/3.pi.R^3
& V1(T) = 4/3.pi.(0.37^0.5)^3.R^3
=>V1(T) / V1(0) = 0.37^1.5 = 0.22
That is, smaller snowball is approx 22.5% its original size.

Posted by David
on 20041021 06:20:16 