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 What's the difference between snowmen and snowwomen? (Posted on 2004-06-14)
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?

 No Solution Yet Submitted by SilverKnight Rating: 4.0000 (2 votes)

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 slightly different result Comment 16 of 16 |

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) = {16-4^(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 2004-10-21 06:20:16

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