cold balls *brrrr*

Originally posted by PBE6
Yeah, just double checked on paper. I meant to divide Vsmall by Vsmall0, not just multiply the (4/3)*pi straight through:

Vsmall = (4/3)*pi*rS0^3 * (1-2(1-b))^3
Vsmall0 = (4/3)*pi*rS0^3

Vsmall/Vsmall0 = (1-2(1-b))^3 = 20.3% (approx)

as stated above.

so you are saying that the bigger snowball loses 50% of it's volume but the smaller snowball loses approx 80% of it's volume even though the surface of each snowball is melting at the same rate?

Originally posted by uzless
so you are saying that the bigger snowball loses 50% of it's volume but the smaller snowball loses approx 80% of it's volume even though the surface of each snowball is melting at the same rate?

According to the question the snowballs melt at a rate proportional to their surface area, so they don't actually melt at the same rate unless the surface areas are equal. They actually melt slower as they get smaller. Of course, this model neglects the effects of heat energy penetrating to the interior of the snowball which would predict an increased rate of melting with time despite the shrinking surface area. But that's the question as posed.

Originally posted by PBE6
According to the question the snowballs melt at a rate proportional to their surface area, so they don't actually melt at the same rate unless the surface areas are equal. They actually melt slower as they get smaller. Of course, this model neglects the effects of heat energy penetrating to the interior of the snowball which would predict an increased rate of melting with time despite the shrinking surface area. But that's the question as posed.

ah yes, the volume...i was thinking volume and diameter shrinkage would be same but now that i've had a few beers and woke up, I see the error of that logic