*k*cm

^{3}/s per cm

^{2}of exposed solid surface (assuming constant T and P in the room).

When will it be at half its original radius? When will it have completely sublimed?

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Started by Yoruichi-san, Aug 18 2012 08:29 AM

8 replies to this topic

Posted 18 August 2012 - 08:29 AM

You have a spherical mothball which sublimes at a rate of *k* cm^{3}/s per cm^{2} of exposed solid surface (assuming constant T and P in the room).

When will it be at half its original radius? When will it have completely sublimed?

When will it be at half its original radius? When will it have completely sublimed?

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Posted 18 August 2012 - 06:04 PM

You would need to know the initial radius R. Simplifying your rate you get k cm/s as the rate at which the radius decreases so it would take 0.5R/k to get to half the radius and R/k to sublime completely.

Posted 20 August 2012 - 05:56 PM

Spoiler for What happened to using spoilers?

Try a right circular cylinder with height = 2R, made of the same material at the same T and P.

Yes, you can assume initial conditions of R

<3 BBC's Sherlock, the series and the man. "Smart

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Posted 20 August 2012 - 06:27 PM

Try a right circular cylinder with height = 2R, made of the same material at the same T and P.

Yes, you can assume initial conditions of R_{0}, t_{0}, etc.

I get the same results with a right cylinder as with the sphere

Spoiler for

Posted 20 August 2012 - 07:55 PM

I get the same results with a right cylinder as with the sphere

Spoiler for

That was the derivation I was looking for.

What happens when we increase the height of the cylinder to H=3R or decreasing to H=R, or general nR, or assume the cylinder is sitting on a base so that it is not exposed?

**Edited by Yoruichi-san, 20 August 2012 - 07:56 PM.**

<3 BBC's Sherlock, the series and the man. "Smart

Chromatic Witch links now on my 'About Me' page! Episode 3 is finally here!

When life hands me lemons, I make invisible ink.

Posted 20 August 2012 - 08:17 PM

You don't need calculus. We just need to know the area of the part of the solid at least t / k cm away from an exposed surface. The radius decreases at the rate of t / k cm per second. The upper surface drops at the same rate, and so does the bottom surface except in the case where the bottom is not an exposed surface. As long as the height is at least 2R(or R if the bottom is not exposed) then the radius decreases linearly down to zero. Otherwise the height goes to zero while the radius approaches a positive limit.

In your first point the limit is a point. With height 3R it approaches a limit that is a line of height R. If the height is R it takes half the time to disappear, and the limit is a 2D disc of radius R, etc.

In your first point the limit is a point. With height 3R it approaches a limit that is a line of height R. If the height is R it takes half the time to disappear, and the limit is a 2D disc of radius R, etc.

Posted 20 August 2012 - 09:54 PM

Um...that* is* calculus ;P.

Calculus is more than just equations with "d/dt" or the integral of something. It's the concepts of rate of change and limits etc. It's from these concepts that the equations were based on, but you don't need to use the equations to be using calculus, just like, say, you're using algebra when you try to figure out how many cobs of 25 cent corn you can buy for $2 even if you don't write the equation for it.

Calculus is more than just equations with "d/dt" or the integral of something. It's the concepts of rate of change and limits etc. It's from these concepts that the equations were based on, but you don't need to use the equations to be using calculus, just like, say, you're using algebra when you try to figure out how many cobs of 25 cent corn you can buy for $2 even if you don't write the equation for it.

<3 BBC's Sherlock, the series and the man. "Smart

Chromatic Witch links now on my 'About Me' page! Episode 3 is finally here!

When life hands me lemons, I make invisible ink.

Posted 23 August 2012 - 11:33 AM

to find the solutions requires some calculus, albeit basic one:

for the sphere to melt down to half its radius requires t = (7/6)*(pi/k)*r^{3} seconds

for the sphere to melt altogether requires t = (4/3)*(pi/k)*r^{3} seconds

for the sphere to melt down to half its radius requires t = (7/6)*(pi/k)*r

for the sphere to melt altogether requires t = (4/3)*(pi/k)*r

Posted 02 September 2012 - 08:56 AM

You have a spherical mothball which sublimes at a rate of

kcm^{3}/s per cm^{2}of exposed solid surface (assuming constant T and P in the room).

When will it be at half its original radius? When will it have completely sublimed?

Spoiler for calculus

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