Yoruichi-san Posted August 18, 2012 Report Share Posted August 18, 2012 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? Quote Link to comment Share on other sites More sharing options...

0 jim Posted August 18, 2012 Report Share Posted August 18, 2012 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. Quote Link to comment Share on other sites More sharing options...

0 Yoruichi-san Posted August 20, 2012 Author Report Share Posted August 20, 2012 It should be apparent that a dimensional analysis is not sufficient. k is dependent on the material the object is made of, and T and P, not geometry. However, the sublimation rate is highly dependent on geometry; geometries with more surface area/volume ratio sublime at a higher rate. 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. Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted August 20, 2012 Report Share Posted August 20, 2012 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 We first assume that the right cylinder holds its cylindrical shape as it sublimates. The Area and Volume of the cylinder are A = 6 pi r^{2} V = 2 pi r^{3} The given differential equation is dV/dt = k * A = k * ( 6 pi r^{2} ) We can compute dV/dr as dV/dr = 6 pi r^{2} We convert dV/dt into dr/dt using the above, and we get dr/dt = k So the change in radius with respect to time is a constant k, which implies that the half-sublimated and fully-sublimated times are (1/2)* R_{0}/k and R_{0}/k, respectively. Quote Link to comment Share on other sites More sharing options...

0 Yoruichi-san Posted August 20, 2012 Author Report Share Posted August 20, 2012 (edited) I get the same results with a right cylinder as with the sphere We first assume that the right cylinder holds its cylindrical shape as it sublimates. The Area and Volume of the cylinder are A = 6 pi r^{2} V = 2 pi r^{3} The given differential equation is dV/dt = k * A = k * ( 6 pi r^{2} ) We can compute dV/dr as dV/dr = 6 pi r^{2} We convert dV/dt into dr/dt using the above, and we get dr/dt = k So the change in radius with respect to time is a constant k, which implies that the half-sublimated and fully-sublimated times are (1/2)* R_{0}/k and R_{0}/k, respectively. 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 August 20, 2012 by Yoruichi-san Quote Link to comment Share on other sites More sharing options...

0 jim Posted August 20, 2012 Report Share Posted August 20, 2012 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. Quote Link to comment Share on other sites More sharing options...

0 Yoruichi-san Posted August 20, 2012 Author Report Share Posted August 20, 2012 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. Quote Link to comment Share on other sites More sharing options...

0 beppe Posted August 23, 2012 Report Share Posted August 23, 2012 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 Quote Link to comment Share on other sites More sharing options...

0 mmiguel Posted September 2, 2012 Report Share Posted September 2, 2012 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? Radius at time t in cm = r(t) dV/dt = -k*A(t) A(t) = 4*pi * r(t)^2 V(t) = 4/3*pi*r(t)^3 dV/dt = 4*pi*r(t)^2 *dr/dt = A(t)*dr/dt = -k*A(t) = -k*4*pi*r(t)^2 dr/dt = -k r(t) = -kt + r(0) Half sublimed is r(0)/2 = r(0) - kt t = r(0)/(2k) Fully sublimed is 0 = r(0) - kt t = r(0)/k Quote Link to comment Share on other sites More sharing options...

## Question

## Yoruichi-san

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?

## Link to comment

## Share on other sites

## 8 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.