# Homework Help: Linear/volumetric expansivity

1. Nov 27, 2008

### moeraeleizhaj

A 1.96-kg steel sphere will not fit through a circular hole in a 0.813-kg aluminum plate, becuase the radius of the sphere is 0.10% larger than the radius of the hole. If both the sphere and the plate are kept at the same temperature, how much heat must be put into the two so that the ball just passes through the hole?

Hi, I'm back again with a new question. This question is just about volumetric expansion and linear expansion right? I would use those 2 equations and equate the 2 radii so that the ball can pass through the hole? Am I correct in thinking this?

2. Nov 28, 2008

### tiny-tim

Hi moeraeleizhaj!
Yup!

(just linear expansion in this case, of course )

3. Dec 1, 2008

### moeraeleizhaj

okay, so the linear expansion formula is D(L)=(alpha)(Lo)(D(T))
Where D(L)=change in lenth
alpha=coefficient of thermal expansion
Lo=original length
D(T)=change in temp

In here I can assume that since it's linear expansion then D(L)=change in diameter for both and the original lenths would be their original radii so the Lo=2r for both. Also:
r1=radius for sphere=1.1*r2 (because it is 10% bigger than r2) <r2=radius of the hole>

So for the steel sphere: (alpha=12x10^(-6))
D(L)=(12x10^(-6))(2)(1.1*r2)(D(T)1) .'. change in radius is that divited by 2

D(r1)=(12x10^(-6))(1.1*r2)(D(T)1) .'. the new radius is

R=1.1*r2+(12x10^(-6))(1.1*r2)(D(T)1)

Same with r2: (alpha=23x10^(-6))
D(r2)=(23x10^(-6))(r2)(D(T2))

Since the sphere passes through the hole then the 2 radii are equal:

(1.1*r2)+(12x10^(-6))(1.1*r2)(D(T)1)=r2+(23x10^(-6))(r2)(D(T2))

After this.. I don't know what to do. it's asking for the heat.
I know the heat formula is Q=mc(D(T))

I thought I'd have to equate the 2 new radii to get a ratio for the change in temperatures and use it for Q=mc(D(T)) but it didn't work out at all.

I'm not sure if what I did is even right x:
Thanks for the help

4. Dec 1, 2008

### moeraeleizhaj

I meant the original length would be the original diameter

5. Dec 1, 2008

### turin

I think you're on the right track, but remember that you have two different materials that have two different masses and two different specific heats. You have the change in temperature, and you need to calculate how much heat is needed to increase the temperature of the steel sphere and how much heat is needed to increase the temperature of the aluminum plate. The sum of these two heats should be the answer.

6. Dec 1, 2008

### moeraeleizhaj

r2[(1.1)+(12x10^(-6))(1.1)(D(T)1)]=r2[1+(23x10^(-6))(D(T2))]

*cancel r2s out

[(1.1)+(12x10^(-6))(1.1)(D(T)1)]=[1+(23x10^(-6))(D(T2))]

12x10^(-6))(1.1)(D(T)1)=1-(1.1)+(23x10^(-6))(D(T2))

D(T)1=[-(0.1)+(23x10^(-6))(D(T2))]/[(12x10^(-6))(1.1)]

so D(T)1=[-(0.1)+(23x10^(-6))(D(T2))]/(13.2x10^(-6))

then

Q=mc(D(T)) => since the 2 energy intakes are equal then equate Qs

(1.96)(452)[-(0.1)+(23x10^(-6))(D(T2))]/(13.2x10^(-6))]=(0.813)(900)(D(T)2)

solve that to get
(D(T)2)=8265.937158

Is that right?
I'm so sorry for wrecking your head >_< but I just want to make sure