# Homework Help: Finding temperature from equal radiant power

1. Nov 11, 2013

### BOAS

1. The problem statement, all variables and given/known data

A solid sphere has a temperature of 785 K. The sphere is melted down and recast into
a cube that has the same emissivity and emits the same radiant power as the sphere.
What is the cube's temperature?

2. Relevant equations

Q = eσT4At

3. The attempt at a solution

I know that the radiant power and emmissivity of the two objects are the same and σ is a constant so I can say that;

Ts4As = T4cAc

Subscript s and c for sphere and cube.

I know the sphere's temperature and can express it's area as 4∏r2 and the area of the cube can be expressed as 6r2. Where r represents one side.

And now, I don't know what to do. Is there a relationship that links the volume of a sphere to the size of cube that can be made from it?

2. Nov 11, 2013

### phyzguy

Since it's the same amount of metal, wouldn't you expect the volumes of the sphere and cube to be equal?

3. Nov 11, 2013

### BOAS

Yes, that was a mistake.

What I meant to say was is there a relationship between the sphere and the cube that can be used to find surface area.

For a sphere of radius r, i'm pretty sure that only one cube can be constructed.

4. Nov 11, 2013

### Staff: Mentor

Yes, in this case there exist a relationship - their volumes are equal. Use that fact to find how rc depends on rs.

Honestly, I have no idea what the problem is.

5. Nov 11, 2013

### phyzguy

So if you know the radius of the sphere (call it r), you can calculate the side of the cube (call it d so you don't get confused) that has the same volume, right? So then you can calculate As and Ac from the formulae you gave before.

6. Nov 11, 2013

### BOAS

I don't actually have any numbers for r or d, but the relationship is this;

d = 3√(4/3 πr3)

T44πr2 = T463√(4/3 πr3)2

Last edited: Nov 11, 2013
7. Nov 11, 2013

### Staff: Mentor

Looks OK to me.

Yes.

Now that you know how d depends on r, you should be able to find how Ac depends on As.

8. Nov 11, 2013

### BOAS

So Ac = 6d2

= 6(4/3 πr3)2/3

T44πr2 = T463√(4/3 πr3)2

T42πr2 = T43(4/3 πr3)2/3

3r6T12 = 27(4/3 πr3)2T12

8πT12 = 48T12

T = 744 K