1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding temperature from equal radiant power

  1. Nov 11, 2013 #1
    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. jcsd
  3. Nov 11, 2013 #2


    User Avatar
    Science Advisor

    Since it's the same amount of metal, wouldn't you expect the volumes of the sphere and cube to be equal?
  4. Nov 11, 2013 #3
    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.
  5. Nov 11, 2013 #4


    User Avatar

    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.
  6. Nov 11, 2013 #5


    User Avatar
    Science Advisor

    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.
  7. Nov 11, 2013 #6
    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

    Am I actually going about this question in a sensible way?
    Last edited: Nov 11, 2013
  8. Nov 11, 2013 #7


    User Avatar

    Staff: Mentor

    Looks OK to me.


    Now that you know how d depends on r, you should be able to find how Ac depends on As.
  9. Nov 11, 2013 #8
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted