0
$\begingroup$

This is somewhat a Physics question, but ends of being more of an algebra/geometry-related question.

Anyway, there is a cube that radiates with a power $P_0$ and it is "cut" into two pieces. These two pieces radiate energy with a power of $P_1$. Find the ratio of $P_1$/$P_0$.

Heat transfer for radiation is defined as P = $\delta$Q/$\delta$T = e $\sigma$ A $T^4$

Everything is constant between the two cubes (except for their total surface area, obviously)

The ratio ends up being $P_1$/$P_0$ = $A_1$/$A_0$

My problem is that it doesn't given any details about the size before or after the cube is cut. I tried every geometric comparison I could think of, I always ended up with 1/2 or 2. The answer supplied is 4/3. Any hints?

$\endgroup$
4
  • $\begingroup$ I kind of makes sense that $P_1 > P_0$ such that the ratio is large than 1. Don't you think? $\endgroup$
    – Fabian
    Apr 26, 2012 at 5:05
  • $\begingroup$ What is $P$? What is $\delta$? What is $Q$? What is $T$? What is e? What is $\sigma$? What is $A$? Don't worry, I know what 4 is. $\endgroup$ Apr 26, 2012 at 5:54
  • $\begingroup$ @Gerry, I only know $T$ is temperature but all that matters is the ratio of areas since everything else is constant. $\endgroup$ Apr 26, 2012 at 5:58
  • $\begingroup$ The answer is only correct when the cube is cut into "boxes." If you slice down along a diagonal of one of the faces, the ratio is $5:3$. And there are many other possibilities for cutting. Some, like cutting off a tiny corner, make no appreciable difference. $\endgroup$ Apr 26, 2012 at 6:21

1 Answer 1

1
$\begingroup$

If it is a unit cube then each of the six faces has unit surface area. When it is cut in two pieces normal to one of its axes, two new unit faces are created... (What if the cut is not normal to an axis?)

$\endgroup$
1
  • $\begingroup$ "What if the cut is not normal to an axis?" Then I could lop a tiny bit off one corner and not change the total surface area very much at all. $\endgroup$
    – user856
    Apr 26, 2012 at 6:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .