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1. Sep 28, 2015

### annedi

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

How much heat is radiated in water per sq cm, from a block of copper at 250 C and 1200 .
2. Relevant equations
e= 0.6

Hr= (Area)(e)(5.67x10^-8 W/m^2K^4)(temperature)^4
3. The attempt at a solution
i'm sorry, i don't know how to solve this one...

2. Sep 28, 2015

### stockzahn

Do you know what this formula says (what the result of it is) and what the dimension of the result could be?

3. Sep 28, 2015

### annedi

The answer must be in joules per second

4. Sep 28, 2015

### stockzahn

Correct, J/s (=W). So it is a heat flux. If you plug in the temperature(s) and the the emissivity from the statement (there is no area), what heat flux have you calculated?

5. Sep 28, 2015

### annedi

is this correct. i computed the Hr as a function of area

132.89 W Area /m^2
70543 W Area/m^2

6. Sep 28, 2015

### stockzahn

According to the statement you posted the temperatures of the copper are given in °C. The formula only works with the absolute temperature, so the values are not correct.

What do you mean with "W Area/m^2"? If you don't know a value for the area and you calculate the formula without it, what would the dimension of the result be?

7. Sep 28, 2015

### annedi

I'm sorry, I forgot to convert the temperature

what i meant by W (area)/ m^2

whatever the area is, you'll plug it in the equation.. so W (x)m^2/m^2 will cancel out thus leaving W as the unit

8. Sep 28, 2015

### stockzahn

Okay, then if you plug in all the values you have, what heat flux would you have calculated? (I mean from where to where or from which object to which object)

9. Sep 28, 2015

### annedi

the heat flux from copper to water is 2548.23 W/m^2 at 250C.. is this what you mean..

10. Sep 28, 2015

### stockzahn

That is basically what I meant, but it is not the heat flux to the water. It would be the heat flux, if all the objects around the copper wouldn't emit radiation at all (at which temperature would that be?). All the objects around, also emit radiation and you need to calculate the difference of the exchanged heat fluxes. Is there any other information about the water?

11. Sep 28, 2015

### annedi

there is no more information stated here.. thanks for the help stockzahn. I really appreciated it. this was due today

12. Sep 28, 2015

### stockzahn

If that should be the solution, then sorry. I obviously misunderstood the statement. Just one more thing: I think you have to calculate the heat flux for one cm^2 not m^2. That has to be still transformed.

13. Sep 28, 2015

### annedi

oh yeah. i just realized that just now.. THANKS AGAIN :)