Calculating Heat Transfer by Radiation: Copper Block at 250 C and 1200 sq cm

[annedi]

Homework Statement

How much heat is radiated in water per sq cm, from a block of copper at 250 C and 1200 .

Homework Equations

e= 0.6

Hr= (Area)(e)(5.67x10^-8 W/m^2K^4)(temperature)^4

The Attempt at a Solution

i'm sorry, i don't know how to solve this one...

 

[stockzahn]

Hr= (Area)(e)(5.67x10^-8 W/m^2K^4)(temperature)^4

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

 

[annedi]

The answer must be in joules per second

 

[stockzahn]

The answer must be in joules per second

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?

 

[annedi]

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

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

 

[stockzahn]

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

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

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?

 

[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

 

[stockzahn]

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

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)

 

[annedi]

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

 

[stockzahn]

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

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?

 

[annedi]

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

 

[stockzahn]

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

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.

 

[annedi]

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