The discussion revolves around calculating heat transfer by radiation from a copper block at 250°C with a surface area of 1200 square centimeters. The participants are exploring the application of the Stefan-Boltzmann law and the implications of emissivity in their calculations.
The discussion is ongoing, with participants providing insights and corrections regarding the calculations and assumptions. Some participants have offered guidance on the importance of absolute temperature and the need to consider the surrounding radiation when calculating heat flux.
There is a noted lack of additional information about the water involved in the heat transfer, which may affect the calculations. Participants are also working under a time constraint, as the problem is due soon.
annedi said:Hr= (Area)(e)(5.67x10^-8 W/m^2K^4)(temperature)^4
annedi said:The answer must be in joules per second
annedi said:is this correct. i computed the Hr as a function of area
132.89 W Area /m^2
70543 W Area/m^2
annedi said: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
annedi said:the heat flux from copper to water is 2548.23 W/m^2 at 250C.. is this what you mean..
annedi said:there is no more information stated here.. thanks for the help stockzahn. I really appreciated it. this was due today