How to Calculate for Thermal Energy

Blackbody Radiation

What is the emmisivity of "oxidized cast iron"?

.63

I came up with 9.355 watts,,,, am i even close ?

As you probably know by now......no you are not close

I found some old stuff using units you have probably never heard of.......here goes:

E=C *(T/100)^4 kcal/m^2*hour

When you put your numbers in and use your 0.63 figure it get to look like this:

E=4.96*0.63((373.15/100)^4)*4.19*1000*15/3600=10576.91 Watt

I hope I did it right......the figure looks a little more likely at least

good luck

far

Put it in the context of 100W light bulbs.

105 with a little left over.......it would be a bright spot and very warm too

Although you asked for "radiated" energy the total dissipated energy would also include thermal conduction into the surrounding medium.

The total energy flux would be proportional to the radiated power + the conductive dissipation.

I would explore Stefan's law regarding the rate of radiated energy and thermal conductivity regarding the conducted energy. I would probably end up guessing it as:

Power = radiated + conducted = (emissivity*Stefan's Constant * T1^4 *area) + (Thermal Conductivity of the surrounding medium * (T1-T2)*area.

Where T1 is the absolute Temperature of the body and T2 is the temperature of the surrounding medium.

I am going to guess it as P watts.

Gav

Heat exchange by thermal radiation between two bodies will also depend on the extent to which the two bodies see each other geometrically.

There's only one body in the problem.

The thermal radiation emitted by a body per unit area of its surface is called emissive power which is proportional to the fourth power of its absolute temperature. For an ideal radiator or black body the emissive power is given by Stefan- Boltzmann law. When we consider thermal radiations emitted by two bodies of area A1 and A2 at temperatures T1 and T2 then the geometry & orientation of the bodies need to be considered.

We aren't considering two bodies.

The amount of thermal energy radiated will depend on the duration of radiation.

The energy radiated per unit time (power) can be calculated.

Other posters have given you enough info for you to be able to search this on the internet. Start with "emissive power" as suggested by mrswamy.

Beware of the temperature units.

theo2,

For the calculation, you would use the Stefan-Boltzmann Law. The sticky part is determining the emissivity of the surface. For iron, it varies widely, depending on the amount of oxidation, but some common data is available here. You must be careful that the surface area is the same area as though this were a sphere, that is, the normal area. If the surface is convoluted, or folded back on itself in any way, some of that surface area would not be counted. This sounds very much like a textbook problem, so you can probably use the 15 m². As noted above, Stefan-Boltzmann only gives you power, or energy per second. If you want energy, you have to provide a time.

Hey stranger,

It's good to have you back in the rotation again.

Hope all is well with you.

I keep trying to leave, but St. Peter has a sign with a picture of the rear end of an ancient horse with mange, and a legend saying, "You must be prettier than this to enter." So, I'm stuck on this mortal coil.

I've missed posting here.

You just have to get up every morning when the light of day shines into your eyes, and do the best you can with the cards you have been dealt for that day.

There's nothing more you can do.

Take care and best of luck to you.

lyn

• What is the speed of the wind blowing over it?
• What is the altitude, which affects barometric pressure (assuming the International Standard Atmosphere applies)? Or is this surface in a vacuum?
• What is the nature and thickness of the insulation on the obverse side of this surface?
• What is the angle of this surface to the horizontal?
• Is it raining or snowing onto this surface?
• Is there any object standing on or close to this surface?
• Why is there such a high dissipation? Of what practical use is this dissipation? Can the dissipation be better used for process heating elsewhere?
• Etc., etc.

Solution to your problem is very simple. You don't have to touch your ear with your hand crossing your back head. There are many laser thermometer in the market that can measure temperature of red hot metal. Use that.

the answer to this one was " A " 10,394 watts,, Thank You all for your input and time on this question,, you have helped me get a little better understanding about this ! Thank You all again ! I did miss one on the test, it was " Calculate the temperature coefficient of resistance (TCR ) of copper in ohms per ohm per C degree,,, I came up with 0.00358,,, wrong answer !! How do i give you all a good answer rate?

0.00393

If the ambient temperature is also <...100 degree C...> then the answer will be zero.

You only have to measure watts in over time(btuh) and subtract losses....