# Radiative Cooling of metal object

We have this experiment where a metal bar is heated and then we have to make a model for the cooling that occurs. We get numbers for how long it takes the metal bar to cool from 200 to 100 degrees Celsius, and we have to calculate how long it takes for the object to cool to 50 degrees. Now, I want to calculate what the cooling would be like if it was only radiative. I know that $T(t)=T_0+(T_s−T_0)e^{−kt}$ is the formula for cooling by convection, so I'm wondering what the equivalent of that is for radiative cooling. In other words, what is the formula for radiative cooling?

• hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html – valerio May 16 '16 at 18:04
• I need to take the air into account. The website was good, but they don't take that into account. – Silentwarrior May 16 '16 at 18:40
• They do. They call it $T_{ambient}$. – Gert May 16 '16 at 18:50
• You said that you wanted to calculate $T(t)$ in the situation in which the cooling is only radiative...anyway, just sum the two contributions: $\frac{dE}{dt} = - (a (T-T_e)^4 + b (T-T_e))$, where $T_e$ is the temperature of the environment and $a,b$ are constant coefficients. – valerio May 16 '16 at 18:59
• Even if the cooling is only radiative, dosen't the temperature of the air still make a difference? – Silentwarrior May 16 '16 at 19:05

If you assume that the object radiates as a blackbody, then the Stefan-Boltzmann Law tells you that the power radiated by the object will be

$P = A \sigma T^4$

where $A$ is the surface area, $\sigma$ is the Stefan-Boltzmann constant, and $T$ is the temperature. If your object is not a blackbody, you can parameterize the above equation with an additional constant, $\epsilon$, called the emissivity.

From this you should be able to figure out the temperature as a function of time by knowing that the thermal energy is proportional to temperature and that the power is equal to the time derivative of energy.

• That loss function is for radiative losses in vacuum. In air some gets radiated back. – Gert May 16 '16 at 18:51
• That's true, although I would imagine that that would only result in a change of the effective emissivity of the object. It would be unlikely to change the temperature dependence of the power by much. – J. O'Brien Antognini May 16 '16 at 19:21
• hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html If OP wants a proper solution then it has to be done properly. – Gert May 16 '16 at 19:23
• I understand that, but the ambient temperature does not enter into the temperature dependence of the cooling time, unless you are assuming that the ambient temperature is changing due to the heat loss from the object. But that's going to be a second-order effect and I've neglected it in my answer. – J. O'Brien Antognini May 16 '16 at 19:30
• It changes the DV for the problem quite drastically. For very hot objects you could neglect ambient $T$ but then it becomes an approximation. It depends what one wants to achieve, of course. – Gert May 16 '16 at 19:36

Use the heat loss function for radiative cooling and plug it into the differential equation I set up for you in the answer. Then integrate.

• Yes, that's what I did at first, but I ran into some trouble into integrating. What I got was $$\Large{tεσA4(T_{air}^4)/mc_v=ln\frac{T_{metal,t}+T_{air}}{-T_{metal,t}+T_{air}}-ln\frac{T_{metal,0}+T_air}{-T_{metal,0},+T_air}+2arctan\frac{T_{metal,t}}{T_{air}}-2arctan\frac{T_{metal,0}}{T_{air}}}$$ which does not seem correct. Maybe it is, but I doubt it, that's why I came back here. – Silentwarrior May 16 '16 at 18:56
• That's difficult to judge w/o the derivation. – Gert May 16 '16 at 19:11
• What I did was basically just switch $$Φ=hA(T_{metal}-T_{air})$$ with $$Φ= εσA(T_{metal}^4-T_{air}^4)$$ from the previous question and then integrate. This is what I got. – Silentwarrior May 16 '16 at 19:17
• Like I said: difficult to judge w/o seeing how you did it. I'm not a mind reader and that integral is not easy. Why not put the derivation in your question? – Gert May 16 '16 at 19:20
• wolframalpha.com/input/… gives the main form of the difficult integral. – Gert May 16 '16 at 19:31