Cooling of a hot metal object in air 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. At first I thought of just using Newtons law of cooling (T(t)=T0+(Ts-T0)e^(-kt)), but I'm wondering if you can use this other way.
$$Φ= εσA(T_{metal}^4-T_{air}^4)$$
(Where $T_{metal}$ is temperature of metal in kelvin, $T_{air}$ for air, $A$ is area of the metal bar, $σ$ is a constant and $ε$ is about 0.1 for metals. All of these are known.) 
Then with the help of the formula: 
$∆Q = mc_v∆T = Φt$ ($m$ is mass of metal bar, $c_v$ is heat capacity and $∆T$ difference in temperature and $t$ time in seconds, all but $t$ is known)
We get:
$t = \frac{mc_v∆T}{εσA(T_{metal}^4-T_{air}^4)}$
Is this correct, or am i missing something?
 A: Firstly you must understand the difference between the two models. Newton's law of cooling concerns itself with purely convective cooling. Your second model assumes purely radiative cooling. In most cooling situations both modes of cooling play a part but at relatively low temperatures (such as yours) the prevalent mode is convective. So Newton's law is more applicable here.
Secondly, let's say you adopt some model where the heat loss is a function of $T_{metal}$, so:
$$Φ= Φ(T_{metal})$$
Now you CANNOT, as you do, state:
$$∆Q = mc_v∆T = Φt$$
Why not? Because the object is cooling down, so $\Delta T$ is also a function of $t$!
The correct statement is in the form of a differential equation:
$$-mc_v\frac{dT_{metal}}{dt}=Φ= Φ(T_{metal})$$
(The negative sign is needed because $\frac{dT_{metal}}{dt}$ is negative)
Or:
$$-mc_vdT_{metal}=Φdt$$
Rework this to:
$$-mc_v\frac{dT_{metal}}{Φ(T_{metal})}=dt$$
Integrated between applicable boundaries that then gives you an expression for $t$:
$$\Large{-mc_v\int_{T_{metal,0}}^{T_{metal,t}}\frac{dT_{metal}}{Φ(T_{metal})}=t}$$
In the case Newton's law applies:
$$Φ=hA(T_{metal}-T_{air})$$
