# 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?

• That law is for radiative cooling. In air the cooling will mostly be convection, especially at these temperatures – Martin Beckett May 13 '16 at 15:27
• Ok, so can I exchange the formula for radiative cooling with the one for convection? In other words : ∆Q = mcv∆T =hA(Ta-Tb)? – Silentwarrior May 13 '16 at 15:48
• In order to match your experimental data, you may also need to include the heat transfer from the metal bar to its support (heat conduction). – user115350 May 13 '16 at 16:25
• We used a ceramic plate as support, do we still need to take that into consideration? – Silentwarrior May 13 '16 at 16:32

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:

$$Φ=hA(T_{metal}-T_{air})$$

• Just to clarify, f(Tmetal) is εσA(T^4metal−T^4air) for radiative cooling, but hA(Tmetal−Tair) for convection? Also, I thought Newton's cooling law was T(t)=T0+(Ts-T0)e^(-kt). Is that not right? – Silentwarrior May 13 '16 at 16:17
• First question: correct. Second question: $T(t)=T_0+(T_s-T_0)e^{-kt}$ is the general form the penultimate equation takes for Newton's cooling, provided you apply the right $k$ and the right temperatures. – Gert May 13 '16 at 16:52
• When we use a ceramic plate as support for the metal bar, do we calculate the area the same or subtract the side of the metal bar that is resting on the ceramic plate? Btw, thanks for your help, was really helpful. – Silentwarrior May 14 '16 at 9:35
• No. The heat transfer through the bottom of the bar into the ceramic plate isn't the same as through the side exposed to air. Ideally you should suspend the bar at both ends, using thin, non-conductive supports, so that the entire surface area of the bar is exposed to air. – Gert May 14 '16 at 14:57