Hey guys I really need help on this problem.

A ceramic slab of dimentions 5cm x 10 cm x .25 cm has to be heated to $177\,^{\circ}{\rm C}$. The ceramic slab travels on a conveyor belt traveling at $.1 \frac{m}{s}$. The slab is initially at a temperature of $18\,^{\circ}{\rm C}$ when it enters a radiant heater on the conveyor belt. The heater surrounds the whole slab and emits heat as a black body at $1000\,^{\circ}{\rm C}$. Since the heat surrounds the slab, all the energy that leaves the heater is absorbed by the slab. The slab has an emissivity of .85. Find the amount of time it takes to heat the slab to $177\,^{\circ}{\rm C}$. The slab has properties $c_{p}=1500\frac{J}{Kg K}$ and $\rho=900\frac{kg}{m^{3}}$ and $k=1 \frac{W}{mK}$

I started off by creating an energy balance on the slab as it's in the heater. I assumed convection that is caused by the conveyor belt traveling at such a slow speed to be insignificant to all calculations. $$\dot{E}_{in}- \dot{E}_{out}+\dot{E}_{g}=\dot{E}_{st}$$ $$A\alpha\sigma T^{4}_{heater}-A\epsilon\sigma T^{4}=\rho c_{p}V\frac{dT}{dt}$$ $$A\sigma(T^{4}_{heater}-\epsilon T^{4})=\rho c_{p}V \frac{dT}{dt}$$

I then plugged in the values and got: $$44.185-1.428x10^{-11} T^{4}=\frac{dT}{dt}$$

I realized that this comes out to be a VERY nasty integral so I tried to use matlab and numerically integrate it using Euler's method. My graph for that came out to be linear which I knew was not right and I came here straight away to figure out what I was doing wrong

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You don't need Euler's method, this can be integrated analytically. Use separation of variables and partial fractions. – Raskolnikov Apr 21 '12 at 8:20
Raskolnikov is right, but also: how do you know that the graph is not supposed to be linear? What exactly is giving you trouble? You know how we are about "check my work" questions... – David Z Apr 21 '12 at 8:24
@DavidZaslavsky: it is a "check my work" question but I've answered anyway. That's because I think there is an important point for budding physicists here i.e. when you're doing a calculation you should always look for ways to sanity check your results. In this case 5 minutes with Excel provide an adequate sanity check. – John Rennie Apr 21 '12 at 8:46
We also prefer titles that tell us about the post, not the poster – Manishearth Apr 21 '12 at 9:14
So here's my problem now. The problem states that "since the heater surrounds the slab, all the heat that leavces the heater is absorbed by the slab". I originally thought that this meant that $\alpha=1$ but now I'm thinking that this just means that the view factor between the heater and the slab is 1. So can I assume that the object is diffuse-grey and says $\alpha=\epsilon$ even though the problem doesnt state that the object is diffuse-grey – Greg Harrington Apr 21 '12 at 20:11

I haven't checked your numbers, but I too get an equation that looks like:

$$\frac{dT}{dt} = a - bT^4$$

I used your figures and graphed dT/dt in Excel over the range 18 to 177C, and it does indeed remain almost constant over this range. That's because the $1.428x10^{-11} T^4$ term is small compared to 44.185. Physically this means the temperature difference between the slab and the oven is approximately constant.

However I think there's an error in your numbers. If you set T = 1000C (i.e. 1273K) shouldn't dT/dt come out as zero? With your numbers I get when T = 1273 dT/dt = 6.68.

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I don't believe it would come out to zero when $T=1273K$ because the emissivity and the absoprtivity are different. The problem states that "the heater surrounds the slab so all the heat that leaves the heater is absorbed by the slab." I assumed this mean that this statement makes the absorptivity 1 while the emissivity is .85. But it can also mean that the angle factor between the heater and the slab is 1, but then what would the slab be? It doesn't say whether or not it is diffuse-grey or not... – Greg Harrington Apr 21 '12 at 17:41
Suppose the emissivity and the absorptivity are different, then the slab would heat to a higher temperature than it's environment. I could connect a heat pipe between the centre of the slab and the environment, and presto I have a perpetual motion machine. As Pygmalion says, $\alpha$ must be the same as $\epsilon$. – John Rennie Apr 22 '12 at 6:16
I never thought of it that way. Thanks for helping me understand that – Greg Harrington Apr 22 '12 at 6:28

The general expression for power exchange between two bodies due to radiation is

$$\dot{Q}_{12} = \frac{\sigma (T_1^4-T_2^4)}{\frac{1-\epsilon_1}{A_1 \epsilon_1} + \frac{1}{A_1 \epsilon_{12}}+ \frac{1-\epsilon_2}{A_2 \epsilon_2}}$$

Even if geometry factor between bodies is $\epsilon_{12} = 1$, and one of the bodies is perfectly black body $\epsilon_1 = 1$, this does not reduce to your expression.

(And there is a good reason for that. Not all the heat that leaves heater is absorbed by the slab due to the fact that its absorptivity is not 1!)

Edit: It is also obvious that the heat that goes from slab back to heater is practically negligible. Which means that slab's temperature should rise about constantly. If that is your conclusion too, you're right.

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it states in the problem that the heat surrounds the slab so all the heat that leaves the heater is absorbed by the slab. Now I was just making assumptions at this point but do you think that's saying that the absoptivity of the slab is 1? Or is it just saying that the angle factor between the slab and heater is 1? In that case, what would I assume the absorptivity to be? – Greg Harrington Apr 21 '12 at 17:38
"All the heat that leaves the heater is absorbed by the slab." That only means that geometry factor between bodies is 1. Of course all heat that leaves heater is absorbed by the slab, but this heat is smaller than $\sigma A T^4$. The expression above is universal, and considered as a standard for civil engineers (e.g. calculating heat transfer between glasses in the window, all books for civil engineering physics include it), but maybe this details are not relevant for mechanical engineers? Check up your book... – Pygmalion Apr 21 '12 at 17:43
Ok I figured that the view factor was going to be 1. But now what can I can about the absorptivity? The problem statement does not say that I can assume the slab to be diffuse-grey. In that case I would be able to say $\alpha = \epsilon$ – Greg Harrington Apr 21 '12 at 17:56
$\alpha = \epsilon$ is always true, non-negotiable! – Pygmalion Apr 21 '12 at 17:59
There are only perfect black bodies, grey bodies and white bodies. See en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation. See also en.wikipedia.org/wiki/Thermal_radiation. There you can find the formula I gave above. – Pygmalion Apr 21 '12 at 18:09