# How much power to keep surface of aluminum plate at given temperature? [closed]

I want to heat one side of an aluminum plate enough to hold the other side of that plate $100K$ above ambient. I'm willing to assume that the heated side of the plate is "well" insulated (along with the edges, if it's significant). The plate is $500cm^2$ and is open to air in a normal room.

I think the formula is simple enough: $P = \frac{\Delta T}{R_{pa}}$ Where $R_{pa}$ is the thermal resistance of the plate to air in $\frac{K}{W}$. The problem I'm having finding the thermal resistance is that most heat-sink-like applications would care about the bottom of the plate in this case: the goal would be to cool the bottom of the plate. I don't actually care what the temperature of the bottom is (within reason): I want to hold the top at a specific temperature.

If I attack it from first principles and consider that the the thermal resistance of aluminum is irrelevant (since it's the top surface I care about) then all I need is the thermal resistance of the air above it. The guessing part comes in when trying to figure out how thick a slab of air to consider. If I guesstimate 3cm I get:

$$T_{air} = 0.025 \frac{W}{mK}$$

$$T = \frac{0.05m^2}{0.03m}T_{air} = 0.04 \frac{W}{K}$$

$$R_{pa} = \frac{1}{T}$$

$$P = \frac{100K}{R_{pa}} = 100K \cdot 0.04 \frac{W}{K} = 4W$$

The 4W result feels about an order of magnitude too low. Of course if I change my guess of the air thickness to consider to 3mm I get 40W which seems more reasonable. But instead of calculating the wattage I need I'm just guessing in different units.

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## closed as off-topic by tpg2114, MAFIA36790, Asher, CuriousOne, GertJun 18 at 1:58

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Convection? Radiation? – Peter Shor Mar 19 '11 at 21:47