homework and exercises - What is a simple calculation to figure out how many watts needed to maintain a hot piece of tungsten?

The specific heat of tungsten is $.13 (\text{kJ /(kg K)})$.

1 cubic cm of tungsten is .0193 kg

And the melting point of stainless steel is $1900 \deg \text{K}$ conservatively (giving it plenty of heat in the liquid state).

So I calculate $.13 \times .0193 \times 1900 \approx 4.8 \text{k J}$.

So how many watts of energy input into a system of tungsten that touches only air and ceramic alumina do I need to maintain $4.8 \text {k J}$ in a steady state?

I'm not sure how much heat gets dissipated.

I'm hoping to maintain it with a $1000 \text{W}$ induction heater, assuming 100% efficiency.

In your answer feel free to use formulas / more abstract-looking math.

Answer

As stated, there is no way to determine what you're asking. Apparently you want to heat 1 cc of tungsten to the melting point of stainless steel using an induction heater. If you do this, you will have to apply a net energy of $4.8 kJ$, as you have calculated. Apparently the tungsten will be supported on an alumina surface or in an alumina crucible. And then you want to maintain the tungsten at that temperature.

First, stainless will melt in a range of 1325 to 1530 C, or 1600 to 1800 K, so I don't know where you got 1900 K from.

At 1900 k the tungsten will radiate as a black body per the Stefan-Boltzmann law but you have not specified the area of tungsten not in contact with the alumina. A sphere will radiate a good deal less power (and take less power to maintain its temperature) than a large thin disk of the same mass.

Similarly, the alumina will radiate but in a constrained way. If you know the radiating area of the alumina and its thickness, you can use its thermal conductivity to determine the outer temperature when the outer surface is radiating. This will be complicated by the fact that the emissivity of alumina is not 1, and does not vary with temperature in a nice way. At any rate, the thermal conductivity of the alumina will depend on its dimensions and geometry, which you have not provided.

Finally, your sample/support combination will also lose heat to convection and conduction, both of which are, again, dependent on dimensions and geometry.