How to find emissivity of tungsten from ln graph?

[dinospamoni]

Homework Statement

I'm doing a lab where I'm trying to find the emissivity of a tungsten filament. I measured voltage and current of a bulb with a tungsten filament and found the power. Then I plotted Power vs Temperature^4 and used the slope along with surface area and Boltzman's constant to find the emissivity. But I am expected to plot ln(P) vs ln(T) and find the emissivity and I can't quite figure out how to do that.

Homework Equations

I plotted it bas on this equation

ln(P) = ln(a) +n ln(T)

which was derived from the equation

P = aT^n

The Attempt at a Solution

I'm lost for this part. I've found everything else I'm supposed to except for this. i found n to be close enough to 4 and can find a from the intercept, but I'm not really sure what a stands for. Is that the emissivity?

 

[haruspex]

P = aT^n

Looks incomplete. At http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html it gives P = AeσT⁴, where σ is a constant, A is area and e is emissivity.

 

[RTW69]

It looks like a=Aeσ. Finding the Surface area of the tungsten element might be tricky. Perhaps you need to measure the resistance of the element, estimate length and determine surface area.

 

[dinospamoni]

Ah of course that makes tons of sense. I have the surface area so now I have everything I need. Thanks!

 

[Nickie]

I would recommend using the Stefan-Boltzmann law to find the emissivity of tungsten from an ln graph. This law states that the power (P) emitted by a blackbody is directly proportional to the temperature (T) raised to the fourth power, with the constant of proportionality being the Stefan-Boltzmann constant (σ). Therefore, the equation can be written as P = σT^4.

To find the emissivity (ε) of tungsten from an ln graph, you can rearrange the equation to ln(P) = ln(σ) + 4ln(T) + ln(ε). By plotting ln(P) vs ln(T), the slope of the line will be equal to 4, and the y-intercept will be ln(σ) + ln(ε). Therefore, you can find the emissivity by subtracting ln(σ) from the y-intercept.

It is important to note that the value of a in your equation, ln(P) = ln(a) + n ln(T), is not the emissivity. It is a constant that takes into account the surface area, distance from the source, and other factors. Therefore, it is not appropriate to use this value to find the emissivity.

In summary, to find the emissivity of tungsten from an ln graph, you can use the Stefan-Boltzmann law and plot ln(P) vs ln(T). The slope of the line will be 4, and the y-intercept can be used to find the emissivity by subtracting ln(σ) from it.