Plotting a Suitable Graph to find Emissivity of Tungsten Filament

[Invertedzero]

Plotting a suitable graph to find emissivity of tungsten
It is given that:

Q = pσ2πal(T^4 - t^4)

Where Q is the Energy Loss Rate, p is Emissivity and T,t are the wire and room temperature. Other symbols are constant.

I have data for Q and (T^4- t^4), which for simplification purposes i'll call Θ.


I need to find p, which is the best way of plotting data to do this?

1/ Plotting Q against Θ -> Linear Line of Best Fit -> Gradient dQ/dΘ = pσ2πal -> (p = 0.61)

2/ Plotting lnQ against lnΘ -> Linear Line of Best Fit -> Intercept = ln(pσ2πal) -> (p = 0.052)


I may be overlooking something, but i don't understand why they get different values and which is correct, or if there's a silly step in the algebra. Any help would be greatly appreciated, as I'm a bit stuck. If it could also be explained why one is more appropriate, and why the other isn't, if it isn't, that's also be helpful.

 

[zachzach]

Plotting a suitable graph to find emissivity of tungsten
It is given that:

Q = pσ2πal(T^4 - t^4)

Where Q is the Energy Loss Rate, p is Emissivity and T,t are the wire and room temperature. Other symbols are constant.

I have data for Q and (T^4- t^4), which for simplification purposes i'll call Θ.I need to find p, which is the best way of plotting data to do this?

1/ Plotting Q against Θ -> Linear Line of Best Fit -> Gradient dQ/dΘ = pσ2πal -> (p = 0.61)

2/ Plotting lnQ against lnΘ -> Linear Line of Best Fit -> Intercept = ln(pσ2πal) -> (p = 0.052)I may be overlooking something, but i don't understand why they get different values and which is correct, or if there's a silly step in the algebra. Any help would be greatly appreciated, as I'm a bit stuck. If it could also be explained why one is more appropriate, and why the other isn't, if it isn't, that's also be helpful.

Method 1 is right since you have $$Q = A\theta$$ and thus is in the form y = mx + b, b = 0 and A = the slope of the line. There is no reason to take the natural log of both sides, it just makes your job harder. Also,
$$lnQ = \ln(A\theta) = lnA + ln\theta$$ Which is not the same as y = mx + b.

 

[Invertedzero]

Method 1 is right since you have $$Q = A\theta$$ and thus is in the form y = mx + b, b = 0 and A = the slope of the line. There is no reason to take the natural log of both sides, it just makes your job harder. Also,
$$lnQ = \ln(A\theta) = lnA + ln\theta$$ Which is not the same as y = mx + b.

Hi, thanks for your response. So does 0.6 seem like a resonable value for the emissivity of a tungsten filament? Because online I've seen quite a range of values, but none as high as 0.6, and I'm pretty much certain the rest of the calculations are correct.

Also could you please explain to me why

$$lnQ = \ln(A\theta) = lnA + ln\theta$$ Which is not the same as y = mx + b.

because i would have though lnA would be essentially the lnm from y =mx + b form. Hence if i ignore the ln Q and ln theta, the intercept of this new graph should be the log of the gradient of the first? I tried it with some basic examples like y=2x and y=x^3 and it seemed to work, but with the experimental calculations I've made the two results of dQ/dtheta and the log method don't correlate, and I'm trying to figure out which is wrong.