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Homework Help: Plotting a Suitable Graph to find Emissivity of Tungsten Filament

  1. May 8, 2010 #1
    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.
  2. jcsd
  3. May 8, 2010 #2
    Method 1 is right since you have [tex] Q = A\theta [/tex] 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,
    [tex] lnQ = \ln(A\theta) = lnA + ln\theta [/tex] Which is not the same as y = mx + b.
  4. May 8, 2010 #3
    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

    [tex] lnQ = \ln(A\theta) = lnA + ln\theta [/tex] 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.
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