Solving a Tungsten Filament Temperature Problem

[mr_coffee]

Hello everyone I'm stuck on this problem it says:
A common flashlight bulb is rated at 0.30 A and 2.9 V (the values of the current and voltage under operating conditions). If the resistance of the bulb filament at room temperature (20°C) is 1.4 , what is the temperature of the filament when the bulb is on? The filament is made of tungsten.
I found the Resisivity, p of tungsten to be: 5.25E-8;
Temperature Coeffeicent of Resisivity, $$\delta$$ to be: 4.5E-3;
Here is the formula to find Temperature...
$$P-Po = Po\delta(T-To)$$;
The book doesn't explain the formula at all, all i got out of it was it doesn't matter if you use Kelvin or celsius for the Temps...
I know To is the Initial temp 20 C, I also found P = V/A = 2.9/.30 = 9.67;
So I think i have everything I need, but the answer is wrong...
$$P-Po = Po\delta(T-To)$$;
5.25E-8 - 9.67 = 9.67*4.5E-3(T-20);
183.7 = .0435T
T = 4272 C huge number, its wrong
I also tried switching the P's and Po's and got an even bigger number which was wrong, do u konw what I'm doing wrong? Thanks.

 

[HallsofIvy]

Hello everyone I'm stuck on this problem it says:
A common flashlight bulb is rated at 0.30 A and 2.9 V (the values of the current and voltage under operating conditions). If the resistance of the bulb filament at room temperature (20°C) is 1.4 , what is the temperature of the filament when the bulb is on? The filament is made of tungsten.
I found the Resisivity, p of tungsten to be: 5.25E-8;
Temperature Coeffeicent of Resisivity, $$\delta$$ to be: 4.5E-3;
Here is the formula to find Temperature...
$$P-Po = Po\delta(T-To)$$;
The book doesn't explain the formula at all, all i got out of it was it doesn't matter if you use Kelvin or celsius for the Temps...

Yes, that's clearly true- since you are subtracting two temperatures, the additive difference between K and C cancels.

I know To is the Initial temp 20 C, I also found P = V/A = 2.9/.30 = 9.67; So I think i have everything I need, but the answer is wrong...
$$P-Po = Po\delta(T-To)$$;
5.25E-8 - 9.67 = 9.67*4.5E-3(T-20);
183.7 = .0435T
T = 4272 C huge number, its wrong
I also tried switching the P's and Po's and got an even bigger number which was wrong, do u konw what I'm doing wrong? Thanks.

You say that you found that the resistivity of tungsten to be 5.25E-8

Are you aware that "resistivity" and "resistance" are not the same? I'm no expert but when I googled "resistivity" I got units like "ohm-m" and "ohm-cm".

 

[mr_coffee]

I know that, i looked it up in the book and it said Resistivity (P) of tungsten is 5.25E-8 ohm-m
And THe Reistance of the bulb filament at room temperature (20C) is 1.4 ohm;
I didn't use that 1.4 ohm in any of my calculations, maybe that's where I messed up? But it didn't seem like i needed it, they gave me the majority of everything and what I didn't find in the equation, i found in the book table of Reistivies of some mateirals at room temperature.

 

[rockerzzz10]

I know this was posted in 2005, but I'm doing the same problem now just little different values and I figured it out; hopefully this helps people in future.

Question: A common flashlight bulb is rated at 0.30 A and 2.9 V (values of current and voltage under operations.) If resistance of the bulb filament at room temperature (20 deg C) is 1.1 ohm. What is the temperature of the filament when the bulb is on? The filament is made of tungsten.

First, key thing: Based on the formula R = p (L/A) notice that R and p are known but L and A aren't but wait L/A in any temperature should be the same! So R and p only changes.

Second, Find the R when light bulb is on; R operating = V operating / Current Operating
R operating = 0.30 A / 2.9 V = 9.667 Ohm

Third, Find L/A value; Bulb @ 20 deg C has; R = 1.1 ohm and (TUNGSTEN) p = 5.25 * 10^-8 ohm-m
R = p (L/A)
(L/A) = R / p = 1.1 ohm / 5.25 * 10^-8 ohm-m = 2.0952 * 10^7 (1/m)

Fourth, p when operating which is p operating = R operating / (L/A)

p operating = 9.667 ohm / (2.0952 * 10^7 1/m) = 4.6138*10^-7 ohm-m


FINALLY TO find the temp when operating; Formula:

$$P-Po = Po\delta(T-To)$$

p = resistivity when operating
p0 = reference resistivity @ 20 deg C for tungsten (5.25 * 10^-8 ohm-m)

delta = resistivity constant for tungsten (4.5 * 10^3 K^-1)

T = Temperature needed to solve
T0 = reference temperature (20 deg C)

So T = (((p-P0)/p0)+ (20*delta))/(delta) <---- PLUG N CHUGGGGG...

I got T = 1750.7 deg celsius

Hope that helps. :)

<ROCKERZZZ10>

 

[Nadialy]

Thanks Rocker, Thats the right answer... I am also trying to work through this question. My question would be... how did you calculate the p=5.25 * 10^-8 ohm-m for tungsten and find the resistivity constant?

 

[NascentOxygen]

I'm just wondering aloud here, so ignore me. :)
Is L/A something to do with the dimensions of the metal filament? Don't the dimensions of the filament change quite a lot when it changes from cold to white hot? So would L/A for the filament change as the filament expands? The filament coil seems to expand perhaps by 33% to me. Is this ignored in the calculations here? Should it be taken into account?

 

[Sangwoo]

Thank you for your interesting discussion.
Could I know the equation`s name.
and Where can I read more detail about this equation?

And I would like to ask you a favor for any suggestion

If I want to design the tungsten heating circuit, what should I do?

I want to heat the tungsten to 800 degree Celsius.
by using the tungsten filament size of 0.1 mm diameter and 5 cm length

Could you please give me some advice.

Thanks in advance,