only part a The emissivity f of tungsten is only 0.3 for the wavelengths that it
ID: 1376968 • Letter: O
Question
only part a
The emissivity f of tungsten is only 0.3 for the wavelengths that it would emit if heated to 2000 K. (a) (2 points) If we have a tungsten sphere with a radius of 2 cm and a temperature of 2000 K, what would we have to raise the temperature to in order to get as much power emitted as we'd get from a perfect blackbody (e = 1) of the same size at a temperature of 2000 K? In other words, what temperature would we have to raise it to in order to compensate for the fact that it isn't producing as much energy as a perfect blackbody? (b) (2 points) If we decided to instead stick with a sphere at T = 2000 K, and compensate for the reduced output by increasing the radius, what should be the radius of the tungsten sphere that we use instead?Explanation / Answer
a.
rate of energy transer H = dQ/dt = Ae (sigma) T^4
where A = area = 4pir^2, e= emissitivity = (1 for black bdoy)
sigma = constant = 5.67*10^-8
T = temperatrue in kelvin
so here P2 = P1
A1 e1 T1^4 = A2 e2 T2^2
T2^4 = T1^4* e1/e2
T2^4 = 2000^4 * 1/2
T2^4 = 8 e 12
T2 = 1681.79 K -------------<<<<<<<<<<<<Answer
----------------------------------------------------------------------------------------------------------------
A2/A1 = T1^2/T2^2
A2 /A1 = 2000^4 /1681/79^4
A2 = 2 A1
r2^2 = 2* r1^2
r2 = 1.414 * r1
r2 = 1.414 times of r1
r2 = 1.414* 2
r2 = 2.828 cm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.