Question

the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament is a blackbody radiator? The filament of a particular electric lamp can be considered as a 90%blackbody radiator. calculate the energy per second radiated when its temperature is 2000k if its surface area is 10∧-6 m²

Answer 1

Answer:

(a) $\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

$P = \sigma AT^4$

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

$\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b)

Now, for 90% radiator blackbody at 2000 K:

$P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4$

P =  0.816 Watt