Please answer all parts. Included is an equation sheet for reference. Some answe

Question

Please answer all parts. Included is an equation sheet for reference. Some answers you should get are: a) 4.13 cm^2 c) 0.202 eV d) 539 degrees C Please still answer these parts. I just included them for reference. 3) (20 points) A resistor inside a vacuum chanber dissipates an electrical power of 1 W. When the temperature of the vacuum chamber is 20 oC, the resistor reaches an equilibrium temperature of 200 °C. Assuming the resistor acts as a blackbody and there is no heat conduction: a) calculate the surface area of the resistor; b) calculate the wavelength at which the resistor radiates the most power; c) From the answer to part b, calculate the photon energy in eV. d) if the resistor is coated with a silvery paint with an emissivity of 0.1, calculate the new equilibrium temperature of the resistor

Explanation / Answer

part a:

as per blackbody radiation principle,

power =stefen boltzman constant*area*(blackbody temperature^4-sourrounding temperature^4)

==>1=5.67*10^(-8)*area*((273+200)^4-(273+20)^4)

==>area=1/(5.67*10^(-8)*((273+200)^4-(273+20)^4))=4.1319*10^(-4) m^2

=4.1319 cm^2

[note: temperature values are converted to kelvin]

part b:

as per wein displacement law,

wavelength=b/T

where b=wein's constant=2.8977*10^(-3)

so here wavelength=2.8977*0.001/(273+200)=6.1262 um

part c:

photon energy=h*c/wavelength

where h=planck's constant

c=speed of light

photon energy=6.626*10^(-34)*3*10^8/(6.1262*10^(-6))=3.2448*10^(-20) J =0.2028 eV

part d:

let new equilibrium temperature be T.

then 5.67*10^(-8)*4.1319*10^(-4)*0.1*(T^4-(273+20)^4)=1

==>T^4=(273+20)^4+(1/(5.67*10^(-8)*4.1319*10^(-4)*0.1))

==>T=811.76 K=811.76-273 degree celcius=538.76 degree celcius

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