A heat lamp emits infrared radiation whose rms electric field is Erms = 2800 N/C

Question

A heat lamp emits infrared radiation whose rms electric field is Erms = 2800 N/C. (a) What is the average intensity of the radiation? (b) The radiation is focused on a person's leg over a circular area of radius 4.4 cm. What is the average power delivered to the leg? (c) The portion of the leg being irradiated has a mass of 0.36 kg and a specific heat capacity of 3500 J/(kg·C°). How long does it take to raise its temperature by 2.2 C°? Assume that there is no other heat transfer into or out of the portion of the leg being heated.

Explanation / Answer

given that

Erms = 2800 N/C

r = 4.4 cm = 0.044 m

part(a)

we know that average intensity of radiation is

|S| = (1/2)*e0*c*E²   where

e0 is vacuum permittivity = 8.8542*10^(-12)   C^2*N^(-1)*m^(-2)

c speed of light, E amplitude of the electric field

and Eo = Erms/(sqrt(2))

So the average intensity of the radiation is:

|S| = (8.85*10^(-12)) * (3*10^8)* ((2800)^2)

|S| = 20815.2 W/m^2

|S| = 2.08*10^4 W/m^2

part (b)

power = intensity *area

P = |S| A = |S| * pi * r^2

P = (2.08*10^4 )*(3.14*(0.044)^2)

P = 0.012644 * 10^4 W

P = 126.44 W

part (c)

deltaQ/delta t = P

we know that

delta Q = m*C * delta T so

m*C*deltaT / delta t = P

where C is heat capacity which is given 3500 J/(kg*degree celcius).

m = 0.36 kg.

delta T = 2.2 degree C.

delta t = (m*C*delta T) / P

delta t = (0.36* 3500*2.2)/126.44

delta t = 21.92 s

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