Question

This problem is about how strongly matter is coupled to radiation, the subject with which quantum mechanics began. For a simple model, consider a solid iron sphere 2.00 cm in radius. Assume its temperature is always uniform throughout its volume. (a) Find the mass of the sphere. (b) Assume the sphere is at 20.0°C and has emissivity 0.860. Find the power with which it radiates electromagnetic waves. (c) If it were alone in the Universe, at what rate would the sphere’s temperature be changing? (d) Assume Wien’s law describes the sphere. Find the wavelength \[\lambda_{max} \] of electromagnetic radiation it emits most strongly. Although it emits a spectrum of waves having all different wavelengths, assume its power output is carried by photons of wavelength \[\lambda_{max}\]. Find (e) the energy of one photon and (f) the number of photons it emits each second.

Answer 1

@JoannaBlackwelder

Answer 2

For a you just need the density of iron.

Answer 3

well now i feel dumb in regards to part (a)

Answer 4

No worries. :)

Answer 5

I don't remember how to do the rest of it though. :(

Answer 6

I can't even find relevant formulas...

Answer 7

for quantum physics? there's:
Wien's displacement law:
\[\lambda_{max} T=2.898 X10^{-3} m K\] (T is temperature, m is in meters, and K is in kelvins)

Stefan's Law:
\[P=\sigma A e T^4\] (sigma is the Stefan-Boltzmann constant, A is the surface area, e is emissivity, and T is temperature in Kelvins)

Then there's the Rayleigh-Jean's Law and Planck's wavelength distribution function.

Answer 8

Thanks! So, for b, don't you just need the surface area of a sphere with radius 2cm?

Answer 9

And use Stephan's Law? I've never taken quantum physics, but it seems reasonable...

Answer 10

*Stefan

Answer 11

I tried. so I applied Stefan's Law
\[P = (5.670 X10^{-8} W/ m^2 K^4)(5.027 X10^{-3} m^2)(0.860)(293K) = 7.182 X10^{-8} W\]

Answer 12

but the answer is apparently incorrect

Answer 13

the book's answer says 1.81W

Answer 14

Don't forget to do T^4

Answer 15

OH!! Well that was a silly mistake...
Thanks!

Answer 16

No worries. :)

Answer 17

I'm actually stuck on part c. There's no equation/formula I can think of to give me the rate of change of temperature over time.

Answer 18

Unless it's one from thermodynamics, which I took awhile back.

Answer 19

Intensity is given as
\[I = dP/d\lambda\]
do you think there's a way to modify that to get \[dT/dt\]?

Answer 20

Would dP/dt = 4σAeT^3 when A and e are constants?

Answer 21

I mean dP/dT

Answer 22

(c) -0.0153 degrees C/s=-0.919 degrees C/min.
If we can get that, we're perfect

Answer 23

now the question is, is it possible to go from Power per unit temperature to temperature per unit time

Answer 24

I think I'm out of my league... @LastDayWork any ideas?

Answer 25

he's not on
@douglaswinslowcooper

Answer 26

Ok. @wolfe8 how is your quantum physics?

Answer 27

LOL

Answer 28

Rate at which the sphere's temperature would be changing is its loss of heat
compared with its internal heat energy
U = (mass)(specific heat)(temperature)
dU/dt = (mass)(specific heat)(dT/dt)
dU/dt = power radiated, Stefan's law

I think.

Answer 29

dU/dt is an alternate version of Planck's function, there are two of them
one is dP/dt, the other is dU/dt

Answer 30

Need to look up Wien's Law and find wavelength of max emission
Then get frequency, f from speed of light c= f lambda
then get photon energy from E= h f

Answer 31

My earlier post assumed uniform T for sphere, which is probably ok approxn.

Answer 32

let me post the two versions of planck's function, maybe it will help you guys help me

Answer 33

\[\huge {dU\over d\lambda }= {8\pi Vhc\over {\lambda ^5({e^{hc\over \lambda k_B T}-1})}}\]
\[\huge {dP\over d\lambda }= {2\pi hc^2\over {\lambda ^5({e^{hc\over \lambda k_B T}-1})}}\]

Answer 34

hmm, my mistake, I said dU/dt was an alternate version of Planck's function. it's actually dU/d lambda

Answer 35

the specific heat of iron is given in Joules kg/ degrees C, how do I convert that into Kelvins?

Answer 36

You convert your temp to Celsius.

Answer 37

I'm not using temperature though since it's on the other side of the equation

Answer 38

The answer is given in deg Celsius/time.

Answer 39

\[U=mc\Delta T\]
\[{dU\over dt} ={mc \frac {dT} {dt}}\]
\[{dT\over dt}= {1\over mc}{dU\over dt}\]

Answer 40

It doesn't work...I just tried the substitution

Answer 41

@douglaswinslowcooper
ok, parts (d) and (e) worked out just fine.
I'm having issues with part (c) and how would I do part (f)?

Answer 42

I get -0.0172 degC/s using that formula. That is mighty close!

Answer 43

I got .014 so we're both close, but no cigar

Answer 44

not that I smoke but yeah...

Answer 45

That sounds like a difference in rounding, not a wrong method to me. And yeah, I don 't smoke either. :)

Answer 46

I'm not too sure. I mean, I don't think dU/dt would refer to Stefan's law. Dimensionally it works, dU/dt would be J/s which is Watts, but I'm just not sure.

Answer 47

@wolfe8 do you have any possible ideas?

Answer 48

Seems like the power divided by the energy of one photon would give you the number of photons emitted per second.

Answer 49

._.;; Not really. I had the same idea as above. Sorry.

Answer 50

@wolfe8 @douglaswinslowcooper
You guys were right, it worked out. I'm not sure what I did wrong. initially.

\[P={dE\over dt}\]
\[E=mc\Delta t => dE=mcdT\]
\[P=mc{dT\over dt}\]
so
\[{dT\over dt}={P\over mc}\]

Answer 51

Glad we could help. Thanks for your praise.

Answer 52

Glad you got it figured out. :)