Bharat Rakshak

jhujarji wrote:( Interesting and long article but not in simple English)

FWIW, this, by Mahopatra et el is simpler English and easier to read..

http://arxiv.org/pdf/1107.2378v3.pdf

This presentation does capture some of the ongoing activities in Astroparticle physics, with some nice view graphs.

http://www.aspera-eu.org/images/stories ... _india.pdf

^^^ Thanks.

Mort may find this story interesting ..
(See previous message here )

The physics of betrayal energize 'Copenhagen'

Just saw a problem in a contest. It reminded me of a few posts in Solar dhaga where people were discussing the color of solar panel, and its effect on temperature of the environment. (IMO, some posts there were some what silly using (fake) scientific words and drawing inaccurate conclusion ... but I digress...

Here is the problem:

There are three very large plates, out on L2 point (just means in outer space - unaffected by other things close by). All three plates are black (perfect black bodies) and highly conductive. All are placed parallel to each other, a small distance apart, but not touching. First plate is kept at a constant temperature of 200 degrees (centigrade), the third plate is kept at a temperature of 300 degrees (centigrade).
In steady state, what will be the temperature of the second (middle) plate.

Don't be shy..Give an answer .. (Just put the numeric value, but not the formula, method etc so that others can try the problem... again give numeric value, say up to 5 significant figures ..without reveling the formula)


(If you are designing a solar panel or any thing close to that, such questions are are easy to answer - I will be interested if people make a guess and still put down a number)

4°T

rsinghji - As I requested above, please, at least at this time do not give rational, logic, hint or a way of solving ... let others think too... you will spoil it for others if you give away your answer.

Can you please edit the above, and just put your answer (4 T) ... May be in a week we can open it for comments and explanation and I will give my understanding then.

Regards.

So, in spite of people talking about Reflectivity/Emissivity , Milankovich cycles, orbital changes and other mumbo jumbo's are not willing to take a stab at a simple problem I posted just above . I note that there is only one answer until now.

Alright....no cash prizes, all I am looking for is a certificate of participation.
257.035 C (530.185 K). [But if this problem were approached a different way,
the answer seems to be 300C. Not sure yet, which one].

SriKumar wrote:Alright....no cash prizes, all I am looking for is a certificate of participation.

So now we have three answer up til now. 4, 257, 300.
(All nook experts and solar experts have been quiet )

Meanwhile in a prestigious journal (PRL) .. Walking with coffee: Why does it spill?

263.8975964 degrees centigrade ?

Amber G. wrote:

SriKumar wrote:Alright....no cash prizes, all I am looking for is a certificate of participation.

So now we have three answer up til now. 4, 257, 300.
(All nook experts and solar experts have been quiet )

Meanwhile in a prestigious journal (PRL) .. Walking with coffee: Why does it spill?

Errrrr it is 4°T in pure lawhori parle et is absolut tapman sar ie -269° C

^^^ Sorry, so now we have the following values/answers (All in degrees centigrade)
(-269, 257,300, and 264)

Meanwhile wrt to spilling coffee:
http://physics.aps.org/synopsis-for/10. ... .85.046117

Just so I can throw in a random answer, how about 3 degrees kelvin, i.e. -270 C (which is the approx. temp of outer space). Should I throw in some reasoning in small font?

ArmenT wrote:Just so I can throw in a random answer, how about 3 degrees kelvin, i.e. -270 C (which is the approx. temp of outer space). Should I throw in some reasoning in small font?

Hint (AmberG ji, are we allowed to give hints?) If true, earth would be a cold place

^^^ I think you have a point in.

If true, earth would be a cold place..

I think we had this for a week already so now hints and solutions should be okay..

(If one can solve the plate problem, one can also deduce, at least, roughly, the (average) temperature of earth (though earth not being a perfect black-body and other complications may make the calculation a bit difficult).. or temperature of Chandrayaan - which I remember calculating in one of the tread, when it was having a overheating problem)

For heat to transmit, things need not be touching .(or have a medium like air between.

Here you go. while Boltzmann would be happy, his compatriot - Stefan - would be unhappy bcos his constant gets cancelled out

One more thing conservation of energy always applies

one more thing ,fot calculating make lord kelvin happier than celsius...poor kelvin does not have a degree..so stefan and boltzman used him in their kalaa far-mullah...since lord celsius has a degree , no kalaa farm-mullah was given to him

Amber G. wrote:^^^ Sorry, so now we have the following values/answers (All in degrees centigrade)
(-269, 257,300, and 264)

My answer. 530K/257 .

The disc in the middle is in thermal equilibrium with the discs at the ends. If it weren't the temperature would rise. It receives half the heat radiated by the discs in the ends and it in turn radiates out exactly the same energy.

ie E2 = 1/2 *( E1 + E3) E1 + E3.. And since E is proportional to T^4, plugging in values for E1 and E3 (in Kelvin of course as always) and doing the arithmetic, we get 257C /530K. 357.3C/630.3K

matrimc wrote:

ArmenT wrote:Just so I can throw in a random answer, how about 3 degrees kelvin, i.e. -270 C (which is the approx. temp of outer space). Should I throw in some reasoning in small font?

Hint (AmberG ji, are we allowed to give hints?) If true, earth would be a cold place

Right. It is not cold because heat is trapped by atmosphere saar. No need to have Phd in Physics for this. I can elaborate on this but for the moment let's wait.

gakakkad wrote:one more thing ,fot calculating make lord kelvin happier than celsius...poor kelvin does not have a degree..so stefan and boltzman used him in their kalaa far-mullah...since lord celsius has a degree , no kalaa farm-mullah was given to him

Mea culpa. You are right. Srikumar ji wins then (and Vina ji's 1st answer).

Rsingh ji, how about day side of the moon? No atmosphere but it is still very hot - max reaches 373 K.

Amber G. wrote:^^^ I think you have a point in.

If true, earth would be a cold place..

(If one can solve the plate problem, one can also deduce, at least, roughly, the (average) temperature of earth (though earth not being a perfect black-body and other complications may make the calculation a bit difficult).. or temperature of Chandrayaan - which I remember calculating in one of the tread, when it was having a overheating problem)

For heat to transmit, things need not be touching .(or have a medium like air between.

Amber ji

Actually since we can measure the average surface temperature of earth, we can actually deduce the average surface temperature of Sun, since almost all the energy earth receives is from Sun to a first approximation.

(OK misunderstood that post by AmberG in the other thread)...

Rsinghji - Other than conduction and convection heat can also travel by radiation. So one does not have to touch a sun, or be in its atmosphere. Away, (separated by vacuum), say on a space craft (or earth), the temperature depends on how far it is from the sun.

The 4T (actual value, I think is something like 2.7K) is due to residual energy left from big bang.

When you are outside on a sunny day.. the heat comes mainly from sun (even if you are surrounded by warm air... the air got its warmth by the sun's radiation).

vina wrote:

Amber G. wrote:^^^ Sorry, so now we have the following values/answers (All in degrees centigrade)
(-269, 257,300, and 264)

My answer. 530K/257 .

The disc in the middle is in thermal equilibrium with the discs at the ends. If it weren't the temperature would rise. It receives half the heat radiated by the discs in the ends and it in turn radiates out exactly the same energy.

ie E2 = 1/2 *( E1 + E3) E1 + E3.. And since E is proportional to T^4, plugging in values for E1 and E3 (in Kelvin of course as always) and doing the arithmetic, we get 257C /530K. 357.3C/630.3K

Slightly confused ... is it 357 or 257 ?

I see vina saar got the right answer.

matrimc wrote: Actually since we can measure the average surface temperature of earth, we can actually deduce the average surface temperature of Sun, since almost all the energy earth receives is from Sun to a first approximation.

To nit pic a small technical point .. please check out <this>
Actually about 44 TW ( Yes, it is still quite small compared to the total energy received from sun, so you are correct ) comes from radioactivity and fission reactor ..

For perspective, this is a few times the total power production in the world.

There is quite a lot of Uranium/Thorium in the earth's crust.. and not to forget, about 4,000 GW comes from decaying K40... (remember all those discussion about radioactivity in a banana! )

Bade wrote:I see vina saar got the right answer.

257 answer was given by Sri Kumar quite a while ago too..

Amber G. wrote:To nit pic a small technical point .. please check out <this>

ah ha, earth core generates more than 1/2. My "First approximation" is incorrect then. I take it back. Of course, one can take this into account to compute Sun's surface temp.

Bade saar, which of Vina saar's answers is correct? My calculation gives me 257.033 C

Assuming Black body behaviour for all 3 disks, it should total impinging radiation on middle disk which then leads to 357.

It is T1^4 + T2^4 = T^4

I do not see how a factor of 2 comes in (as per vina and then corrected later)

T1=200+273 K
T2 =300+273 K

^^^ Bade sir, are you sure?

matrimc wrote: ah ha, earth core generates more than 1/2. My "First approximation" is incorrect then. I take it back. Of course, one can take this into account to compute Sun's surface temp.

No you were correct the first time. (in the sense, that virtually all the energy we get comes from sun)...The "half" part is NOT with respect to the total energy but rather narrow part of residual heat vs fission vs radioactivity etc... 44 TW may be a larger amount but still it is of the order of 1/5,000 of what is being received from sunlight ...!

Interesting discussion!

For Bade Sir: (If T1=T2 then per your (Vina's) formula do we get T>T1.. right? Could it be that bread in a toaster be hotter than the filaments?

Amber_G, seems to caught me sleeping.

Stefan-Boltzmann law states the energy radiated per unit area per second. The area that radiates out for the center disk is twice, and hence the factor of 2.

To recap - (First posted by Sri Kumarji
The question I was told was one from IIT (Jee?) entrance ...

Left plate - total radiation = k(T1)^4 - half goes left , half goes right..
Right plate - total radiation = k(T2)^4 ...half goes right, half goes left ..
Assuming middle one is at (steady state) T
then kT^4 = (1/2) k(T1^4+T2^4)
(Basically it is not exactly average but rather a power mean (with 4th power)

This problem can be extended to one with a sphere of the same radius as the disks located at the center. It might remove the confusion on the area of the radiating surface, like a disk (thin) does.

A=pi * r^2 for the disks, and 4*pi*r^2 for the sphere, so another factor of 2.

Okay - some open ended questions for fun .. (All answers/insights are welcome

- Why U235 is fissile but not U238 (at least with slow neutrons)?
- Why (almost) all fissile materials are of odd mass (eg U235 .. Pu239 etc.. but even atomic numbers?
- Why you don't have any odd atomic number nuclear fuel ?
(You have Th(90), U(92), Pu(94) ... etc.. but not atomic numbers Pa(91),Np(93) etc..)

- Radioactivity from your own body (Primary from K40 and C14)
Does not cause cancer (because it is natural) / Is responsible for more than half of the cancers in humans / Help reduce natural cancer (aka radiation therapy) / something else ...
- The dangerous radioactivity in a fallout comes from:
neutrons / neutrinos / gamma rays / something other?

Amber G. wrote:To recap - (First posted by Sri Kumarji
The question I was told was one from IIT (Jee?) entrance ...

Left plate - total radiation = k(T1)^4 - half goes left , half goes right..
Right plate - total radiation = k(T2)^4 ...half goes right, half goes left ..
Assuming middle one is at (steady state) T
then kT^4 = (1/2) k(T1^4+T2^4)
(Basically it is not exactly average but rather a power mean (with 4th power)

Alright.....so the solution has been declared! I followed a thermal 'free-body' diagram route and set up "Heat in = Heat out" for the 2 surfaces of the middle plate. The middle plate has 2 surfaces. If we assume the temperatures of the first and second surface to be T1 and T2 respectively; we have (using Stefan-Boltzman eqn.):
Face1: Heat in = q = const*(200+273)^4 ; Heat out= const*(T1)^4
Face2: Heat in = const*(300+273)^4 ; heat out = const*T2^4
Setting 'total heat in' = 'total heat out'; gives: (473^4+573^4)= (T1^4 + T2^4).
Since plate was designated 'highly conductive', I assumed 'perfect conductor', and therefore T1 = T2 = T
So, 2*T^4 = (473^4+573^4); T = 530K (257 C)

Amber G. wrote: Slightly confused ... is it 357 or 257 ?

Sorry. It is 530/257 . The "formulation in English" was correct, and I had initially solved it correctly, but made a dog's meal of it when I corrected it.Dont know what I was thinking when I did that.

Amber G. wrote:Okay - some open ended questions for fun .. (All answers/insights are welcome

- Why U235 is fissile but not U238 (at least with slow neutrons)?
- Why (almost) all fissile materials are of odd mass (eg U235 .. Pu239 etc.. but even atomic numbers?

For the second, dislodging electrons/protons requires lot more energy than dislodgin the extra neutrons (odd mass) (?)

^ why would dislodging electrons require more energy than dislodging something from inside a nucleus ?