A Black Body Is Kept At A Temperature Of 2500K. If Volume Of Body Is Reduced Reversibly And Adiabatically To 1/100 Of Its Initial Volume What Will Be The Final Temperature?

[Sainani]

A black body is kept at a temperature of 2500K. If volume of body is reduced reversibly and adiabatically to 1/100 of its initial volume what will be the final temperature?

[Old_Geezer]

How about you have a bash at your homework first, then present it and ask where you went wrong, if you did ?

[rowanwitch]

We may well have members who will try answering this for you but better to try to find a solution first then asking for help checking your methodology.

[Barsel]

I'm wondering why it's specifically a Black body?

[Old_Geezer]

That's racist that is !

[AuntPollyGrey]

I'm sure some people actually think this is an Answer Bank. How quaint. ;-)

[dannyk13]

//I'm sure some people actually think this is an Answer Bank. How quaint//
That is hardly an answer. is it?

[naomi24]

This is a question and answer site and no one should be telling anyone not to ask questions - regardless of the reason. If a poster wants help with homework, so be it. Don't answer if you object.

[JimF]

25K.

[dannyk13]

Well said Naomi.

[bhg481]

Barsel - defining it as a "Black Body" defines how it radiates heat or electromagnetic radiation.

[Old_Geezer]

It is in the OP's interest to learn how to answer homework questions. It is a much better answer to encourage them to try first.

[Peter Pedant]

Hi Jim
your answer 25K must be wrong because ....when you compress something it must get hotter

think bicycle pump

I punt something like - 2500 to the power of y/(y-1)
( or its reciprocal) - which ever is gtr than 2500
er I explain my workings and historial background above

[JimF]

Yes, but you're ignoring thermal equilibrium and entropy, which will assume a constant value as the temperature approaches absolute zero; the zeroth law of thermodynamics doesn't apply to a single system.

[Peter Pedant]

yeah I am allowed to ignore entropy because it is adiabatic

[Now, since the process is adiabatic, so the heat transfer is zero and so the entropy change is zero through heat transfer. And if the process is also reversible, then there is no entropy generated inside the system and the entropy change due to entropy generation is also zero.]

and 2500K is nowhere near zero ( 25k is tho)

[jim360]

In a reversible adiabatic change the guiding equation is T_1 V_1^(ga-1) = T_2 V_2^(ga-1), where ga = 5/3.

So I'd get that

T_2 = T_1 (V_1/V_2)^(2/3) =T_1 (100)^(2/3)

Which is closer to 54000 K. I think this makes sense because although the volume decreases the pressure will grow massively as a result of the adiabatic change.

My thermodynamics is a little rusty so I could be wrong of course, but I've checked that it's consistent with the ideal gas law (pressure increases to 2154 = 100^(5/3) times the initial value).

[jim360]

As regards the meta-debate of whether or not we should answer or encourage to try first: how do we know that the OP hasn't already tried a few times?

[Old_Geezer]

We know they've not shown any attempts nor asked where they're going wrong, they've simply asked for the answer; so it is the default position/view to take to minimise such assumptions and maximise help.

[Peter Pedant]

thx Jim
yeah that other equation
gives the answer