Thermodynamics - know process, work and change in entropy. Final temperature?

AI Thread Summary
A 200-gram sample of dry air is heated isobarically, resulting in an entropy increase of 19.2 J/K and work done by expansion of 1.61x10^3 J. The discussion revolves around calculating the final temperature using the relationship between heat transfer (dq), internal energy change (du), and work done (dw). Participants express confusion about how to derive the final temperature without an initial temperature provided. The solution involves using the equation dφ = dq/T and integrating to find the temperature difference. Ultimately, the final temperature can be determined through the work done and the change in entropy.
HHveluj
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So, the problem is:
A 200-gram sample of dry air is heated isobarically. Its entropy increases by 19.2 J/K and the work done by expansion is 1.61x103 J. Solve for the final temperature of the air.

Okay, since p=const => dq=CpdT thus instead of dq=du+dw we can write
dq=dq(Cv/Cp)+dw
can find dq=460.46
So i know dq,du,dw,d\phi, but how do I get final temperature? All I can find is dT but it's useless sine I don't know initial temperature.
And where do I put mass and how to use the d\phi=dq/T formula? Looks like it will be needed here...

Thanks a lot in advance!
 
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nevermind. solved it.
 
Aw. I have this problem too and I'm stuck on it. :(
 
Get Tfinal - Tinitial through work, then you can solve integral dphi=dq/t and get the answer for Tfinal.
 
I have this problem too and i still do not understand how you can solve for T final if there is not an initial temp given.
 

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