[bluefairy_nebul]

A brass rod is in contact thermally with a heat reservoir at temperature 150 C at one end and a heat reservoir at 35 C at the other end.Compute the total change in the entropy arising from the process of conduction of 2100J of heat through the rod. Does the entropy of the rod change in the process?

[harum]

The entropy of the rod does not change if the temperature profile of the rod does not change during heat transfer. Change in the entropy of the entire system (the rod and both heat reservoirs) is:

dS = Q(1/Tc - 1/Th); Tc = 35 deg, Th = 150 deg (you will have toswitch to Kelvin from Celcius), Q is the amount of conducted heat.

Note that the total amount of heat in the system does not change, but the entropy increases.

[bluefairy_nebul]

hi harum, thanks for your help. I've solved it in the following way. I guess it's a bit different from yours one. Can you be kind enough to check it again?

let,

Th=150+273=423k
Tc=35+273=308k

let the absorbed heat from the source is Qin=2100J
let the heat released into the sink is Qout=?
entropy change due to absorption of heat by the brass rod from the source dS1=Qin/Th=2100/423 J/k=4.96J/k
we know,
Qin/Th=Qout/Tc
Qout=Qin/ThxTc=2100/423x308=1529J/k
change in entropy due to release of heat into the sink,dS2=Qout/Tc=(-1529/308)=-4.9

so, total change of entropy of the process is dS=dS1+dS2=4.96-4.96=0

so the entropy of the process does not change.

[bluefairy_nebul]

oh.. I got it... in this process heat is not wasted or being converted into work. So 2100J of heat is being conducted to the other end... so here Qin=Qout=Q. I was thinking about carnot's cycle... but it's not that... many many thanks.

[harum]

You are making a wrong assumption: "Qin/Th=Qout/Tc". This is not true! Qin = Qout! Because of energy conservation. The entropy does change, but the energy does not.

[bluefairy_nebul]

Hmm... I badly need to read the whole chapter again... I'm getting trouble with this second law of thermodynamics and entropy...