Re: Thermo 101 - 2nd Law Clarification

Makes a lot of sense! Cheers Greg.

Although I'm still trying ot get my head around the previous post with
the pot, it sounds cool. I guess thinking of your problem another way
since heat always flows from a region of high temp. to low temp. and DS
= Q/T then the entropy of the heat source giving off heat at a constant
temperature will be negative but more importantly its magnitude will
ALWAYS be lower than that of the heat recipient at constant temp. since
the T of the recipient MUST be lower than that of the source and T is
in the denominator of the DS equation. When the two are added together
you are left with a NET POSITIVE result which represents the entropy of
that closed system or taken further the universe. Therefore the
entropy of the universe always increases. Do I make any sense?

Greg Hansen wrote:
Maybe a sample problem will help?

Two blocks with initial temperatures T1 and T2, and heat capacities C1
and C2, are thermally connected and come to a final temperature Tf.
What is the change of entropy of block 1, block 2, and the two-block system?

dS = dQ/T, dQ = C dT

dS = C dT/T

Integrate both sides,

delta S1 = C1 ln(Tf/T1)

delta S2 = C2 ln(Tf/T2)

delta S = delta S1 + delta S2

All of thermodynamics is nominally for an isolated system. But
sometimes that means "the rest of the universe" is a reservoir. Draw a
boundary around your system of interest and track heat, work, entropy,
etc. coming in and going out. If heat goes out, it will carry entropy
away with it. The heat and entropy will go somewhere and the laws of
thermodynamics will be satisfied, but you don't always care where they
go. E.g. if you're analyzing a jet engine you don't really care what
happens to the exhaust once it's clear of the nozzle.

.

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