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fayled
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Homework Statement
A mass m of water cools down from 50degc to 10degc (the temperature of the surrounding environment). Calculate the entropy increase of the system (the water). The water has specific heat capacity c.
Homework Equations
dS=dQrev/T
S is entropy, Q is heat added to the system, T is temperature.
The Attempt at a Solution
So I can get the answer by doing
S=∫dQrev/T
dQrev=mcdT
S=mc∫dT/T from T=323K to T=283K
S=mcln(323/283)J/K.
However I don't exactly understand why this works - as above, entropy is defined in terms of the heat added to the system in a reversible process. However, this cooling of water is not a reversible process, so why am I allowed to simply set dQrev=mcdT if mcdT is not the heat added to the system in a reversible change?
Thankyou :)