Maximum entropy and thermal equilibrium

[sapiental]

3) An object of mass m1, specific heat c, and temperature T1 is placed in contact with a second object of mass m2, specific heat c2 and temperature T2>T1. As a result, the temperature of the first object increases to T and the temperature of the second object decreases to T'.

a) Show that the entropy increase of the system is

deltaS = m1c1 ln(T/T1) + m2c2ln (T'/T2)

b) Show that energy conservation requires that

m1c1(T-T1) = m2c2 (T2-T')

c)Show that the entropy change in S, considered as a function of T, is a maximum if T' = T, which is just the condition of thermodynamic equilibrium.

 

[quasar987]

step by step with the variables...

 

[sapiental]

This is what I get

a) This is a calculation of an entropy change for an irreversible process. Since entropy is a state function, ΔS is independent of path. All we have to do is imagine a reversible path which will effect the same change and calculate the entropy change for the reversibly path.

ΔS_total = ΔS_cold + ΔS_hot


ΔS_total = m1c1 integral T1 to T (dT/T) + m2c2 integral T2 to T' (dT/T)

ΔS_total = m1c1 ln(T/T1) + m2c2 ln(T'/T2)

b) C = Q/dT

therefore m1(Q/dT)(T-T1)=m2(Q_2/dT)(T2-T')


Due to conservation of energy, these two equations must be equal. Because the Heat capacities differ, T and T' are different as well.
I know the specific heat is somehow supposed to cancel out the temp change but I am not sure what dt = for each side. somebody help

C) no clue what do do here, maybe integrate from T to T'?

 

[quasar987]

c) you must get the total differential of S considered as a function of T (i.e. an equation of the form dS = f(T)dT +...

And since a max in entropy means dS=0, what condition does this set on f(T)?

 

[sapiental]

total differential would be f(t)dt+f(t1)dt+f(t2)dt+f(t')dt?

sorry, I'm very bad at calculus.