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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 75
whereD is an integration constant that is determinded applying the bound-
ary conditions T = Ti at t = 0. Hence, D = ln(Ti − Tsur), and therefore
ln( T − Tsur
Ti − Tsur
) = −αt .
(b) �e above equation is rearranged to give an expression for T
T = Tsur + (Ti − Tsur) × e−α t
Given that S(T) − S(Ti) = C ln(T/Ti), it follows that
S(t) = S(T(t)) = S(Ti) + C ln(
T(t)
Ti
)
= S(Ti) + C ln(
Tsur + (Ti − Tsur) × e−α t
Ti
) .
P3B.8 (a) Because enthalpy is a state function, ∆H between the initial and �nal
states is the same irrespective of the path taken.�us the overall process
can be broken down into steps that are easier to evaluate. First consider
condensation of vapour. Given that all the vapour turns to liquid water,
the heat released is the opposite of that of vaporization
∆H1 = n(−∆vapH−○).
�e newly formed liquid water is initially at the boiling point. �e next
step is the water cooling from the boiling to �nal temperatures. �e en-
thalpy change associated with a temperature change is given by [2B.6b–
49], ∆H = Cp∆T .�us
∆H2 = Cp∆T = nCp ,m(H2O(l))(Tf − Tb)
�e �nal step is for the metal block to come to thermal equilibrium with
the water
∆H3 = Cp∆T = m
M
Cp ,m(Cu(s))(Tf − Ti)
where Ti is the initial temperature of the block.
Because the system is insulated there is no overall change in the enthalpy
of the system as a whole, ∆Hsys = ∆H1+∆H2+∆H3 = 0. Putting together
the expresssions of enthalpy change at each step gives
n(−∆vapH−○)+nCp ,m(H2O(l))(Tf −Tb)+
m
M
Cp ,m(Cu(s))(Tf −Ti) = 0
Hence
Tf×(nCp ,m(H2O(l)) + m
M
Cp ,m(Cu(s))) =
= n∆vapH−○ + nCp ,m(H2O(l)) × Tb +
m
M
Cp ,m(Cu(s)) × Ti