Page 174 - Physical chemistry understanding our chemical world

THE TEMPERATURE DEPENDENCE OF ENTROPY 141

Aside

This argument is oversimplified because it is expressed in terms of jam:

(1) Jam, in comprising mainly water and sugar, will contain more moles per gram
than does the pastry, which contains fats and polysaccharides, such as starch in
the flour. Jam can, therefore, be considered to contain more energy per gram
from a molar point of view, without even considering its liquid state.
(2) The jam is more likely to stick to the skin than does the pastry (because it is
sticky liquid), thereby maximizing the possibility of heat transferring to the
skin; the pastry is flaky and/or dusty, and will exhibit a lower efficiency in
transferring energy.

Worked Example 4.6 Calculate the entropy change S caused by heating 1 mol of
sucrose from 360 K to 400 K, which is hot enough to badly burn the mouth. Take C p =
−1
425 J K −1 mol .

Because the value of C p has a constant value, we place it outside the integral, which
allows us to rewrite Equation (4.9), saying

T 2
S = C p ln (4.10)
T 1
We insert data into Equation (4.10) to obtain
400 K

−1
−1
S = 425 J K mol × ln
360 K
so
S = 425 J K −1 mol −1 × ln(1.11)
and

S = 44.8J K −1 mol −1

SAQ 4.3 We want to warm the ice in a freezer from a temperature of
−15 Cto0 C. Calculate the change in entropy caused by the warming
◦
◦
(assuming no melting occurs). Take C p for ice as 39 J K −1 mol −1 .[Hint:
remember to convert to K from C.]
◦

Aside

C p is not independent of temperature, but varies slightly. For this reason, the approach
here is only valid for relatively narrow temperature ranges of, say, 30 K. When determin-
ing S over wider temperature ranges, we can perform a calculation with Equation (4.9)