PASCO ET-8499 Energy Transfer Calorimeter - Manual (Page 18)

Energy Transfer Calorimeter Teacher’s Notes
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Part A:
Q
gained
= ((30.2)(4.186)+(22)(.90))(14) = 2.05 kJ
Q
Lost
= (31.9)(4.186)(15.2) = 2.03 kJ
Difference of 1 %, is acceptable, therefore it is OK to proceed to part B.
Part B:
∆
S= (31.9)(4.186) ln (297.1/283.1) = + 7.06 J/K
This is the value for the cup and cold water. The change in entropy is positive for an increase in
temperature. The molecules are moving faster, moving around more, and occupying more space.
This is an increase in "disorder".
∆
S= ((30.2)(4.186) + (22)(.9)) ln (297.1/312.3) = - 6.66 J/K
This is the value for the hot water. The change in entropy is negative for a decrease in temperature
and is a decrease in "disorder".
Total
∆
S= +7.06 + (-6.66) = +0.40 J/K
Note that the sign is positive. For any real, spontaneous process, the total change in entropy of the
entire system must be greater than zero.