Untitled - Kelly Walsh High School

104 CHEMISTRY FOR THE UTTERLY CONFUSED
Don’t Forget!
The presence of water implies that we may need some properties of water. In
this case, we will need the specific heat of water (4.18 J/gC).
The presence of water in any problem may require various unspecified properties
of water. The most common unspecified values are the specific heat of water, as
in this problem, and the density of water.
Extracting the information from the problem gives:
C6H12O6(s) 6 O2(g) l 6 CO2(g) 6 H2O(l) 1.5886 g 3.682C
3.562 kJ/C
1000. g H2O 4.18 J/gC
We will begin by finding the quantity of energy involved in the reaction. The
energy produced did two things: part of the energy warmed the calorimeter and
the remainder of the energy warmed the water. We need to determine these two
values separately and then combine them to determine the total energy change.
The amount of energy absorbed by the calorimeter is:
H calorimeter a
The amount of energy absorbed by the water is:
H water a
4.18 J
b (1000. g) (3.682C) 15390.76 J (unrounded)
gC
The total amount of heat energy will be the sum of these values (H calorimeter
H water). However, before we can sum these values, we must make sure the units
match. We can either convert the kilojoules to joules or the joules to kilojoules.
In this case we will convert the joules to kilojoules and then add the values.
H total H calorimeter H water 13.1153 kJ 15390.76 Ja 1kJ
10 3 J b
28.50606 kJ
3.562 kJ
b(3.682 C) 13.1153 kJ (unrounded)
C