Page 129 - Physical chemistry understanding our chemical world

96 ENERGY AND THE FIRST LAW OF THERMODYNAMICS

where the ‘m’ means ‘molar’. The negative sign arises from the conventions above,
since heat is given out if the temperature goes up, as shown by T being positive.
It is wise first to calibrate the calorimeter by determining an
U (combustion) for ben- accurate value of C (overall) . This is achieved by burning a com-
zoic acid is −3.2231 MJ pound for which the change in internal energy during combustion
mol −1 at 298 K. is known, and then accurately warming the bomb and its reservoir
with an electrical heater. Benzoic acid (I) is the usual standard of
choice when calibrating a bomb calorimeter.

O

OH
(I)

The electrical energy passed is q, defined by

q = V × I × t (3.10)

where V is the voltage and I the current of the heater, which operates for a time of
t seconds.

SAQ 3.3 A voltage of 10 V produces a current of 1.2 A when applied
across a heater coil. The heater is operated for 2 min and 40 s. Show that
the energy produced by the heater is 1920 J.

Worked Example 3.5 A sample of glucose (10.58 g) is burnt com-
We can assume C p is
pletely in a bomb calorimeter. What is the change in internal energy
constant only if T is U if the temperature rises by 1.224 K? The same heater as that in
small. For this reason,
SAQ 3.3 is operated for 11 240 s to achieve a rise in temperature of
we immerse the ‘bomb’ 1.00 K.
in a large volume of
water. This explains
why we need to oper- Firstly, we calculate the energy evolved by the reaction. From
ate the heater for a Equation (3.10), the energy given out by the heater is q = 10 V ×
long time. 1.2A × 11 240 s = 134 880 J.
Secondly, we determine the value of C (overall) for the calorimeter,
saying from Equation (3.9)

energy released 134 880 J
C = =
change in temperature 1.00 K
so
−1
C = 134 880 J K .