Question 6.PS.9: Measuring Energy Change with a Bomb Calorimeter A 3.30-g sam......

– 2810 kJ

Strategy and Explanation When the glucose burns, it heats the calorimeter and the water. Calculate the heat transfer from the reaction to the calorimeter and the water from the temperature change and their heat capacities. (Look up the heat capacity of water in Table 6.1.) Use this result to calculate the heat transfer from the reaction. Then use a proportion to find the heat transfer for 1 mol glucose.

ΔT =  (34.1 –  22.4)°C = 11.7 °C

\begin{matrix} Energy  transferred  from \\ system  to  bomb \end{matrix} = – heat capacity  of  bomb × ΔT

= – \frac{847  J}{°C} × 11.7 °C = – 9910  J = – 9.910  kJ

\begin{matrix} Energy  transferred  from \\ system  to  water \end{matrix} = – ( c × m × ΔT)

= – (\frac{4.184  J}{g  °C} × 850. g × 11.7 °C) = – 41,610  J = – 41.61  kJ

 ΔE = q_V = – 9.910  kJ –  41.61  kJ =  – 51.52  kJ

This quantity of energy transfer corresponds to burning 3.30 g glucose. To scale to 1 mol glucose, first calculate how many moles of glucose were burned.

3.30  g  C_6H_{12}O_6 × \frac{1  mol}{180.16  g} = 1.832 × 10^{-2}  mol  C_6H_{12}O_6

Then set up this proportion.

\frac{- 51.52  kJ}{1.832 × 10^{-2}  mol} = \frac{ΔE}{1  mol}      ΔE = 1  mol × \frac{- 51.52  kJ}{1.832 × 10^{-2}  mol} = – 2.81 × 10^3  kJ

Reasonable Answer Check  The result is negative, which correctly reflects the fact that burning sugar is exothermic. A mole of glucose (180 g) is more than a third of a pound, and a third of a pound of sugar contains quite a bit of energy (many Calories), so it is reasonable that the magnitude of the answer is in the thousands of kilojoules.