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

Measuring Energy Change with a Bomb Calorimeter

A 3.30-g sample of the sugar glucose, $C_6H_{12}O_6(s)$, was placed in a bomb calorimeter, ignited, and burned to form carbon dioxide and water. The temperature of the water and the bomb changed from 22.4 °C to 34.1 °C. If the calorimeter contained 850. g water and had a heat capacity of 847 J/°C, what is ΔE for combustion of 1 mol glucose? (The heat capacity of the bomb is the energy transfer required to raise the bomb’s temperature by 1 °C.)

Step-by-Step
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– 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.

 Table 6.1    Specific eat Capacities for Some Elements, Compounds, and Common Solids Substance Specific Heat Capacity $( J g^{-1} °C^{-1})$ Elements Aluminum, $Al$ 0.902 Carbon (graphite), $C$ 0.720 Iron, $Fe$ 0.451 Copper, $Cu$ 0.385 Gold, $Au$ 0.128 Compounds Ammonia, $NH_3(\ell)$ 4.70 Water (liquid), $H_2O(\ell)$ 4.184 Ethanol, $C_2H_5OH(\ell)$ 2.46 Ethylene glycol (antifreeze), $HOCH_2CH_2OH(\ell)$ 2.42 Water (ice), $H_2O(s)$ 2.06 Carbon tetrachloride, $CCl_4(\ell)$ 0.861 A chlorofluoro- carbon $(CFC), CCl_2F_2(\ell)$ 0.598 Common solids Wood 1.76 Concrete 0.88 Glass 0.84 Granite 0.79

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