Question

When 0.963 g of glucose, C₆H₆, is burned in a bomb calorimeter, the temperature of the calorimeter increases by 8.39°C. The bomb has a heat capacity of 784 J/°C and is submerged in 925 mL of water. How much heat was produced by the combustion of the glucose sample?

Short answer

Heat produced by combustion of glucose sample = -39.0 kJ.

Step-by-step solution

Formula

The heat which is produced by the combustion reaction is primarily absorbed by the water and the bomb calorimeter. This is shown by

\({{\bf{q}}_{{\bf{rxn}}}}{\bf{ = - (}}{{\bf{q}}_{{\bf{water}}}}{\bf{ + }}{{\bf{q}}_{{\bf{bomb}}}}{\bf{)}}\)

Density of water is 1.0 g/mL; so 925 ml water = 925 g water.

Calculation of heat

The known values are substituted in the above equation. We have

\(\begin{array}{c}{{\rm{q}}_{{\rm{rxn}}}}{\rm{ = - (}}{{\rm{q}}_{{\rm{water}}}}{\rm{ + }}{{\rm{q}}_{{\rm{bomb}}}}{\rm{)}}\\{\rm{ = - }}\left[ {\left( {{\rm{4}}{\rm{.184 J/g ^\circ C}}} \right){\rm{ \times }}\left( {{\rm{925 g}}} \right){\rm{ \times }}\left( {{\rm{8}}{\rm{.39 ^\circ C}}} \right){\rm{ + 784 J/^\circ C \times }}\left( {{\rm{8}}{\rm{.39 ^\circ C}}} \right)} \right]\\{\rm{ = - }}\left( {{\rm{32471 + 6577}}} \right)\\{\rm{ = - 39048 J\; = \; - 39}}{\rm{.0 kJ}}\end{array}\)

So, -39.0 kJ heat is produced by the combustion of 0.963 g of glucose.