91Ó°ÊÓ

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.

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.