Question

what temperature increase would be expected if the same 1.200 g sample of benzoic acid were combusted when the calorimeter contained 1.00 kg of water?
\\( q_{rxn} = -31.66 \space kj \\) \\( c_{cal(dry)} = 2.39 \space kj/^\circ c \\)
\\( -q_{rxn} = q_{cal} = \left c_{cal,dry}(\delta t) + c_{cal,h_2o}(\delta t) \
ight \\)
\\( q_{cal} = \left c_{cal,dry}(\delta t) + (m_{h_2o})(c_{h_2o})(\delta t) \
ight \\)
combine like terms using \\( \delta t \\).
\\( 31.66 \space kj = \left 2.39 \space kj/^\circ c + (1.00 \space kg \times 4.184 \space kj/kg^\circ c) \
ight (\delta t) \\)
\\( \delta t = ? \space ^\circ c \\)

Explanation:

Step1: Calculate the term inside the brackets

First, calculate \((1.00\space kg\times4.184\space kJ/kg^{\circ}C)\) which is \(4.184\space kJ/^{\circ}C\). Then add it to \(2.39\space kJ/^{\circ}C\):
\(2.39\space kJ/^{\circ}C + 4.184\space kJ/^{\circ}C = 6.574\space kJ/^{\circ}C\)

Step2: Solve for \(\Delta T\)

We have the equation \(31.66\space kJ = 6.574\space kJ/^{\circ}C\times\Delta T\). To find \(\Delta T\), divide both sides by \(6.574\space kJ/^{\circ}C\):
\(\Delta T=\frac{31.66\space kJ}{6.574\space kJ/^{\circ}C}\)
\(\Delta T\approx4.82\space ^{\circ}C\)

Answer:

\(4.82\)