The combustion enthalpy of lactose $C_{12}H_{22}O_{11}(s)$ (combustion of 1 mole) equals $-5652\;kJ$ $4\;g$ of this substance are burnt in a bomb calorimeter of specific heat capacity = $1630\frac{J}{K}$ and containing $1.35\; kg$ water. The initial temperature was $24.58^oC$. Calculate the final temperature!
$C_{12}H_{22}O_{11}(s)$ $+$ $12O_2(g)$ $\longrightarrow$ $12CO_2(g)$ $+$ $11H_2O(l)$ $\Delta H$ $=$ $-5652 \cdot \frac{4}{12\cdot 12+22\cdot 1+11\cdot 16}$ $\approx$ $ -66.1\;kJ$ $\Delta U$ $=$ $\Delta H $ $- $ $\Delta n RT$ $=$ $-66.1-(12-12)\cdot 8.3\cdot10^{-3}\cdot 298$ $=$ $-66.1\;kJ $
Law of calorimetry: $ Q $ $=$ $-4184\cdot m_{eau}\cdot (\theta_f$ $-$ $\theta_i)$ $ -$ $ C \cdot (\theta_f$ $-$ $\theta_i)$ $ -66100 $ $=$ $ -4184\cdot 1,35\cdot (\theta_f$ $-$ $24,58)$ $ -$ $ 1630 \cdot (\theta_f$ $-$ $24,58)$ Solution of this equation: $\theta_f$ $=$ $33,66^oC$