Cho rằng $\rm d_{H_2O}=1.00(g/mL)$

`=>`$\rm q$ $\rm =mC\Delta t$

$\rm =200 \times 4.2 \times (100-25)$

$\rm =63000(kJ)$

$\textbf{Có:}$

$\rm n_{C_2H_5OH}$

$\rm =\dfrac{q}{|\Delta H_{combustion}|} \times \dfrac{100}{H}$

$\rm =\dfrac{63000}{|-1365|} \times  \dfrac{100}{80}$

$\rm =\dfrac{750}{13}(mol)$

$\textbf{Suy ra:}$

$\rm m_{C_2H_5OH}$

$\rm =n \times M$

$\rm =\dfrac{750}{13} \times 46$

$\rm =\dfrac{34500}{13}(gam)$

$\textbf{Từ đó, ta tính được thể tích rượu tinh khiết:}$

$\rm V_{C_2H_5OH}$

$\rm = \dfrac{m_{C_2H_5OH}}{d_{C_2H_5OH}}$

$\rm =\dfrac{\dfrac{34500}{13}}{0.8}$

$\rm =\dfrac{43125}{13}(mL)$

$\textbf{Do đó, ta dễ dàng tính được thể tích rượu }96^o:$

$\rm V_{\text{Rượu }}96^o$

$\rm = \dfrac{\text{Độ rượu} \times V_{C_2H_5OH}}{100}$

$\rm =\dfrac{96 \times \dfrac{8635}{26}}{100}$

$\rm = \dfrac{41400}{13}(mL)$

$\rm \approx 3184(mL)$