Question

A volume of $120ml$ of drink (half alcohol + half water by mass) originally at a temperature of ${25}^{\circ }C$ is cooled by adding $20gm$ ice at $0^{\circ }C$. If all the ice melts, find the final temperature(in ${}^{o}C$) of the drink. (density of drink $=0.833gm/cc,$ specific heat of alcohol $=0.6cal/g{m}^{\circ }C$)

Answer 1

Mass of drink $m_{drink}=v_{drink}\times d_{drink}=120\times 0.833$20 gms of ice $0^{o}C$ added melts.
Heat absorbed by ice $Q_{ice}=20\times 80=1600calories$ .....(1)
$d_{drink}=\frac{v_{alcohol}{d}_{alcohol}+v_{water}{d}_{water}}{v_{alcohol}+v_{water}}$

$\therefore 0.833=\frac{60\times d_{alcohol}+60\times 1}{60+60}$
$\therefore d_{alcohol}=0.833\times 2-1=0.666$
$m_{alcohol}=60\times 0.666=40.2$
$m_{water}=60\times 1=60$
Heat release by the drink $Q_{drink}=(m_{alcohol}\times s_{alcohol}+m_{water}\times s_{water})\times (25-T)=(40.2\times 0.6+60\times 1)(25-T)=84.12\times (25-T)$ .....(2)
Equating (1) and (2) $1600=84.12\times (25-T)$
$\therefore T=25-\frac{1600}{84.2}=25-19=6^{o}C$