Question

A piece of ice of mass mice = 80g, which is initially at -15℃ is dropped into an isolated container with negligible heat capacity. The container contains 150g of water at 80℃. Determine the final temperature of the system? 

(Ci = 2093 J/kg℃, Cw = 4186 J/kg℃, Lf = 333 KJ/kg)

Answer 1

Heat require to reach to temperature of ice at $0 ^oc$

$Q= m s\Delta T$

$=0.080\times 2093\times (15)$

$=2511.6J$

Heat require to melt the ice

$Q_1=mL$

$=0.080\times 333000 J$

$=26640J$

Hence, total energy require to melt the ice $=Q+Q_1$

$=26640+2511.6$

$=29151.6J$

Thermal energy of 150g of water at $80^\circ$

$=0.150\times 4200\times 80J$

$=50400J$

Hence we can say that ice will get melt completely.

$\Rightarrow 29151.6J=0.08\times 4200(t)+0.15\times 4200 \times (80-t)$

$\Rightarrow 336t+50400-630t=29151.6J$

$\Rightarrow 294t=21248.4J$

$\Rightarrow t=\frac{21248.4}{294}^\circ$

$\Rightarrow t= 72,27^\circ C$