Question

Ice of mass 52.5 g at -10.7° C is added to 220 g of water at 15.4° C in a 110 g glass container of specific heat 0.200 cal/g-°C at an initial temperature of 25.1° C. Find the final temperature of the system.

Answer 1

mi = 52.5 g
Ti = -10.7 C
mw = 220 g
Tw = 15.4 C
mg = 110 g
Tg = 25.1 C
Cg = 0.2 cal/g C = 0.837 J/gm C
Cw = 4.186 J/gm C
Ci = 2.108 J/gm C
Li = 333.55 J/gm

Heat needed to completely melt the ice,
Q = mi*Ci*10.7 + mi*Li
Q = 52.5 * 2.108 * 10.7 + 52.5 * 333.55
Q = 18695.5 J

Heat given by water & glass ,
Q = mw*Cw* 15.4 + mg*Cg*25.1
Q = 220*4.186*15.4 + 110*0.837*25.1
Q = 16493.1 J

As we see the heat given by water and glass is not sufficient to completely melt the ice, therefore
Final Temperature of the system = 0^o C