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

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**

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