Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.200 kg and a temperature of 76.0 ?C .

How much ice at a temperature of -20.4 ?C must be dropped into the water so that the final temperature of the system will be 33.0 ?C ?

Take the specific heat of liquid water to be 4190 J/kg?K , the specific heat of ice to be 2100 J/kg?K , and the heat of fusion for water to be 3.34×105 J/kg .

Answer 1

Mass of water M = 0.200 kg

Initial temperature of water t = 76.0o C

Mass of ice m = m

Initial temperature of ice t ' = -20.4o C

The final temperature of the system T =33.0o C

Specific Heat of liquid water C = 4190 J/kg/oC

Specific Heat of ice c =2100 J/kg/oC

Latent Heat of Fusion of water L = 3.35 x 10^5 J/kg

Heat lost by water = heat gain by ice

MC(t-T) = mc(0-t ') + mL + mC(T-0)

36034 = m[42840 + 334000 +138270 ]

From this mass of ice m = 0.06995 kg

= 69.95 g