Temperature & Thermal Dynamics?

Question

An insulated beaker with negligible mass contains a mass of 0.345 kg of water at a temperature of 60.4 deg. Celsius.

How much ice at a temperature of -11.2deg. Celsius must be dropped in the water so that the final temperature of the system will be 20.2deg. Celsius?

Take the specific heat for water to be 4190 J/(kg*K), the specific heat for ice to be 2100 J/(kg* K), and the heat of fusion for water to be 334 kJ/kg.

Answer 1

Let the mass of ice required = x kg.

Now, heat lost by 0.345 kg water when its temperature falls from 60.4 to 20.2 degrees
= 0.345 *1000*( 60.4 - 20.2 ) = 13869 cal.

Heat gained by x kg of ice when its temperature rises from - 11.2 to 0 degree
= x*1000*0.5*11.2 = 5600x cal.

Heat gained by x kg of ice at 0 degree to melt to water at 0 degree
= x*1000*80 = 80000x cal

Heat gained by x kg of water when its temperature rises from 0 to 20.2 degrees
= x*1000*1*20.2 = 20200x cal.

Total heat gained by ice
= ( 5600 + 80000 + 20200 ) x cal.
= 105800 x cal.

Heat gained by ice = heat lost by water

105800x = 13869

x = 0.131 kg.

Answer 2

0.345*4190(60.4-20.2) = 58,100 J

334x + 2100*11.2x + 4190*20.2x = 58,100
334x + 23500x + 84600x = 58100
108000x = 58100
x = 0.538 kg

all answers to 3 sig figs