Heat

Q. Heat

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. 
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.