A calorimeter of heat capacity 83.72J K^(-1) contains 0.48 Kg of water at 35^(0)C. How much mass of ice at 0^(0)C should be added to decrease the temperature of the calorimeter to 20^(0)C.

A calorimeter of heat capacity 83.72J K^(-1) contains 0.48 Kg of water

1.	A calorimeter of heat capacity 83.72J K^(-1) contains 0.48 Kg of water at 35^(0)C. How much mass of ice at 0^(0)C should be added to decrease the temperature of the calorimeter to 20^(0)C.
Answer» Solution :Heat CAPACITY of the calorimeter `= 83.72 J K^(-1)` , Fall in TEMPERATURE of the calorimeter `= (35^(0)C - 20^(0)C) = 15^(0)C` Heat lost by the calorimeter `= 83.72 xx 15 = 1255.8J` , Mass of water = 0.45 Kg Specific heat of water `= 4186 J Kg^(-1)K^(-1) ,` Fall in temperature of water `= 35^(0)C -20^(0)C = 15^(0)C` Heat lost by water `= 0.48 xx 4186 xx 15 = 30139.2J ,` LET .m. be the mass of ice added, Heat gained by ice during melting `= mL = m xx 0.335 xx 10^(6)J` Heat gained by ice (water) during rise in temperature `= m xx 4186 xx (20 - 0) = m xx 83720J` `therefore` Heat gained by the ice = Heat lost by the calorimeter and water `m xx (0.335 xx 10^(6) + 83720)` = 1255.8 +30139.2 = 31395 = , m[335000+83720] = 31395 `implies m = (31395)/(418720)=0.07498 Kg.`