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

A calorimeter of heat capacity 83.72JK^(-1) contains 0.48 kg of water

1.	A calorimeter of heat capacity 83.72JK^(-1) contains 0.48 kg of water at 35^(@)C. How much mass of ice at 0^(@)C should be added to decrease the temperature of the calorimeter to 20^(@)C.
Answer» Solution :Let .m. be the mass of ICE added, Heat gained by ice during melting = ML = `mxx0.335xx10^(6)J` Heat gainerd by molten ice (water) during rise in temperature = `mxx4186xx(20-0)=mxx83720J` `therefore` Heat gained by the ice = Heat LOST by the CALORIMETER and water `mxx[0.335xx10^(6)+83720]=1255.8+30139.2=31395J` `m[335000+83720]=31395` `m=31395/418720=0.07498kg`.