An insulated beaker with negligible mass contains a mass of

An insulated beaker with negligible mass contains a mass of 0.250 kg of water at a temperature of 83.8 C.

How many kilograms of ice at a temperature of 10.2 C must be dropped in the water to make the final temperature of the system 29.5 C? Take the specific heat for water to be 4190 J/(kgK) , the specific heat for ice to be 2100 J/(kgK) , and the heat of fusion for water to be 334 kJ/kg .

Solution

The unknown mass of ice is M.

Q(heat energy)=(mass)(specific heat)(delta T)

The amount of heat lost by the 0.250 kg of water in the beaker to the ice is easily calculated:

Q(beaker liquid) = (0.250)x(4190)x(83.8 - 29.5) = 56879.25 J

The amount of heat gained by the ice will be:

Q(ice) + Q(fusion) + Q(liquid) for the unknown mass M

Q(ice)=Mx2100x(0-(-10.2)) = = 21420 M

Q(fusion)=Mx3.34x10^5=(3.34x10^5)M

Q(liquid)=Mx4190x(29.5-0)= 123605M

Heat lost by beaker liquid = Heat gained by ice when equilibrium is reached.

56879.25 = 21420M + (3.34x10^5)M + 123605M

33421.5 = (29820 + 3.34x10^5 + 1.5503x10^5)M

33421.5 = (518850)M

M=0.1187 kg