Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.235 kg and a temperature of 66.5 ∘C .

How much ice at a temperature of -18.4 ∘C must be dropped into the water so that the final temperature of the system will be 20.0 ∘C ?

Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×10⁵ J/kg .

Answer 1

The unknown mass of ice is M.
Q(heat energy)=(mass)(specific heat)(delta T)

The amount of heat lost by the 0.235 kg of water in the beaker to the ice is easily calculated:
Q(beaker liquid) = (0.235)x(4190)x(66.5-20) = 45786.2 joules

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-(-18.4))=(38640)M
Q(fusion)=Mx3.34x10^5=(3.34x10^5)M
Q(liquid)=Mx4190x(20-0)=(83800)M

Heat lost by beaker liquid = Heat gained by ice when equilibrium is reached.
45786.2= (38640)M + (3.34x10^5)M + (83800)M

M=0.10035 kg