Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.350 kg and a temperature of 70.9 ∘C . How much ice at a temperature of -23.4 ∘C must be dropped into the water so that the final temperature of the system will be 35.0 ∘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 0.350 kg of water in the beaker to the ice is easily calculated:

Q(beaker liquid) = (0.350)x(4190)x(70.9-35) = 52647.35joules

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-(-23.4))= (49140)M

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

Q(liquid)=Mx4190x(35-0)=(1.46x10^5)M

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

52647.35 = (49140)M+ (3.34x10^5)M + (1.46x10^5)M

M= 0.0994 kg

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