Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.345 kg and a temperature of 76.5 ∘C .

How much ice at a temperature of -19.9 ∘C must be dropped into the water so that the final temperature of the system will be 27.0 ∘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 3.34×10^{5} J/kg .

Answer #1

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

The amount of heat lost by the 0.300 kg of water in the beaker to
the ice is easily calculated:

Q(beaker liquid) = (0.345)x(4190)x(76.5-27) = 71554.725
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-(-19.9))=(41790)M

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

Q(liquid)=Mx4190x(27-0)=(1.1313x10^5)M

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

71554.725 = (41790)M + (3.34x10^5)M + (1.1313x10^5)M

71554.725 = (41790 + 3.34x10^5 + 1.1313x10^5)M

71554.725 = (488920)M

71554.725/488920 = M

M=0.1463 kg

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