Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.230 kg and a temperature of 72.0 ∘C .

**Part A**

How much ice at a temperature of -13.8 ∘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 3.34×10^{5} J/kg .

Answer #1

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

The amount of heat lost by the 0.230 kg of water in the beaker to
the ice is easily calculated:

Q(beaker liquid) = (0.230)x(4190)x(72-20) = 50112.4 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-(-13.8))=(28980)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.

50112.4 = (28980)M + (3.34x10^5)M + (83800)M

50112.4 = (28980 + 3.34x10^5 + 83800)M

50112.4 = (446780)M

50112.4/446780 = M

M=0.1121 kg

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