Question

# An insulated beaker with negligible mass contains liquid water with a mass of 0.290 kg and...

An insulated beaker with negligible mass contains liquid water with a mass of 0.290 kg and a temperature of 72.4 degrees C.
How much ice at a temperature of -11.2 degrees C must be dropped into the water so that the final temperature of the system will be at 33.0 degrees C?
Take the specific heat of liquid water to be at 4190 J/kg • K, the specific heat of ice to be 2100 J/kg • K, and the heat of fusion for water to be 3.34x10^5 J/kg

Solution-

Let M be the unknown mass of ice.

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

Amount of heat that is lost by the 0.290 kg of water in the beaker to the ice can be calculated as
Q(beaker liquid) = (0.290)x(4190)x(72.4-33) = 47874.94 joules

Now the amount of heat gained by the ice will be:
Therefore Q(ice) + Q(fusion) + Q(liquid) of the unknown mass M
Q(ice)=Mx2100x(0-(-11.2))=(23520)M
Q(fusion)=Mx3.34x10^5=(3.34x10^5)M
Q(liquid)=Mx4190x(33-0)=(138270)M

Now the heat lost by beaker liquid = Heat gained by ice when equilibrium is reached.
47874.94 = (23520)M + (3.34x10^5)M + (138270)M
42612.3 = (23520 + 3.34x10^5 + 138270)M
42612.3 = (495790)M
42612.3/495790 = M
M=0.086 kg

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