Equilibrium Temperature Of Water Mixed With Ice
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Equilibrium Temperature Of Water Mixed With Ice

40 grams of ice at 0oC is mixed with 100 grams of water at 60oC.
Determine the final temperature of the water after equilibrium has been established.
Heat of fusion of water (H2O) = 80 cal/gram
Heat capacity/Specific Heat of water = 1 cal/gram degrees C.

The strings:

S7P5A51 (change - physical change)

The math:
Pj Problem of Interest is of type change. The melting of the ice in the water changed the water temperature from 60oC to an equilibrium temperature. It is this final temperature due to temperature change that is of interest eventhough heat energy is operative.

Heat Of Vaporization: the quantity of heat required to vaporize 1 gram of a liquid substance at its boiling point at constant temperature.
Heat Of Fusion: the quantity of heat required to liquefy 1 gram of a solid substance at its melting point at constant temperature.
Specific Heat: number of calories required to raise the temperature of 1 gram of liquid by 1 degree.

Heat accounting:
(1) Ice needs heat to melt
(2) The cold water from the melted ice needs heat to warm up to the equilibrium temperature.
(3) The sum of the heat from (1) and (2) is total heat absorbed.
(4) The water at 60oC is the source of the heat needed for (3). Therefore it will loose that amount of heat.

Heat required to melt 40 grams of ice = 40 x 80 = 3200 cal.
Amount of heat absorbed by melted water before attaining final temperature, t (equilibrium temperature):
= 40 x specific heat of water x t = 40 x 1 x t = 40t.
So, total heat absorbed = 3200 + 40t.

Temperature change of water from 600C to toC = 60 - t
So, amount of heat lost by water at 60oC
= weight of water x specific heat of water x temperature change
= 100 x 1 x (60 - t) = 100(60 - t) = 6000 -100t

Heat absorbed = heat lost
So, 3200 + 40t = 6000 - 100t
So, 2800 = 140t
So, t = 20oC.
So, final temperature (equilibrium temperature) = 20oC.
Temperature of water dropped by 40oC.

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