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An insulating cup contains water at 25 ∘C. 30 grams of ice at 0 ∘C is...

An insulating cup contains water at 25 ∘C. 30 grams of ice at 0 ∘C is placed in the water. The system comes to equilibrium with a final temperature of 14∘C. How much water in grams was in the cup before the ice was added? (specific heat of ice is 2090 J/(kg LaTeX: ^\circ∘C), specific heat of water is 4186 J/(kg LaTeX: ^\circ∘C), latent heat of the ice to water transition is 3.33 x10^5 J/kg)

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Homework Answers

Answer #1

Suppose 'm' kg of water was in the cup then

Using energy conservation:

Energy released by water = energy absorbed by ice

Q1 = Q2 + Q3

Q1 = energy released by water from 25 C to 14 C = m*Cw*dT1

Q2 = energy absorbed by ice during phase change = mi*Lf

Q3 = energy absorbed by ice from 0 C to 14.0 C = mi*Cw*dT3

So,

m*Cw*dT1 = mi*Lf + mi*Cw*dT3

mi = mass of ice = 30 gm = 0.030 kg

Lf = 3.33*10^5 J/kg, Cw = 4186 J/kg.C

dT1 = 25 - 14 = 11 C, dT3 = 14.0 - 0 = 14.0 C

So,

m = [mi*Lf + mi*Cw*dT3]/(Cw*dT1)

m = [0.030*3.33*10^5 + 0.030*4186*14]/(4186*11.0)

m = 0.255 kg = 255 gm

mass of water required = 255 gm

Let me know if you've any query.

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