Question

How much ice at -14.2 C must be placed in 0.150 kg of water at 18.4 C to cool the water to 0.0 C and melt all of the ice?

_____ kg

Answer 1

Suppose mass of ice is 'm'

then Using energy conservation:

Energy absorbed by ice = Energy released by water

Q1 = Q2

Q1 = Energy absorbed from -14.2 C to 0 C + Energy absorbed in phase change

Q1 = Mi*Ci*dT1 + Mi*Lf

Mi = mass of ice = m

Ci = Specific Heat capacity of ice = 2090 J/kg-C

dT1 = 0 - (-14.2) = 14.2

Lf = latent heat of fusion = 3.34*10^5 J/kg

So, Q1 = m*2090*14.2 + m*3.34*10^5

Now Energy released by water = Q2 = Mw*Cw*dT2

Mw = mass of water = 0.150 kg

Cw = 4186 J.kg-C

dT2 = 18.4 - 0 = 18.4

So,

Q2 = 0.150*4186*18.4 = 11553.36 J

Now Since

Q1 = Q2

m*2090*14.2 + m*3.34*10^5 = 11553.36

m = 11553.36/(2090*14.2 + 3.34*10^5)

m = 0.0318 kg = Mass of ice

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