Question

Assume a glass that is insulated that contains 0.45 kg of water at 20℃. You would like to bring its temperature down to 0 ℃ by adding cubes of ice. The enthalpy of fusion of the ice h_melting = 333.5 KJ/Kg (this is the variation of the enthalpy of the substance per unit mass or the amount of heat needed to change the state of the water from a solid to a liquid, at constant pressure). How much ice (in kg) should be added to the glass of water?

Answer 1

Answer 2

To bring the water from 20°C down to 0°C, you'll need to calculate the heat transfer required to cool the water to its freezing point (0°C) and then further to convert it into ice at the same temperature.

The heat transfer required to cool the water can be calculated using the specific heat capacity of water (c) and the formula:

Q1 = m * c * ΔT

Where: Q1 = Heat transfer required to cool the water to 0°C m = Mass of water (0.45 kg) c = Specific heat capacity of water (4.186 J/g°C) ΔT = Temperature change (20°C - 0°C = 20°C)

Q1 = 0.45 kg * 4.186 J/g°C * 20°C = 377.74 J

Now, you need to calculate the heat transfer required to convert the water at 0°C into ice at the same temperature. The enthalpy of fusion of ice (h_melting) is given as 333.5 KJ/kg, which is equivalent to 333,500 J/kg.

Q2 = m * h_melting

Where: Q2 = Heat transfer required to convert water at 0°C into ice at the same temperature m = Mass of ice (in kg) h_melting = Enthalpy of fusion of ice (333,500 J/kg)

You want to bring the water to 0°C, so you don't need to further cool it. The heat required is the same as the heat released when water freezes.

Q1 = Q2

377.74 J = m * 333,500 J/kg

Now, solve for m (mass of ice):

m = 377.74 J / 333,500 J/kg ≈ 0.00113 kg

So, you should add approximately 0.00113 kg (or 1.13 grams) of ice to the glass of water to bring it from 20°C to 0°C.