How much ice (in grams) be added to lower the temperature of 355 mL of water from 25 ∘C to 5 ∘C ? (Assume the following: the density of water is 1.0 g/mL, the ice is at 0 ∘C, the heat of fusion of ice is 6.02 kJ/mol, and the volume of water produced by the melting ice is negligible.)

Answer 1

First, let's determine how much heat energy (Q) will be required to lower the temperature of water from 25°C to 5°C.

The formula to calculate heat energy is:

Q = m * c * ΔT

Where m is the mass of water, c is the specific heat capacity of water, which is 4.18 J/g°C, and ΔT is the change in temperature.

First, we need to convert the volume of water (355 mL) to mass. Since the density of water is 1.0 g/mL, the mass of water is:

m = 355 mL * 1.0 g/mL = 355 g

Now, let's find the change in temperature:

ΔT = T_final - T_initial = 5°C - 25°C = -20°C

Now we can calculate the heat energy:

Q = m * c * ΔT = 355 g * 4.18 J/g°C * (-20°C) = -29666 J

Now, we need to determine how many grams of ice will be required to absorb this heat energy.

We know the heat of fusion of ice is 6.02 kJ/mol. First, let's convert this to J/g. The molar mass of water is 18.015 g/mol, so:

Heat of fusion = (6.02 kJ/mol) * (1000 J/kJ) / (18.015 g/mol) = 334.19 J/g

Now we can calculate the mass of ice:

m_ice = Q / Heat of fusion = -29666 J / 334.19 J/g ≈ 88.8 g

So, you would need to add approximately 88.8 grams of ice to lower the temperature of 355 mL of water from 25°C to 5°C.