Question

What is the minimum mass of ice at 0.0 C that must be added to 1.00 kg of water to cool the water from 28.0 C to 12.0 C? (Heat of fusion = 333 J/g; specific heat capacities: ice = 2.06 J/gK, liquid water = 4.184 J/gK)

Answer 1

(4.184 J/g·°C) x (1000 g) x (28.0 - 12.0)°C = 66944 J required

(66944 J) / ((333 J/g) + ((4.184 J/g·°C) x (12.0 - 0)°C)) = 175 g ice

Answer 2

Answer:

175g of ice

Explanation:

The heat (Q) required to cool an amount of water is:

Q = C×ΔT×m

Where C is specific heat capacity of water (4,184J/g°C), ΔT is the change in temperature (28,0°C-12,0°C = 16,0°C) and m is mass of water, 1000g of water.

Replacing:

Q = 4,184J/g°C × 16,0°C × 1000g = 66944 J of energy.

This energy is obtained from ice that requires an energy for the fusion process and an additional energy increasing its temperature from 0.0°C to 12,0°C. That could be written as:

66944 J = 333 J/g×m + 4,184J/g°C×12,0°C×m

Where m is mass of ice.

66944 = 333m + 50.2m

66944 = 383.2m

m = 175g of ice

I hope it helps!