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)

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!

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.