If 5.40 kcal of heat is added to 1.00 kg of ice at 0⁰C, how much liquid water at 0⁰C is produced?

q = mass ice x heat fusion.

Plug in q (in cal) and heat fusion (in cal) and calculate mass ice that melts.

To determine how much liquid water is produced when heat is added to ice at 0°C, we need to calculate the amount of heat required to convert the ice into liquid water.

The process of converting ice at 0°C to liquid water at the same temperature is known as the heat of fusion. The heat of fusion for ice is 334 J/g.

First, let's convert the given heat of 5.40 kcal to joules:
1 kcal = 4184 J
So, 5.40 kcal = 5.40 * 4184 J = 22615.2 J

Next, we need to find the mass of the ice. Given 1.00 kg of ice, we know that 1 kg = 1000 g.

Now we have all the information we need to calculate the amount of liquid water produced.

The heat required for the ice to melt (in joules) can be calculated using the following formula:

Heat of fusion = mass of ice * heat of fusion

Heat of fusion = (mass of ice in grams) * (334 J/g)

Substituting the given values:

22615.2 J = (mass of ice in grams) * 334 J/g

Now we can solve for the mass of ice:

mass of ice in grams = 22615.2 J / 334 J/g

mass of ice in grams = 67.79 g

Since the density of water is 1 g/mL, 67.79 g of ice is equivalent to 67.79 mL of water.

Therefore, when 5.40 kcal of heat is added to 1.00 kg of ice at 0°C, it will produce 67.79 mL of liquid water at 0°C.