A quantity of ice at 0.0 degrees C was added to 33.6 of water at 21.0 degree C to give water at 0.0 degrees C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g * degrees C)

To determine the amount of ice that was added, we can use the principle of conservation of energy. The energy lost by the water as it cools down to 0.0 degrees Celsius should be equal to the energy gained by the ice to melt it.

First, let's calculate the energy lost by the water using the equation:

Q = mcΔT

Where:
Q = heat lost or gained
m = mass of the water
c = specific heat capacity of water
ΔT = change in temperature

Given:
m = 33.6 g
c = 4.18 J/(g * degrees C)
ΔT = 21.0 degrees C

Q = (33.6 g) * (4.18 J/(g * degrees C)) * (21.0 degrees C)
Q = 2973.2 J

Next, let's calculate the heat gained by the ice using the equation:

Q = nΔHf

Where:
Q = heat lost or gained
n = number of moles of ice
ΔHf = heat of fusion of water (per mole)

Given:
ΔHf = 6.01 kJ/mol (note that this value is in kilojoules, so we need to convert it to joules)
1 kJ = 1000 J
ΔHf = 6.01 kJ/mol * 1000 J/1 kJ
ΔHf = 6010 J/mol

Now, we need to convert the energy lost by the water (in joules) to the heat gained by the ice (in joules/mol). To do that, we need to divide the energy lost by the water by the heat of fusion of water:

n = Q / ΔHf
n = 2973.2 J / 6010 J/mol

n ≈ 0.494 mol

Finally, since each mole of ice corresponds to 18.02 g (molar mass of water), we can calculate the mass of the ice added:

mass of ice = n * molar mass of water
mass of ice = 0.494 mol * 18.02 g/mol
mass of ice ≈ 8.90 g

Therefore, approximately 8.90 grams of ice was added to the water.

Do you know that q = mass x specific heat x delta T?

And q = mass x heat fusion?
What do you not understand about how to use these equations?