A quantity of ice at 0.0 °C was added to 33.6 g of water at 45.0 °C to give water at 0.0 °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•°C).

The ice at zero is to melt and remain at zero. So the heat fusion must be enough to lower the liquid water from 45.0 to zero C. How much heat must be removed from the water. That's

q = mass water x specific heat H2O x (Tfinal-Tinitial). That gives you q in J.

Now convert heat fusion from kJ/mol to J/g. That's approx 334 J/g but you need to look up the exact number.
Then how much ice is needed to furnish that q.
q for melting is mass ice x heat fusion. Solve for mass ice to produce that q.

To solve this problem, we can use the equation:

Q = mcΔT

Where:
Q = heat transferred
m = mass
c = specific heat
ΔT = change in temperature

First, we need to calculate the heat transferred when the water cools down to 0.0 °C.

Q1 = mcΔT = (33.6 g)(4.18 J/(g•°C))(45.0 °C - 0.0 °C)
Q1 = 6278.4 J

Now, we need to calculate the heat released when the ice melts at 0.0 °C.

Q2 = nΔHf
n = number of moles of water = mass/Molar mass of water
Molar mass of water (H2O) = 2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol

n = 33.6 g / 18.02 g/mol = 1.8621 mol

Q2 = (1.8621 mol)(6.01 kJ/mol)(1000 J/1 kJ)
Q2 = 11182.62 J

Since energy is conserved, the heat transferred (Q1) is equal to the heat released (Q2).

6278.4 J = 11182.62 J

However, we need to convert the heat released to negative since it was released.

-6278.4 J = -11182.62 J

Now, we can calculate the mass of ice added using the equation:

Q = mcΔT

-6278.4 J = m(2.09 J/(g•°C))(0.0 °C - (-5.09 °C))
Simplifying,
-6278.4 J = m(2.09 J/(g•°C))(5.09 °C)
-6278.4 J = m(10.6251 J/g)
-6278.4 J / 10.6251 J/g = m

m ≈ 590.3 g

Therefore, approximately 590.3 grams of ice were added.

To solve this problem, we need to calculate the amount of heat gained by the ice and the amount of heat lost by the water. Since the final temperature is 0.0 °C, we know that the heat gained by the ice is equal to the heat lost by the water.

First, let's calculate the amount of heat lost by the water. We can use the formula:

Q = m * c * ΔT

where Q is the heat lost, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature.

Given:
m = 33.6 g (mass of water)
c = 4.18 J/(g•°C) (specific heat of water)
ΔT = (0.0 °C - 45.0 °C) = -45.0 °C (change in temperature)

Q = 33.6 g * 4.18 J/(g•°C) * (-45.0 °C)
Q = -6776.64 J

Since the heat gained by the ice is equal to the heat lost by the water, the heat gained by the ice is also -6776.64 J.

Next, let's calculate the amount of heat gained by the ice using the heat of fusion of water.

Q = n * ΔH

where Q is the heat gained, n is the number of moles of ice, and ΔH is the heat of fusion of water.

We need to convert the mass of water to moles using the molar mass of water (H2O):
molar mass of water = 2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol

moles of water = mass of water / molar mass of water
moles of water = 33.6 g / 18.02 g/mol
moles of water = 1.865 mol

Since the heat gained by the ice is equal to the heat lost by the water, we can equate the two equations:

-6776.64 J = n * 6.01 kJ/mol (since 1 kJ = 1000 J)
-6776.64 J = n * 6.01 * 10^3 J/mol
n = -6776.64 J / (6.01 * 10^3 J/mol)
n ≈ -1.13 mol

The negative sign indicates that the ice is losing heat, which means it is being heated. Since we can't have a negative number of moles, we take the absolute value of n:

|n| = 1.13 mol

Therefore, approximately 1.13 moles of ice were added.