The molar heat of fusion for water is 6.01 kJ/mol. How much energy must be added to a 75.0-g block of ice at 0°C to change it to 75.0 g of liquid water at 0°C?
1.44 kJ
10.18 kJ
12.48 kJ
25.04 kJ
q = mass x heat fusion = ?
q = 75 g x (1 mol H2O/18 g) x 6.02 = ? kJ.
12.48
To calculate the amount of energy required to change a certain mass of ice to liquid water at the same temperature, you can use the formula:
Energy = mass * molar heat of fusion
First, convert the mass of the ice from grams to moles. To do this, divide the mass by the molar mass of water (18.015 g/mol):
moles of ice = 75.0 g / 18.015 g/mol ≈ 4.16 mol
Next, multiply the moles of ice by the molar heat of fusion to find the energy required:
Energy = 4.16 mol * 6.01 kJ/mol ≈ 24.98 kJ
Rounding this value to two decimal places gives us:
Energy ≈ 25.04 kJ
Therefore, the correct answer is 25.04 kJ.