Final answer:
To convert 15.1 g of ice to water at 0.00°C, 5.04 kJ of heat is required. This is calculated by determining the number of moles in the ice and then multiplying by the heat of fusion for water.
Step-by-step explanation:
The quantity of heat required to convert 15.1 g of ice to water at 0.00°C can be calculated using the heat of fusion. The heat of fusion (ΔHfus) for water is 6.01 kJ/mol. First, we need to determine the number of moles in 15.1 g of ice:
To find moles, we divide the mass in grams by the molar mass of water (H2O), which is approximately 18.0 g/mol:
Moles of ice = (15.1 g)/(18.0 g/mol) = 0.839 mol
Then, we multiply the moles of ice by the heat of fusion to find the quantity of heat in kJ:
Heat required (Q) = Moles of ice × ΔHfus
Q = 0.839 mol × 6.01 kJ/mol
Q = 5.04 kJ
Therefore, 5.04 kJ of heat would be required to convert 15.1 g of ice to water at 0.00°C.