Question

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?

Answer 1

25.0 kJ

Step-by-step explanation:

This is correct on edge.

Answer 2

Answer is: 25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.