To compute the energy, Q, needed to melt a certain amount of ice, n, in moles, we have
[tex] Q = n \Delta H_{f} [/tex]
where the latent heat of fusion for ice is equal to 6.009 kJ/mol.
Now, since we have a 20.0 lb ice, we must first convert its mass to grams. Thus mass = (20.0)(1000)(0.4536) = 9072 g.
To find the number of moles present, we must recall that the molar mass of water (ice) is equal to 1.00794(2) + 15.9994 ≈ 18.01 g/mol. Hence, we have
[tex] ice_{moles} = (\frac{9072 g}{1})(\frac{1 mol}{18.01 g}) =503.72 moles [/tex]
Now, to compute for the molar heat of fusion, Q,
[tex] Q = (503.72)(6.009) = 3026.9 kJ [/tex]
Therefore, the amount of heat needed to melt the 20-lb bag of ice is equal to 3026.9 kJ.
Answer: 3026.9 kJ
