Question

How much heat energy, in kilojoules, is required to convert 79.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer to three significant figures and include the appropriate units.

The constants for H2O are shown here:

Specific heat of ice: sice=2.09 J/(g⋅∘C)

Specific heat of liquid water: swater=4.18 J/(g⋅∘C)

Enthalpy of fusion (H2O(s)→H2O(l)): ΔHfus=334 J/g

Enthalpy of vaporization (H2O(l)→H2O(g)): ΔHvap=2250 J/g

Answer 1

Q = Q1 + Q2 + Q3

Q1 = heat energy required to convert ice at -18 degreeC into ice at 0 degreeC

Q2 = heat energy required to convert ice at 0 degreeC to water at 0 degreeC

Q3 = heat energy required to convert water at 0 degreeC to water at 25 degreeC

Now,

Q1 = m * s(ice) * dT

Q1 = 79 g * 2.09 J/g-degreeC * (0 -(-18)) degreeC

Q1 = 79 * 2.09 * 18 = 2978 J = 2.97 kJ

Q2 = m*deltaH(fus)

Q2 = 79 g * 334 J/g = 26368 J = 26.37 kJ

Q3 = m * s(water) * dT

Q3 = 79 g * 4.18 J/g-degreeC * (25 - 0) degreeC

Q3 = 79 * 4.18 * 25 = 8255.5 J = 8.26 kJ

So,

Heat energy required (Q) = Q1 + Q2 + Q3 = (2.97 + 26.37 + 8.26) kJ = 37.6 kJ