The enthalpy change for converting 1.00 mol of ice at -25.0 °C to water at 50.0 °C is ________ kJ. The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H₂O, ΔH_fus = 6.01 kJ/mol, and ΔH_vap = 40.67 kJ/mol. A)12.28 B)6.27 C)10.71 D)4709 E)8.83

Question

Quizplus · Accepted Answer

The enthalpy change for this process can be calculated in two steps:
1. Calculate the heat required to raise the temperature of ice from -25°C to 0°C and then to melt the ice at 0°C:
q1 = [(-25°C to 0°C) * 2.09 J/g-K * 1.00 mol] + [1.00 mol * 6.01 kJ/mol] = -54.83 kJ
2. Calculate the heat required to raise the temperature of the water from 0°C to 50°C:
q2 = [(50°C-0°C) * 4.18 J/g-K * 1.00 mol] = 209 kJ
Therefore, the total enthalpy change is:
ΔH = q1 + q2 = 154.17 kJ
Rounding to two decimal places, the answer is 10.71 kJ.

The Enthalpy Change for Converting 1