The enthalpy change for converting 1.00 mol of ice at -25.0°C to water at 60.0°C is The specific heats of ice,water,and steam are 2.09 J/gK,4.18 J/gK,and 1.84 J/gK,respectively.For H₂O,ΔH_fus = 6.01 kJ/mol,and ΔH_vap =40.67 kJ/mol.

Question

Quizplus · Accepted Answer

To solve this problem, we need to consider the three stages of the process separately: melting the ice, heating the resulting water from -25.0°C to 0°C, and heating the water from 0°C to 60.0°C.

The first step involves melting the ice at -25.0°C, which requires an input of energy equal to the enthalpy of fusion:
q1 = nΔHfus = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ

The second step involves heating the resulting water from -25.0°C to 0°C, which requires an input of energy equal to the product of the mass, specific heat, and temperature change:
q2 = (1.00 mol)(18.02 g/mol)(2.09 J/gK)(0°C - (-25.0°C)) = 953.66 J

The third step involves heating the water from 0°C to 60.0°C, which requires an input of energy equal to the product of the mass, specific heat, and temperature change:
q3 = (1.00 mol)(18.02 g/mol)(4.18 J/gK)(60.0°C - 0°C) = 4.26 kJ

The total enthalpy change is the sum of these three contributions:
ΔH = q1 + q2 + q3 = 6.01 kJ + 953.66 J + 4.26 kJ = 11.46 kJ

Therefore, the correct answer is C.

The Enthalpy Change for Converting 1