The enthalpy change for converting 1.00 mol of ice at -50.0 °C to water at 70.0 °C is ________ kJ. The specific heats of ice, water, and steam are 2.09 J g^-1 °C^-1, 4.18 J g^-1 °C^-1, and 1.84 J g^-1 °C^-1, respectively. For H₂O, Δ_fusH = 6.01 kJ mol^-1, and Δ_vapH = 40.67 kJ mol^-1. A)12.28 B)6.41 C)13.16 D)7154 E)9.40

Question

Quizplus · Accepted Answer

To find the total enthalpy change, calculate the energy required for each step: heating ice from -50°C to 0°C, melting ice, and heating water from 0°C to 70°C. Sum these values: (1 mol x 18.015 g/mol x 2.09 J/g°C x 50°C) + (6.01 kJ/mol) + (1 mol x 18.015 g/mol x 4.18 J/g°C x 70°C) = 1.88 kJ + 6.01 kJ + 5.27 kJ = 13.16 kJ.

The Enthalpy Change for Converting 1