# Heating ice - disrupt the intermolecular forces • ∆ H...

This preview shows page 1. Sign up to view the full content.

Heating ice, water and water vapor In the region of the curve where we are not undergoing a phase transition, we are simply changing the temperature of one particular phase of water (either solid, liquid or gas) as a function of heat input The slope of the lines relates temperature to heat input The greater the slope, the greater the temperature change for a given unit of heat input The amount of heat needed to change the temperature of a substance is given by the specific heat or molar heat capacity o Specific heat of ice = 2.09 J/g °K o Specific heat of water = 4.18 J/g °K o Specific heat of water vapor = 1.84 J/g °K In the regions of the curve where we are undergoing a phase transition, the heat energy input is not raising the temperature of the sample, rather it is being used to
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: disrupt the intermolecular forces • ∆ H fus = 6.01 kJ/mol • ∆ H vap = 40.67 kJ/mol Calculate the enthalpy change for converting 2 moles of ice at -25°C to +125°C. • Converting to grams: (2 mol)*(18 g/mol) = 36 g • Heating ice from -25 to 0°C: (25°C)*(2.09 J/g °K)*(36 g) = 1.88 kJ • Fusion of ice to liquid water: (6.01 kJ/mol)*(2 mol) = 12.02 kJ • Heating of water from 0 to 100°C: (100°C)*(4.18 J/g °K)*(36 g) = 15.05 kJ • Vaporization of water to water vapor: (40.67 kJ/mol)*(2 mol) = 81.34 kJ • Heating of water vapor from 100 to 125°C: (1.84 J/g °K)*(25°C)*(36 g) = 1.66 kJ • Grand total: 1.88 + 12.02 + 15.05 + 81.34 + 1.66 = 111.95 kJ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online