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

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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
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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...
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