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Heating Curve at Constant Pressure

Heating Curve at Constant Pressure. Curve is flat during phase changes. Area A: Temperature remains constant until all the solid has become liquid because melting requires energy. Once the energy is no longer required for phase change, kinetic energy again increases.

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Heating Curve at Constant Pressure

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  1. Heating Curve at Constant Pressure Curve is flat during phase changes. Area A: Temperature remains constant until all the solid has become liquid because melting requires energy. Once the energy is no longer required for phase change, kinetic energy again increases. Mullis

  2. Heating Curve at Constant Pressure, cont. Length of horizontal line A is proportional to the heat of fusion. The higher the heat of fusion, the longer the line. Line B is longer than A because heat of vaporization is higher than heat of fusion. Ex: For water: Heat of fusion= 334 J/g Heat of vaporization=2260J/g ΔHfus = 6.02 kJ/mol ΔHvap = 40.7 kJ/mol Mullis

  3. Heating Curve at Constant Pressure, cont. Gas and solid warming slopes are steeper than that for liquid warming. The specific heat of the liquid phase is usually greater than that of the solid or gas phase. Ex: for water: • Solid = 2.09 J/g-ºC • Liquid = 4.18 J/g-ºC • Gas = 2.03 J/g-ºC Mullis

  4. Sample Heat of Fusion Problemq = mcΔT If the temperature change includes a phase change, you must split the problem into steps to separate the phase change. Use the heat of fusion or heat of vaporization instead of specific heat for the phase change. Ex. Calculate the heat that must be absorbed by 10.0 grams of ice (H2O) at -9.00 ºC to convert it to water at 25.0 ºC. 1. Solid warming: -9.0 ºC to 0 ºC: (10.0 g) x (2.09 J/gºC) x (0.0 – (-9.00 ºC) ) = 188 J 2. Phase Change at 0 ºC: (10.0 g) x (334 J/g) = 3340 J 3. Liquid warming: 0 ºC to 25.0 ºC: (10.0 g) x (4.18 J/gºC) x (25.0 - 0ºC ) = 1045 J Total amount of heat absorbed = 4573 J ~ 4570 J Mullis

  5. A Bigger Heat of Fusion Problemq = mcΔT Calculate the heat needed to convert 25.0 grams of ice (H2O) from -30.0 ºC to steam at 150. ºC. 1. Solid warming: -9.0 ºC to 0 ºC: (25.0 g) x (2.09 J/gºC) x (0.0 – (-30.0 ºC)) = 1568 J 2. Phase Change at 0 ºC: (25.0 g) x (334 J/g) = 8350 J 3. Liquid warming: 0 ºC to 100. ºC: (25.0 g) x (4.18 J/gºC) x (100. - 0ºC ) = 10450 J 4. Phase Change at 0 ºC: (25.0 g) x (2260 J/g) = 56500 J 5. Vapor warming: 100.ºC to 100.0 ºC: (25.0 g) x ( 2.03 J/gºC) x (150. – 100.ºC ) = 2537 J Total amount of heat absorbed = 79405 J ~ 79.4 kJ Mullis

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