1Mullis Heating Curve at Constant Pressure Curve is flat during phase changes. Area A: Temperature...
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1 Mullis Mullis Heating Curve at Constant Heating Curve at Constant Pressure Pressure Curve is flat during Curve is flat during phase changes. phase changes. Area A: Temperature Area A: Temperature remains constant remains constant until all the until all the solid has become solid has become liquid because liquid because melting requires melting requires energy. energy. Once the energy is Once the energy is no longer required no longer required for phase change, for phase change, kinetic energy kinetic energy again increases. again increases. S am ple H eating C urve HeatAdded Tem perature Liquid is warm ed G as is warm ed B B oilin g Endothermic Exothermic Solid is warm ed A M eltin g
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Transcript of 1Mullis Heating Curve at Constant Pressure Curve is flat during phase changes. Area A: Temperature...
11 MullisMullis
Heating Curve at Constant Heating Curve at Constant PressurePressure
Curve is flat during phase Curve is flat during phase changes.changes.
Area A: Temperature Area A: Temperature remains constant until remains constant until all the solid has become all the solid has become liquid because melting liquid because melting requires energy.requires energy.
Once the energy is no Once the energy is no longer required for longer required for phase change, kinetic phase change, kinetic energy again increases.energy again increases.
Sample Heating Curve
Heat Added
Tem
per
atu
reLiquid is warmed
Gas is warmed
B
Boiling
Endothermic
Exothermic
Solid is warmed
A
Melting
22 MullisMullis
Heating Curve at Constant Pressure, Heating Curve at Constant Pressure, cont.cont.
Length of horizontal line Length of horizontal line A is proportional to A is proportional to the heat of fusion.the heat of fusion.
The higher the heat of The higher the heat of fusion, the longer the fusion, the longer the line.line.
Line B is longer than A Line B is longer than A because heat of because heat of vaporization is higher vaporization is higher than heat of fusion.than heat of fusion.
Ex: For water:Ex: For water:Heat of fusionHeat of fusion = 334 J/g= 334 J/gHeat of vaporizationHeat of vaporization
=2260J/g=2260J/g
Sample Heating Curve
Heat Added
Tem
pera
ture
Liquid is warmed
Gas is warmed
B
Boiling
Endothermic
Exothermic
Solid is warmed
A
Melting
ΔHfus = 6.02 kJ/molΔHvap = 40.7 kJ/mol
33 MullisMullis
Heating Curve at Constant Pressure, Heating Curve at Constant Pressure, cont.cont.
Gas and solid warming Gas and solid warming slopes are steeper than slopes are steeper than that for liquid warming.that for liquid warming.
The specific heat of the The specific heat of the liquid phase is usually liquid phase is usually greater than that of the greater than that of the solid or gas phase.solid or gas phase.
Ex: for water:Ex: for water:a.a. Solid = 2.09 J/g-Solid = 2.09 J/g-ºCºCb.b. Liquid = 4.18 Liquid = 4.18 J/g-J/g-ºCºCc.c. Gas = Gas = 2.03 J/g-2.03 J/g-ºCºC
Sample Heating Curve
Heat Added
Tem
per
atu
reLiquid is warmed
Gas is warmed
B
Boiling
Endothermic
Exothermic
Solid is warmed
A
Melting
44 MullisMullis
Sample Heat of Fusion ProblemSample Heat of Fusion Problem
q = mcq = mcΔΔTTIf the temperature change includes a phase change, you must split the If the temperature change includes a phase change, you must split the
problem into steps to separate the phase change.problem into steps to separate the phase change.Use the heat of fusion or heat of vaporization instead of specific heat for Use the heat of fusion or heat of vaporization instead of specific heat for
the phase change.the phase change.Ex. Calculate the heat that must be absorbed by 10.0 grams of ice (HEx. Calculate the heat that must be absorbed by 10.0 grams of ice (H22O) O)
at -9.00 at -9.00 ºC to convert it to water at 25.0 ºC.ºC to convert it to water at 25.0 ºC.
1. Solid warming: 1. Solid warming: -9.0 -9.0 ºC to 0 ºC: ºC to 0 ºC: (10.0 g) x (2.09 J/gºC) x (0.0 – (-(10.0 g) x (2.09 J/gºC) x (0.0 – (-9.00 9.00 ºC) ) ºC) ) = 188 J= 188 J2. Phase Change at 0 ºC: 2. Phase Change at 0 ºC: (10.0 g) x (334 J/g) (10.0 g) x (334 J/g) = 3340 J= 3340 J3. Liquid warming: 3. Liquid warming: 0 0 ºC to 25.0 ºC: ºC to 25.0 ºC: (10.0 g) x (4.18 J/gºC) x (25.0 - 0(10.0 g) x (4.18 J/gºC) x (25.0 - 0 ºC ) ºC ) = 1045 J= 1045 J
Total amount of heat absorbed Total amount of heat absorbed = 4573 J ~ 4570 J= 4573 J ~ 4570 J
55 MullisMullis
A Bigger Heat of Fusion ProblemA Bigger Heat of Fusion Problem
q = mcq = mcΔΔTTCalculate the heat needed to convert 25.0 grams of ice (HCalculate the heat needed to convert 25.0 grams of ice (H22O) from O) from -30.0 -30.0 ºC to steam at 150. ºC.ºC to steam at 150. ºC.
1. Solid warming: 1. Solid warming: -9.0 -9.0 ºC to 0 ºC: ºC to 0 ºC: (25.0 g) x (2.09 J/gºC) x (0.0 – (-30(25.0 g) x (2.09 J/gºC) x (0.0 – (-30.0 .0 ºC)) ºC)) = 1568 J= 1568 J2. Phase Change at 0 ºC: 2. Phase Change at 0 ºC: (25.0 g) x (334 J/g) (25.0 g) x (334 J/g) = 8350 J= 8350 J3. Liquid warming: 3. Liquid warming: 0 0 ºC to 100. ºC: ºC to 100. ºC: (25.0 g) x (4.18 J/gºC) x (100. - 0(25.0 g) x (4.18 J/gºC) x (100. - 0 ºC ) ºC ) = 10450 J= 10450 J4. Phase Change at 0 ºC: 4. Phase Change at 0 ºC: (25.0 g) x (2260 J/g) (25.0 g) x (2260 J/g) = 56500 J= 56500 J5. Vapor warming: 100.5. Vapor warming: 100. ºC to 100.0 ºC: ºC to 100.0 ºC: (25.0 g) x ( 2.03 J/gºC) x (150. – 100.(25.0 g) x ( 2.03 J/gºC) x (150. – 100. ºC ) ºC ) = 2537 J= 2537 J
Total amount of heat absorbed Total amount of heat absorbed = 79405 J ~ 79.4 kJ= 79405 J ~ 79.4 kJ
