BELLWORK‐heatflowinpool

Whenyouenteraswimmingpool,thewatermayfeelquitecold.ABerawhile,though,

yourbody“getsusedtoit,”andthewaternolongerfeelsso

cold.Usetheconceptofheattoexplainwhatisgoingon.

The specific heat capacity (c) of anysubstance is the amount of heatrequired to raise the temperature of1gram of that substance by 1°C.

• Every substance has its ownspecific heat.

Water has a high specific heat.c= 4.184 J/(g·°C)

Water can absorb or release alot of heat before changingtemperature

The heat absorbed during a change intemperature is calculated using theequation

Heat(J)

SpecificheatcapacityJ/(g·°C)

Mass ofsample(g)

Changein temp(°C)

ΔT= Tfinal-Tinitial

PracIce

Howmuchenergyisrequiredtoraise50gofwaterfrom25°Cto40°C?

q=cmΔT

q=unknownc=4.18J/(g°C)forwater

m=50gΔT=15°C q = 4.18 x 50 x 15

= 3135 J of energy

∆Hfus = thequanItyofheatabsorbedduringmelIng

=thequanItyofheatreleasedduringfreezing.

For water

H2O(s) H2O(l) ∆Hfus = 6.01kJ/mol

H2O(l) H2O(s) ∆Hfus = -6.01kJ/mol

Howmanygramsoficeat0°Cwillmeltif2.25kJofheatareadded?

Known• iniIalandfinaltempsare0°C

• ΔHfus=6.01kJ/mol

• ΔH=2.25kJ

2.25 kJ x 1mol 6.01kJ

= 0.374 moles wouldbe melted

50gH2Ox1mole18g

Howmuchenergyisneededtomelt50gofice?

Known

• ΔHfus=6.01kJ/mol

• 50g

The heat offusion is PERMOLE soconvert gramsto moles

= 2.8 moles x 6.01 kJ 1mole

=16.8 kJ

ThequanItyofheatabsorbedbyaevaporaIngliquidisexactlythesameasthequanItyofheatreleasedwhenthevaporcondenses=∆Hvap

For water

H2O(l) H2O(g) ∆Hvap = 40.7kJ/mol

H2O(g) H2O(l) ∆Hvap= -40.7kJ/mol

Temperatureisconstantduringaphasechange

Phasechangeoccursataspecifictemperature(akathemelIngpointandboilingpoint)

Duringaphasechangeheatisusedtoovercomeintermoleculara`racIons.

Temperatureisconstantduringaphasechange

∆Hfusistheenergyrequiredtoovercomesomeintermoleculara`racIons.

∆Hvapistheenergyrequiredtoovercomeallintermoleculara`racIons.

TheΔHvapisalwayslargerthanΔHfus

IN YOUR NOTES--

Include a labeled drawing ofthe water curve and all ofthe notes in black and redon the slides to follow.

Tocalculateenergychanges(ΔH)forastatechangemolesxΔHfusORmolesxΔHvap

Energy

Tocalculateenergyrequiredtoincreaseordecreasetemperaturewithinasinglestate

q=cmΔT

Usethespecificheatofice

Usethespecificheatofliquidwater

Usethespecificheatofsteam

TocalculatetheenergychangeovermorethanonesecIonofthecurve…

Calculatetheenergyforeachstepandaddthemtogether.

‐50°C

125°C

CalculatetheΔHassociatedwithchanging100gofsteamat125°Ctoiceat‐50°C.

1.Determineifaphasechangeoccurswithinthetemperaturerange.• waterwillcondenseat100°C• waterwillfreezeat0°C

‐50°C

125°C

CalculatetheΔHassociatedwithchanging100gofsteamat125°Ctoiceat‐50°C.

2.Splitproblemintosteps.125°Csteam100°Csteamsteamwater100°Cwater0°Cwaterwaterice0°Cice‐50°Cice

3.Calculatetheenthalpychangeforeachstep.125°Csteam100°Csteam

q=1.89 J/(g°C)x100gx‐25°C=‐4725J

steamwater100gx1mol/18g=5.55molx40.7kJ/mol=‐226kJ

100°Cwater0°Cwaterq=4.18 J/(g°C)x100gx‐100°C=‐41,800J

waterice5.55molx6.01kJ/mol=‐33.4kJ

0°Cice‐50°Ciceq=2.10 J/(g°C)x100gx‐50°C=‐10,500J

4.Addthevaluesforeachsteptogetthetotalenergychange

=‐4.725kJ

=‐41.8kJ

=‐10.5kJ

‐4725J+ ‐226kJ+‐41,800J+‐33.4kJ+‐10,500J

Whoops!Theunitsaren’tthesame!!

4.Addthevaluesforeachsteptogetthetotalenergychange

‐4.725kJ+ ‐226kJ+‐41.8kJ+‐33.4kJ+‐10.5kJ

ΔHtotal=‐316.4kJ

ThetotalenthalpychangeandthechangesineachstepareallnegaIvevaluesbecausecoolingwaterisanexothermicprocess

EXOTHERMIC STATE CHANGES Heat comes out of substance(system) ∆H is negative Happens in a freezer Freezing(solidification) & condensation

ENDOTHERMIC STATE CHANGES Heat goes into substance ∆H is positive Happens on a stove Melting(fusion), vaporization, sublimation

Theinsulateddeviceusedtomeasureheatchangesinchemicalorphysicalprocessesiscalledacalorimeter.

qwater = - qrxnThe heat change forthe water in thecalorimeter is equalto the heat changeof the reaction butopposite in sign.

CalculaIonsforHonorsOnly