HeatHeatHeatHeat

Solid Liquid Gas

Heat = AMOUNT of internal energy

Temperature = a MEASURE of the average molecular kinetic energy

10 g PbT = 40 °C

100 g PbT = 40 °C

Both blocks are at the same temperature.

Do they both contain the same amount of heat?

Which substance requires more heat to increase the temperature by 5 °C?

Specific heat capacity (Cp): amount of heat(q)

required to raise 1 g of substance by 1 °C

Cp(Pb) = 0.126 J/g°CCp(paraffin) = 2.1 J/g°C

Pb100 g

How much heat is required by the 100 g candle to increase the temperature by 5 °C?

Cp(paraffin) = 2.1 J/g°C q = Cp(mass)(T)

q = (2.1 J/g°C)(100 g)(5 °C)

q = 1050 J

q = Cp(mass)(T)

1050 J = (4.184 J/g°C)(100g)(T)

T = 2.5 °C

If the same amount of heat was used to heat 100 g of water [Cp(liquid water) = 4.184 J/g°C], what would be the T of the water?

For the same amount of heat and mass, T decreasesdecreases as the

specific heat of the substance increasesincreases

For the same amount of heat and mass, T decreasesdecreases as the

specific heat of the substance increasesincreases

Baltimore Shot Towerhttp://www.baltimore.to/ShotTower/

200 ft

100 kg Pb

If the temperature of the lead is 327°C before it hits the water, what is the final temperature of the lead after hitting the water?

If the temperature of the lead is 327°C before it hits the water, what is the final temperature of the lead after hitting the water?

Cp(Pb) = 0.13 J/g°C

Cp (H2O) = 4.18 J/g°C

q = mCpT T = Tf - Ti

Tf = 93 °C

-qPb = qH2O

-(1x105 g)(0.13 J/g°C)(Tf – 327°C) =

(1x104 g)(4.18 J/g°C)(Tf – 20°C)

10 kg H2O

Ti = 20 °C

Melting one 14-gram Al soda can requires 5.55 kJ of energy. What is its molar heat of fusion?

Melting one 14-gram Al soda can requires 5.55 kJ of energy. What is its molar heat of fusion?

105,000 cans are recycled in the US every minute.How many kJ/s are being used in recycling Al cans?105,000 cans are recycled in the US every minute.How many kJ/s are being used in recycling Al cans?

Almol 0.519g/mol 26.982 Alg 14

kJ/mol 10.7 Almol 0.519

can 1x

can 1kJ 5.55

(Al)Hfus Δ

kJ/s 9.72x10 Almole 1kJ 10.7

x can 1

Almol 0.519 x

s 60cans 1.05x10 3

5

That’s equivalent to burning 2300 food Calories/s!

18 g H2O = 1 mole H2O

- 10 °C 90 °C

Experiment: Heat two beakers containing 18 g of water at the same rate, and monitor their temperatures.

Question: Will their temperatures increase at the same rate?

0 °C 100 °C

Experiment: Heat two beakers containing 18 g of ice and water at the same rate, and monitor their temperatures.

Question: Will their temperatures increase at the same rate?

Answer: It takes twice as long to increase the temperature of the liquid water by 10 °C than it does to increase the temperature of the ice by the same amount.

Tem

pera

ture

(°

C)

0

100

Heating curve of water

solid warming

solid + liquid present

liquid warming liquid + gas present

Gas warming

Heat (kJ/s)

Tem

pera

ture

(°

C)

0

100

Heating curve of water

melting/freezing point

boiling/condensation point

Temperature is constant during phase transitions!!

All heat energy goes to changing the state of matter.

Heat (kJ/s)

Heat (kJ/s)

Tem

pera

ture

(°

C)

0

100

Heating curve of water

Hfus = the amount of heat needed to covert a solid into its liquid phase

Hfus

Hvap

(heat of fusion)

(heat of vaporization)

Hvap = the amount of heat needed to convert a liquid into its gaseous phase

Tem

pera

ture

(°

C)

0

100

Heating curve of water

Heat (kJ/s)

H2O: Hfus = 6.01 kJ/mol Hvap = 40.7 kJ/mol

Hfus = 20.2 kJ/mol Hvap = 10.3 kJ/molH2PEw:

A greater Hfus = more time to melt

And vice versa

Heating Curve Wrap Up:

•The specific heat capacity (Cp)of a substance determines the temperature change observed when heat is added or withdrawn from the substance.

•Temperature is INVARIANT during phase transitions.

•The amount of heat required to convert one mole of the substance from one phase to another is its molar enthalpy of transition (Hfus, Hvap, Hsub).

•The amount of heat given off for one mole of a substance during a phase transition while cooling is its molar enthalpy of transition (Hcond, Hsol, Hdep).

•The shape of a heating curve depends upon the heating rate, specific heat capacities of the phases involved, and the enthalpies of transition.

What is the sign for all three?

+H

What is the sign for all three?

-H

2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2

Energy

2 AlBr3 + 3 Cl2

2 AlCl3 + 3 Br2

Hrxn = Heat content of products – heat content reactants

Hrxn < 0 Reaction is exothermic

But how do we determine the heat content in the first

place?

Heat of formation, Hf

• The Hf of all elements in their standard state equals zero.

• The Hf of all compounds is the molar heat of reaction for synthesis of the compound from its elementsHf (AlBr3):2 Al + 3 Br2 2 AlBr3

Hrxn = 2Hf(AlBr3)

Hrxn

2Hf(AlBr3) =

• Since the Hrxn can be used to find Hf, this means that Hf can be used to find Hrxn WITHOUT having to do all of the calorimetric measurements ourselves!!

The Law of Conservation of Energy strikes again!!

Hess’s Law: Hrxn = Hf(products) – Hf(reactants)

6 CO2 (g) + 6 H2O (l) C6H12O6 (s) + 6 O2 (g)

Hrxn = [Hf(C6H12O6) + 6 Hf(O2)] – [6 Hf(CO2) + 6 Hf(H2O)]

From Hf tables: Hf(C6H12O6) = -1250 kJ/mol

Hf(CO2) = -393.5 kJ/mol Hf(H2O) = -285.8 kJ/mol

Hrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]

Hrxn = +2825.8 kJ/molHrxn = +2825.8 kJ/mol

H2O (l) H2O (g)

Energy

H2O (l)

H2O (g)

Hvap = +40.7 kJ/mol

• Water will spontaneously evaporate at room temperature even though this process is endothermic.

• What is providing the uphill driving force?

a measure of the disorder or randomness of the

particles that make up a system

• Water will spontaneously evaporate at room temperature because it allows the disorder of the water molecules to increase.

• The entropy, S, of gases is >> than liquids or solids.

• If Sproducts > Sreactants, S is > 0

Predict the sign of S:

ClF (g) + F2 (g) ClF3 (g) S < 0

CH3OH (l) CH3OH (aq) S > 0

Are all +S reactions spontaneous?

2 H2O (l) 2 H2 (g) + O2 (g)

S is large and positive…

…but H is large and positive as well.

• Gibb’s Free Energy, G, allows us to predict the spontaneity of a reaction using H AND S.

2 H2O (l) 2 H2 (g) + O2 (g)

What is G for this reaction at 25C?

Hrxn = Hf(products) –Hf(reactants)

Hrxn = [2(0) + 0] - 2(-285.83 kJ/mol) = 571.66 kJ/mol

Srxn = [2(130.58 J/molK) + 205.0 J/molK] - 2(69.91 J/molK)

Srxn = Sf(products) –Sf(reactants)

Srxn = 326.34 J/molK = 0.32634 kJ/molK

Grxn = Hrxn – TSrxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK)

Grxn = +474.41 kJ/mol

2 H2O (l) 2 H2 (g) + O2 (g)

Grxn = Hrxn – TSrxn = 571.66 kJ/mol - T(0.32634 kJ/molK)

What is the minimum temperature needed to make this reaction spontaneous?

Set Grxn = 0 to find minimum temperature

0 = 571.66 kJ/mol - T(0.32634 kJ/molK)

T = (571.66 kJ/mol)/(0.32634 kJ/molK) = 1751 K

T > 1479 C