How heat Is

Measured

Some specific heat capacity of substances at 250C

Substance Specific heat (J /g.0C)Water 4.184Aluminum 0.900Copper 0.380Glass 0.753Iron 0.444Mercury 0.1389Nitogen(N2) 1.0376Silver 0.240Steam 1.87 Table salt 0.875

Sand 0.787

Sample Problem 1

A sample of 50g of water was placed in a coffee cup calorimeter at a temperature of 200C. A substance was dissolve in water lowering the temperature to 100C. Calculate the energy exchange, Q, for the water. Specific heat of water = 4.18J/g.0C

Given:m= 50g Tf = 100C Ti = 200C S = 4.184J/g.0C

Find: QSolution : Q = m s ( T) 50g ( 4.184J) ( -100C)

g. C Q = -2092J

Specific heat (s) heat required to raise one gram of a substance by one

degree celcius

The transfer of heat from an object depends on an object’s mass, specific heat and the difference in temperature between the object and the surrounding:

Mathematically: Q = m s TWhere: Q amount of heat m mass s specific heat T change in temperature ( Tf - Ti )

Sample 2

How much heat is needed to raise the temperature of 5kg of water from 100C to 350C?

Given: mass: 5 Kg x 1000g = 5000g 1 Kg T = ( Tf – Ti) = ( 35 – 10) = 250C S water = 4.18 J/g.0C Find: QSolution: Q = m s T = 5000g ( 4.18J/g.0C) ( 250C) = 522500J

Sample #3

A 40g sample of alloy ( a mixture of metals) is heated to 90.800C and then placed in water, where it cools to 23.540C. The amount of heat lost by the alloy is 865J. What is the specific heat of the alloy?

Given: m = 40g T = ( 23.540C – 90.800C) Q = 865J = -67.26Find: sSolution : Q = m s T s = Q = 865J = 0.32J/g.0C m T 40g( -67.260C)

Enthalpy

Enthalpy (H)

Heat content of the systemChange in enthalpy ( H ) The heat transferred between the system and the

surrounding under constant pressure Also represents the difference between the enthalpy of

the system before and after the process and is represented as:

H = Hfinal - Hinitial

Note: If Hf > Hi H is positive ( endothermic)

If Hf < Hi H is negative( exothermic)

Enthalpies of Reaction

Enthalpy change that accompanies a reaction the sum of the enthalpies of the products of the

reaction minus the sum of the enthalpies of the reactants in a reaction.

The equation is given as H rxn =Σ H products - Σ H reactants

Example of a thermochemical equation 2Na(s) + Cl2(g) - 2NaCl(s) H = -822.30kJ

What does the H tell you about the reaction? The reaction release heat (exothermic)

Ex: of a thermochemical equation

2Na(s) + Cl2(g) - 2NaCl(s) H = -822.30kJ

What does the H tell you about the reaction? The reaction release heat (exothermic)

What happens to the enthalpy of the system in an exothermic reaction? Why?

Decreases because the sum of the enthalpies of the products is smaller than the enthalpies of the reactants

Example

N2(g) + O2(g) 2 NO2 H = + 67.8kJWhat does the H tell you about the

reaction? The reaction absorbed heat Endothermic

The enthalpy of the system in an endothermic reaction increases because the sum of the enthalpies of the products is greater than the enthalpies of the reactants

Sample Problem

When 1 mol of methane( CH4) is burned at constant pressure, 890kJ of energy is released as heat. Calculate the H of the process in which 5.8g sample of methane is burned at constant pressure.

Given: q = -890kJ m = 5.8gSolution: 5.8gCH4 x molar mass CH4 x q

5.8gCH4 x 1molCH4 x -890kJ = -320kJ

16.0gCH4 1molCH4

Self-Check10.5 p. 302

The reaction that occurs in the heat packs to treat sports injuries is

4 Fe(s) + 3O2(g) ---> 2Fe2O3(s) H= -1652kJ How much heat is released when 1.00 g Fe(s) is

reacted with excess O2?Solution: molar mass Fe = 55.85g/mol 1.00g Fe x 1mol Fe = 1.79x 10-2 mol Fe 55.85gFe

1.79x 10-2 mol Fe x -1652kJ = -7.39kJ 4 molFe

Objective: Explain the energy involve during the phase change

Change of phase Occurs when a substance change from one state to

another

Note:Change of phase always occur with a change in the

amount of heat of a substance

Heat travels spontaneously from a warmer body to a colder body

What happens during phase change?

During phase change, heat either comes out of the material or goes into the material but its temperature remains the same

Terms for different phase changesSolid to liquid melting (fusion)Solid to gas sublimationLiquid to solid freezing (solidification)Liquid to gas vaporizationGas to liquid condensationGas to solid deposition

vaporization condensation

sublimation deposition

melting freezing (fusion) (solidification)

SOLID

LIQUID

GAS

The amount of heat absorbed by one mole of a substance when it changes from solid to liquid

The quantity of heat absorbed during fusion is equal to the heat released during freezing

The amount of heat released by one mole of a substance to change from liquid to solid.

Molar heat of vaporization

Amount of heat required by one mole of a substance to change from liquid to gas

Molar heat of condensation

Amount of heat released by one mole of a substance to change from gas to liquid

Explain the kinetic energy of the water molecule in the heating curve

When all the ice has turned to liquid and heat continues to flow into the water, the heat goes into increasing the kinetic energy of the molecules resulting in an increase in temperature

a process that occurs by itself. Which means no external force is needed to continue the process

Ex: Sugar dissolving in hot coffee rusting of iron

is a process that occurs when an external force is continuously applied

Ex: riding in a playground swing electrolysis of water

Note:

Most exothermic reactions are spontaneous

Most endothermic reactions are non spontaneous

However there are some endothermic reactions that are nonspontaneous

Ex: the melting of ice at room temperature or above 00C

Consider the melting of ice cube

Melting of an ice cube is spontaneous if the temperature is above the melting point

Nonspontaneous is below the melting point

Degree of freedom that particles have Degree of disorder in a system The greater the degree of randomness or disorder, in a

system, the greater is its entropyWhich has a greater entropy liquid or gas? Gas has greater entropy than pure liquids

Entropy is measured by J/K

How to predict the S using the rules

Ex: C6H12O6(s) ---> 2C2H5OH(l) + 2CO2(g)

What is the sign of the S for the above reaction?positive because the entropy of the final product is

higher

Sometimes predicting the sign of S is not possibleEx: CO(g) + H2O(g) ---> CO2(g) + H2(g)

Predicting S in this case is impossible

Symbol ( S) Is the difference between the final entropy and the

initial entropy of a system. S = Sfinal - Sinitial

S is positive if there is an entropy increase Ex: solid ---> liquid or gas S is negative if there is an entropy decreaseEx: when a precipitate for Ag+(aq) + Cl-(aq) AgCl(s)The entropy of the solid is lower than the original

aqueous ions

Rules to predict the sign of S Generally entropy increases when A reaction breaks up a larger molecule

into smaller molecular fragments A reaction occurs in which there is an

increase in the mole of gas in the product

A process where a solid changes to a liquid or gas or a liquid to a gas

Exercises 2NH4NO3(s) --->2N2(g) + 4H2O(g) + O2(g) 2SO2(g) + O2(g) --> 2SO3(g) C12H22O11(aq) C12H22O11(s) CO2(g) + H20(g) ---> CO2(g) + H2(g)

Hess Law States that if the reaction occurs in

two or more steps, the enthalpy for the reaction is the sum of the enthalpies of the individual steps

Why is it useful? because it allows us to calculate heats

of reaction that are difficult or inconvenient to measure in a calorimeter

Gibbs Free Energy thermodynamic entity that takes into

account both entropy and enthalpy Useful for determining whether a

process is spontaneous or not Defined as G = H – T SWhere: H enthalpy T temperature S entropy

For a process at constant temperature the change in free energy is

G = H - T S G -> represents the maximum energy

that is free to do useful work to the surrounding.

- G means that the system does useful work to

the surrounding

means spontaneous process

Ex: movement of water in a water fall is spontaneous

Why? Because it does a useful work in turning the wheels of a

turbine

+ G indicates that work has to be done on the system for the process to take place means that energy will have to be absorbed from the surrounding means non spontaneous processEx: maintenance of life It requires sustained input of work or energy that we

get from food If G = 0 The reaction is at equilibrium

Example

CoCl2(g) CO(g) = Cl2(g)1. Calculate the delta G at 1270C.

2. Calculate the temperature when the above reaction is at

equilibrium Solution: G = H – T SSolve for H using standard enthalpies of formation (See

Thermodynamics Table) H = [ sum Hfproducts ] – [ sum Hf reactants] = [ 1mol x (-110.5kJ) + 1mol ( 0kJ) ] – [ 1mol x 220kJ ] mol mol mol = -110 kJ – (-220kJ) H= + 110kJ

Determine delta S for the reaction using standard molar entropies and Hess Law of Summation ( See Thermodynamics table)

S = [ sum S products] – [ sum Sreactants]

= [1 mol x 197.5 J + 1 mol x 223J] – [1 mol x 283.7 J] mol.K mol.K mol.K = 420.5J/K - 283.7J/K = 136.8J/K ( convert this to kJ) 136.8J/K x 1kJ = 0.1368kJ/K 1000J S = 0.1368kJ/KSolve G G = 110.5kJ – 400K ( 0.1368kJ/K G =+55.78kJ

2) When delta G = 0, solve TSolution G = H - T S 0 = 110.5kJ – T ( 0.1368kJ/K) (0.1368kJ/K)T = 110.5kJ T = 110.5kJ 0.1368kJ/K T = 807.7K

Calculating Energy Changes:Solid to Liquid ( Chapter14: Solids and Liquids)

Calculate the energy required to melt 8.5g of ice at 00C. The molar heat of fusion of ice is 6.02kJ/mole

Solution: 8.5gH2O x 1mol H2O x 6.02kJ = 2.8kJ 18gH2O 1molH2O

Calculating Energy changes( Liquid to Gas) Calculate the energy in kJ to heat 25g of liquid water from 25oc

to 100oc and change it to steam at 100oC. The specific heat capacity of liquid water is 4.184J/goC and the molar heat of vaporization of water is 40.6kJ/mol.

Solution:Q = ms T (25g)(75oC)(4.184J) = 7.8x103J x 1kJ = 7.8kJ(to boil) goC 1000J25gH2O x 1molH2O =1.4mol H2O x 40.6kJ = 57kJ( vaporization) 18gH2O 1mol H2OThe total energy is the sum of the two steps 7.8kJ + 57kJ = 65kJ

Heating and Cooling Curve

Fig 20.3 Heating curve for water

125 F.100 D . E . E-FWater vapor

75 Liquid-gas C-D50 B-C- melting25 .B .C0 -25 A.

Cooling Curve

Describes the change in temperature and the amount of heat during the cooling process

The slope of the curve is decreasing when heat is removed from the material

Fig 20.4 Cooling Curve of water

1 2 3 4 5

100 Liquid 75 Gas

50 Gas-liquid 25 Liquid-solid solid

0 HoVap

-25 Hofus

Stage1:Gaseous water cools. Stage 2: Gaseous water condenses

Stage3: Liquid water cools. Stage4: Liquid water freezes Stage5: solid water cools