Chapter 6

Thermochemistry: Energy Flow and Chemical Change

Chapter 6: Thermochemistry

6.1 Forms of Energy and Their Interconversion

6.2 Enthalpy: Heats of Reaction and Chemical Change

6.3 Calorimetry: Laboratory Measurement of Heats of Reaction

6.4 Stoichiometry of Thermochemical Equations

6.5 Hess’s Law of Heat Summation

6.6 Standard Heats of Reaction (!H0rxn)

Thermochemistry involves the monitoring of energy transformations that occur with a chemical reaction.

CH4(g) + 2 O2 ==> CO2 + 2 H2O + HEAT

NH4NO3 + H2O + HEAT ==> NH4+ + NO3-

• Reaction gives off heat with rise in temperature in the flask • We say “EXOTHERMIC REACTION” • Heat is written as a product.

• Reaction absorbs heat with decrease in temperature • We say “ENDOTHERMIC REACTION” • Heat is written as a reactant.

--the theory is developed without knowledge of the atoms and molecules.

--The focus is the “bulk properties” of matter (Temperature, pressure, volume,internal energy, enthalpy, heat capacity).

--It makes sense once you learn the language and basic ideas.

Thermodynamics is a formal accounting system based on observations over centuries that tracks energy changes in a chemical system.

Energy is the capacity to do work or produce heat. Energy can be transformed to different forms.

• Radiant energy comes from the sun and is earth’s primary energy source (nuclear fusion)

• Thermal energy is the energy associated with the vibration and rotational motion of atoms and molecules.

• Chemical energy is the energy stored within the chemical bonds of substances.

• Nuclear energy is the energy stored within the neutrons and protons in the atom (E = mc2)

• Kinetic energy: energy associated with moving mass

• Potential energy: energy available by virtue of an object’s position or height above a reference height.

The SI Unit of energy is the Joule (kg m2/s2).

KE = 1/2 mass !v2 = 1/2 (kg)(m/s)2 = kg!m2/s2 = 1 joule

PE = mass gh = (kg)(m/s2)(m) = kg!m2/s2 = 1 jouleg = acceleration gravity (9.8 m/s2), and h is object’s height.

Work = F x d = mass (a) x d = (kg)(m/s2) x (m) = kg!m2/s2 = 1 joule

1 calorie (cal) = 4.18J

1 British Thermal Unit (BTU) = 1055 Joules

1000 calorie (cal) = 1 Cal (food calorie)

Know how to play with these conversions

What is the kinetic energy (in kilojoules) of a 2300 lb car moving at 55 mi/h?

• KE = 1/2 mv2 =?

A single Fritos chip snack has a energy content of 5.0 Cal. How many joules is this?

E = 5 Cal × 1000 cal

1 Cal× 4.18J

1 cal= 2.1× 104J

• 1 Joule = 1 kg"m2/s2

• 1 mile = 1.609 km = 1609 m• 1 lb = 0.454 kg

• Energy is the capacity to do work and exchange heat-- it can take on many many forms.

• Work = Energy = Force x Distance.

• Internal Energy is the kinetic energy of molecules in motion (translational, rotational, and vibrational).

• Heat is thermal energy transferred between two objects at different temperatures. Heat “flows” only when there is a difference in temperature.

• Chemical Energy is energy stored in a chemical bond.

Thermodynamics uses a specialized vocabulary that we need to get used to.

In 1843, James Joule found that heat and work can be interconverted and are manifestations of the same thing, energy. This may seem to be trivial but it’s a actually a significant observation & discovery!

Joule found 1 cal heat = 4.1868 J work

amount of heat required to raise the temperature of 1g of water by 1˚C

W = Force x distance

The 1st Law of Thermodynamics that energy is neither created nor destroyed in any process, it is just transformed to other forms of energy across two boundaries.

!Esystem + !Esurroundings = 0

!Esystem = -!Esurroundings

“The total energy of a system and its surroundings remains constant.”

A process is an event in which a physical or chemical change causes a state property of the system to change. When a process occurs matter, heat or work must cross a boundary between the system and surroundings.

A system is defined when its variable of states (independent variables) are defined.

A path is a specific way to carry out a process from initial state through intermediate states and final state. Work and heat are both path functions.

A system is that part of the universe that exchanges matter and energy with its surroundings across a boundary.

Thermodynamic Vernacular & Definitions

The surrounding that part of the universe which is not the system.

• Reactants and products are the system; everything else is the surroundings.

• We track the energy flow into and out of the system.

• Loss of energy from the system to the surroundings has a negative sign.

• Increase of the system energy from the surroundings has a positive sign.

Thermodynamic Formalism: A system has a bulk property of matter that we call “internal energy”.

-energy in +

surroundings

system

energy out

There are 3-classes of “systems” depending on how each exchanges matter and energy with its surroundings.

Closed System Isolated System

No matter exchangedNo heat or work exchanged

Matter not exchangedHeat or work exchanged

Matter

Adibatic

q q

Open System

Matter exchangedHeat or work exchanged

• Translational kinetic energy.• Molecular rotation.• Bond vibration.• Intermolecular attractions.• Chemical bonds.• Electrons.• Nuclear • Electrochemical

The internal energy, E, of a system is a bulk property of matter that on a microscopic level represents all energy within the system.

On a microscopic scale the internal energy is:

The internal energy of a system can only change by exchange of heat (q) and mechanical work (w) with the surroundings.

Surroundings

Internal Energy ESystem

Rotation, Vibrations, Electronic, translational energy of molecules

+q = heat added to the system

+w = work done on the system

-q = heat lost from the system

-w = work done by the system

!ESys = q + w

MechanicalWorkHeatInternal

energy

SurroundingsAdd Energy Lose Energy

We can not measure the “absolute” internal energy of a system......we only measure changes in internal energy---we call this a “change in state”.

Change in Internal Energy

!ESYS = Efinal - Einitial = q + w

WorkHeat

change in state from initial conditions before chemical reaction to final condition after a chemical reaction.

Eproducts - Ereactants = q + w

1. Mount Everest is king if measured from sea level (8,848 meters)2. Mauna Kea is when measured from the depths of the Pacific Ocean floor. It rises 10,203 meters. 3. Mt. Chimborazo in the Andes triumphs. Although it stands but 6,267 meters above sea level, its peak is the farthest from the earth's core.

Measurements are relative to something. What’s the tallest mountain in the world?

Chemical reaction causes a change in internal energy that is accounted for completely by measuring the exchange of heat (q) and mechanical work (w) with the surroundings.

Initial State Final State

+/-q

+/-w

system system

!ESys = q + w = q - P!V

surroundingssurroundings

w = Force×Distance

w =Force Area

Area× height

w = P × (Area× height)

w = P ×∆V

w = P ×∆V

height

A system can do work on the surroundings or the surroundings can do mechanical work on the system.

The internal energy, U can be changed only by exchanging heat (q) or work (w) with its surroundings.

Surroundings

Surroundings

+q = heat added

+w = work done on the system

-q = heat lost from the system

-w = work done by the system

System!E

energy out

energy in + -

It’s a state function!Mechanical

workHeatInternal energy

!ESys = q + wSOLUTION:

q = - 325 J and w = - 451 J

!E = q + w = -325 J + (-451 J) = -776 J

-776J 103JkJ = -0.776kJ -0.776k4.18k

kcal= -0.185 kcal

Determining A Change in Internal Energy

When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (!E) in J, kJ, and kcal.

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Chapter 4-6

Chemists don’t use internal energy to monitor energy changes in the lab. Instead, we define and use a related quantity called enthalpy.

!Esys = q + w = q - P!V

!E = qV - P!V

At constant volume: !V=0

!E = qV

CASE 2

!E = qP - P!V

At constant pressure:

qP = !E + P!V

CASE 1

!H = qP = !E + P!V

An Exothermic reaction releases heat to the surroundings, !H < 0.

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

H2O (g) H2O (l) + heat!H = <0

heat written as a product (given off)

heat + 2HgO (s) 2Hg (l) + O2 (g)

heat + H2O (s) H2O (l)

An Endothermic reaction absorbs heat from the surroundings into the system, !H > 0.

!H = >0

heat written as a reactant (consumed)

I. In systems containing solids and liquids; there is no volume change (or it is very small); !V = 0 and !H = !E

2. In reactions where the number of moles of gas does not change from reactants to products; no work is done.

N2(g) + O2(g) ==> 2NO(g) !V = 0 and !H = !E

3. Reactions in which the number of moles of gas does change but q is >>> P!V.

!H = qp = !E + P!V There are 3-Cases When !H = !E

P = 5.0 atm !V = (14.5-12.0) L = 2.5 L w = –P!Vw = -(5.0 atm)(2.5 L) = -12.5 L"atm(-12.5 L"atm)(101 J/L"atm) = -1.3 # 103 J = -1.3 kJ

Calculate the work (in kilojoules) done during a reaction in which the volume expands from 12.0 L to 14.5 L against an external pressure of 5.0 atm. Watch the conversion of PV units to joules.

Note PV has units of energy! 101.3 J = 1 L atm

A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?

w = -P !V

(a) !V = 5.4 L – 1.6 L = 3.8 LP = 0 atm

W = -0 atm x 3.8 L = 0 L•atm = 0 joules

(b) !V = 5.4 L – 1.6 L = 3.8 LP = 3.7 atm

w = -3.7 atm x 3.8 L = -14.1 L•atm

w = -14.1 L•atm x 101.3 J1L•atm

= -1430 J

1) Has any work been done? If so, is its sign positive or negative?

2) Has there been an enthalpy change? If so, what is the sign of "H? Is the reaction exothermic or endothermic?

Imagine a chemical reaction that results in a change in both volume and temperature as shown below.

A state function is any math function that depends only on the initial and final states and not the path taken to between the two states.

The potential energy of hiker 1 and hiker 2 is the same even though they took very different paths to get to the mountain top. PE only depends on the initial and final heights only.

PE = mghh

!E = Efinal - Einitial

!P = Pfinal - Pinitial

!V = Vfinal - Vinitial

!T = Tfinal - Tinitial

!H = Hfinal - Hinitial

All are state functions!

0

Final State

Initial State

Why Do We Care About State Functions?• A change in a state function tells us that the function

has one unique value between any two arbitrary states. This fact is why thermodynamics is so powerful and useful.

mgh2

h = 1

!h

Initial State

Final State

Arbitray State

C3H8(g) + 5O2(g) #$ 3CO2(g) + 4H2O(l) !H = -2043 kJ

2) 2043 kJ are released and can be written as productC3H8(g) + 5O2(g) #$ 3CO2(g) + 4H2O(l) + 2043 kJ

A thermochemical equation shows the enthalpy change (!H) of a chemical reaction and also the reactions stoichimetric factors.

1 mol C3H8(g) = -2043 kJ 5O2(g) = -2043 kJ

3CO2(g) = -2043 kJ 4H2O(l) = -2043 kJ

Stoichiometeric factors

3) more useful stoichiometric factors for calculations

1) !H <0 = REACTION IS EXOTHERMIC--gives off heat

Using the Heat of Reaction (!Hrxn) to Find AmountsThe major source of aluminum in the world is bauxite which is mostly aluminum oxide. Its thermal decomposition can be represented by:

If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 103 kJ of heat is is used to decompose the oxide?

Al2O3(s) 2Al(s) + 3/2O2(g) !Hrxn = 1676 kJ

SOLUTION:

g Al = 1000. kJ× 2 mol Al1676 kJ

× 26.98 g Al1 mol Al

= 32.2 g Al

Using the thermochemical equation calculate how much heat is evolved (or what is the enthalpy) when 266 g of white phosphorus (P4) is combusted in air? The molar mass of P4 is 123.9 g/mol.

P4 (s) + 5O2 (g) P4O10 (s) !Hrxn = -3013 kJ/mol

!H = H (products) – H (reactants)

∆Hrxn = 266. g P4 ×1 mol P4

123.9 g P4

× −3013 kJ

1 mol P4

= −6470 kJ

Rules For Manipulating Thermochemical Equations.

• 3. If you multiply both sides of the equation by a factor n, then !H must change by the same factor n.

2H2O (s) 2H2O (l) !H = 2 x 6.01 = 12.0 kJ

• 2. If you reverse a reaction, the sign of !H changes

H2O (l) H2O (s) !H = -6.01 kJ

• 1. The stoichiometric coefficients always refer to the number of moles of a substance---it’s extensive!

H2O (s) H2O (l) !H = 6.01 kJ

• 4. The physical states of all reactants and products must be specified in thermochemical equations.

H2O (l) H2O (g) !H = 44.0 kJH2O (s) H2O (l) !H = 6.01 kJ

Look at the coefficients!

Thermochemistry is concerned primarily with calculating !Hrxn (the heat of reaction).

Use !H˚f data

2-Ways or Problem Types For Hess Law Enthalpy

Calculations

Multiple Thermochemical

Equations To Rearrange--Use Rules!

21

1) standard state2) formation equation3) Calculation of !H°

rxn

!H°rxn

= % ni !Hif°(products) - % mi !Hjf

°(reactants)

• Hess’s Law: The enthalpy change for any reaction is equal to the sum of the enthalpy changes for any individual step in the reaction.

CH4 (g) + 2O2 ==> CO2 + 2H2O !H3 = ? kJCO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O !H2 = - 283 kJ CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 !H1 = - 607 kJ

Because enthalpy is a state function, its value only depends on the final and initial state and not the path. This gives us a powerful way to predict enthalpy of many chemical reactions.

CH4 (g) + 2O2

CO(g) + 2H2O + 1/2 O2(g)

CO2 + 2H2O

∆H1 = - 607 kJ

∆H2 = - 283 kJ

∆H3 = ?

Rules For Manipulating Thermochemical Equations.

• 3. If you multiply both sides of the equation by a factor n, then !H must change by the same factor n.

2H2O (s) 2H2O (l) !H = 2 x 6.01 = 12.0 kJ

• 2. If you reverse a reaction, the sign of !H changes

H2O (l) H2O (s) !H = -6.01 kJ

• 1. The stoichiometric coefficients always refer to the number of moles of a substance---it’s extensive!

H2O (s) H2O (l) !H = 6.01 kJ

• 4. The physical states of all reactants and products must be specified in thermochemical equations.

H2O (l) H2O (g) !H = 44.0 kJH2O (s) H2O (l) !H = 6.01 kJ

Look at the coefficients!

Energy level diagrams are useful pictorials of the thermodynamic transitions that take place during a chemical reaction.

CH4 (g) + 2O2 ==> CO2 + 2H2O !H3 = -890 kJCO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O !H2 = - 283 kJ CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 !H1 = - 607 kJ

Enth

alpy

, H

CH4 (g) + 2O2

CO2 + 2H2O

∆H1 = - 607 kJ

∆H2 = - 283 kJ

∆H3 = -890 kJCO(g) + 2H2O + 1/2 O2(g)

Multiply Equation B by 1/2 and reverse it.

!HB = -90.3 kJCO(g) + 1/2O2(g) CO2(g) !HA = -283.0 kJNO(g) 1/2N2(g) + 1/2O2(g)

!Hrxn = -373.3 kJCO(g) + NO(g) CO2(g) + 1/2N2(g)

Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:

CO(g) + NO(g) CO2(g) + 1/2N2(g) !H = ?

Given the following information, calculate the unknown !H:

CO(g) + 1/2O2(g) CO2(g) !HA = -283.0 kJ

N2(g) + O2(g) 2NO(g) !HB = 180.6 kJ

NO2(g) ! NO(g) + 1/2O2(g) !H = +57.07 kJ

1/2N2(g) + O2(g) ! NO2(g) !H = +33.18 kJ

"N2(g) + O2(g) ! NO + 1/2O2(g) !H = 90.25 kJ

Calculate the enthalpy of the following reaction and draw and energy level diagram that summarize the thermodynamic enthalpies involved in this reaction.

Use the properties of thermochemical equations with Hess’s Law to calculate the enthalpy of the following reaction using the reactions and their associated enthalpies taken from a handbook below.

FeCl2(s) + 1/2 Cl2(g) FeCl3(s) !H = ?

1. Fe(s) + Cl2(g) FeCl2(s) !H = -341.8 kJ/mol

2. Fe(s) + 3/2 Cl2(g) FeCl3(s) !H = -399.5 kJ/mol

Thermochemistry is concerned primarily with calculating !Hrxn (the heat of reaction).

Use !H˚f data

2-Ways or Problem Types For Hess Law Enthalpy

Calculations

Multiple Thermochemical

Equations To Rearrange--Use Rules!

21

1) standard state2) formation equation3) Calculation of !H°

rxn

!H°rxn

= % ni !Hif°(products) - % mi !Hjf

°(reactants)

Physicists and chemists define a standard thermodynamic “standard state” and denote with a superscript degree sign, !H˚

• 1 atmosphere pressure• 25˚C = 298.15K• 1 Molar concentration for solutions

Thermodynamic Standard State Means:

!Hrxn !H˚rxn

The superscript indicates thermodynamic standard state: P = 1 atm, T = 298, [1M]

Conditions of P, T, [] are given in problem and stated

The Standard Enthalpy of Formation, !Hf°, is the

enthalpy change associated with the formation of 1 mole of product from its naturally occurring elements under standard state conditions.

H2(g) + O2(g) #$ H2O(l ) !Hf° = -285.8 kJ/mol

3C(s) + 4H2(g) #$ C3H8(g) !Hf° = -103.85 kJ/mol

Ag(s) + 1/2Cl2(g) #$ AgCl(s) !Hf° = -127.0 kJ/mol

•Are familar molecular gases as O2, N2, F2, H2 etc as at 25 °C•Elemental Carbon exists as solid graphite C(graph) at 25 °C.•Elemental Sulfur exits as S8 as a solid at 25˚C•Water is H2O(l ) in its standard state (not ice or water vapor).

•Examples: Note: 1 mol product!

Note: fractional coefficients allowed

It is not possible to measure the “absolute enthalpy”. We define an arbitrary scale with the standard enthalpy of formation (!Hf

°) as a reference point.

1) The standard enthalpy of formation, !Hf˚ of any element in its most stable form is zero.

The standard enthalpy of formation, !Hf˚ of many compounds is tabulated in many handbooks. This gives us predictive capability and is someone useful in the real world.

!H0 (O2) = 0f

!H0 (O3) = 142 kJ/molf

!H0 (C, graphite) = 0f

!H0 (C, diamond) = 1.90 kJ/molf

!H0 (S8, rhombic) = 0 kJ/molf

1) Writing Formation EquationsWrite balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include !H0

f.

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

---Use the table of heats of formation for values.

---Remember 1 mole of product is formed and fractions are allowed as coefficients.

(c) Hydrogen sulfide, H2S, a gas at standard conditions.

Writing Formation EquationsWrite balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include !H0

f.

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Use the table of heats of formation for values.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

SOLUTION:

!H0f = -127.0 kJ(a) Ag(s) + 1/2Cl2(g) AgCl(s)

!H0f = -1206.9 kJ(b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s)

!H0f = 33.9 kJ(c) O2(g) + 1/2N2(g) NO2(g)

where % means “the sum of”

ni is the respective stoich coefficient for ith product mi is the respective stoich coefficient for each ith reactant

METHOD 2: The enthalpy of any reaction, !H°rxn can be

obtained by using !H°f and the following equation:

!H°rxn

= " ni !Hif°(products) - " mi !Hjf

°(reactants)

!H°rx = [c!H°f (C) + d!H°f (D)] – [a!H°f (A) + b!H°f (B)]

!H°rx = !H°f (Products) – !H°f (Reactants)

aA + bB cC + dDExample: Suppose !H°rx = ?

Calculate the heat of decomposition for the following process using standard enthalpies of formation foundin Appendix. What kind of reaction is this? Is energy given off or needed to make this reaction proceed?

CaCO3 (s) CaO (s) + CO2 (g)

]!H0rxn !H0 (CO2)f!H0 (CaO)f= [ + ] - !H0 (CaCO3)f[

!H0rxn = [ -635.6 + -393.5 ] – [ -1206.9 ] = 179 kJ

!H°rxn

= % ni !Hif°(products) - % mi !Hif

°(reactants)

Reaction is decomposition and endothermic. Energy must be absorbed to proceed.

!H0rxn = ?

PLAN:

Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate !H0rxn from !H0

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Look up the !H0f values in Appendix and use Hess’s Law to

find !Hrxn.

!H0f NH3(g) = -46.3 kJ/mol

!H0f H2O(g) = -241.8 kJ/mol !H0

f NO(g) = 90.4 kJ/mol

!H0f O2(g) = 0

Calculate !H0rxn from !H0

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

SOLUTION: !Hrxn = % m!H0f (products) - % n!H0

f (reactants)

!Hrxn = [4(!H0f NO(g) + 6(!H0

f H2O(g)] - [4(!H0f NH3(g) + 5(!H0

f O2(g)]

= (4 mol)(90.4 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-46.3 kJ/mol) + (5 mol)(0 kJ/mol)]

!Hrxn = -904 kJ

Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

!H0rxn n!H0 (products)f= % m!H0 (reactants)f

%-

!H0rxn 6!H0 (H2O)f12!H0 (CO2)f= [ + ] - 2!H0 (C6H6)f[ ]

!H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

q = C !T = s m !T

n�

i=1

qi = 0

Physics of heat transfer: the heat absorbed by a substance depends on the material’s specific heat capacity, mass and temperature change.

Enthalpies (!H) of chemical reactions can be determined using the physics of heat transfer a lab method called calorimetry.

No heat enters

Hot Cold

TlowThighHeat Transfer

Block BBlock A

• Heat transfer only occurs when there is a temperature difference between two bodies

• Heat is transferred occurs through conduction, convection or radiation.

Observed Physics: The amount of heat, q, transferred from an object at higher temperature to an object at lower temperature is proportional to the difference in temperature of the two objects!

Principle 1: The amount of heat, q, transferred from an object at higher temperature to an object at lower temperature is proportional to the difference in temperature of the two objects!

C = heat capacity

q = C !T

q ∝ !T

ThermometerThermometer

T1T1 T2

Object 1 Object 2Heat Q

Heat transferred = q & (Tfinal - Tinitial) q = C !T

C has units of Energy per degree T (J/°C, J/K, cal/°C, cal/K)

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a quantity of the substance by one degree Celsius or Kelvin (units of J/˚C or J/K or cal/˚C).

material dependent constant “heat capacity”

temperature difference between objects

!TC =q

The heat capacity, C, of any material is found to be proportional to mass of a substance under question.

C ∝ m

C = m s

Units of s: J/g °C, J/g K

specific heat

mass

s is called the specific heat (s) of a substance. It is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. (How thermally sensitive a substance is to the addition of energy!)

Heat (q) absorbed or released:

q = s m !T

q = C !T C = m s

Substituting (2) into 1 gives:

(1) (2)

The molar heat capacity of a substance is the amount of heat (q) required to raise the temperature of one mole of the substance by one degree Celsius.

Heat (q) absorbed or released:

q = s m !T

Units: J/mol °C or J/mol K Units: moles

Units: ˚C or K

KNOW THIS EQUATION AND HOW TO USE IT!Heat (q) absorbed or released:

q = s m !T

How much heat, q, is required to raise the temperature of 1000. kg of iron and 1000 kg of water from 25 to 75˚C? Note that the specific heat capacity of iron is 0.45 J/g ˚C

qH2O = 1.00× 103 kg × 1000 g

1 kg× (75◦ − 25◦)× 4.18J

◦C g= 2.09× 108J

qFe = 1.00× 103 kg × 1000 g

1 kg× (75◦ − 25◦)× 0.450J

◦C g= 2.25× 107J

Determine the specific heat capacity of lead when 150.0 grams of lead pellets at 100.0˚C are added to 50.0 ml of water held at 22.0˚C. The final temperature of the insulated calorimeter is 28.8˚C. Assume the container and thermometer do not absorb heat.

1. 150.0 g Pb @ 100.0˚C

2. Add pellets to 50.0 ml H2O @ 22.0˚C

3. What’s Up?

28.8˚C

sp. heat ?

qwater = mc#T = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C = 1.40 x103 J

qlead = mc!T = (150.0 g)(c)(28.8 - 100.0)°C = -1.40 x103 J

clead = 0.13 Jg-1°C-1

1. Use the law of conservation of energy pertains to ALL OF THE PARTICIPANTS!

mPb cPb !TPb = -mwat cwat !Twat

clead = mwat cwat !Twat / mlead !Tlead

qPb + qwater + qcontainer = 0

qPb = - qwater

n�

i=1

qi = 0

1. Conservation of energy is the key fundamental physics!

A 237 g piece of molybdenum, initially at 100.0 oC, is dropped into an insulated container with 244 g of water at 10.0 oC. When the system comes to thermal equilibrium, the temperature of the system is 15.3 oC. What is the specific heat of molybdenum?

!

qmb + qwater = 0

-qmb = -smb mmb !T = swtr mwtr !T = qwater

n�

i=1

qi = 0