Chapter 6: Thermochemistry. Chapter 6. CH4(g) + 2 O2 ==> CO2 + 2 H2O + HEAT NH4NO3 + H2O + HEAT ==> NH4 + + NO3 -

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1 Chapter 6 Chapter 6: Thermochemistry Thermochemistry: Energy Flow and Chemical Change 6.1 Forms o Energy and Their Interconversion 6.2 Enthalpy: Heats o Reaction and Chemical Change 6.3 Calorimetry: Laboratory Measurement o Heats o Reaction 6.4 Stoichiometry o Thermochemical Equations 6.5 Hess s Law o Heat Summation 6.6 Standard Heats o Reaction (!H 0 rxn) Thermochemistry involves the monitoring o transormations that occur with a chemical reaction. CH4(g) + 2 O2 ==> CO2 + 2 H2O + HEAT Reaction gives o heat with rise in temperature in the lask We say EXOTHERMIC REACTION Heat is written as a product. NH4NO3 + H2O + HEAT ==> NH4 + + NO3 - Reaction absorbs heat with decrease in temperature We say ENDOTHERMIC REACTION Heat is written as a reactant. Thermodynamics is a ormal accounting based on observations over centuries that tracks changes in a chemical. --the theory is developed without knowledge o the atoms and molecules. --The ocus is the bulk properties o matter (Temperature, pressure, volume,internal, enthalpy, heat capacity). --It makes sense once you learn the language and basic ideas. Energy is the capacity to do work or produce heat. Energy can be transormed to dierent orms. Radiant comes rom the sun and is earth s primary source (nuclear usion) Thermal is the associated with the vibration and rotational motion o atoms and molecules. Chemical is the stored within the chemical bonds o substances. Nuclear is the stored within the neutrons and protons in the atom (E = mc 2 ) Kinetic : associated with moving mass Potential : available by virtue o an object s position or height above a reerence height.

2 The SI Unit o is the Joule (kg m 2 /s 2 ). KE = 1/2 mass!v 2 = 1/2 (kg)(m/s) 2 = kg!m 2 /s 2 = 1 joule Work = F x d = mass (a) x d = (kg)(m/s 2 ) x (m) = kg!m 2 /s 2 = 1 joule PE = mass gh = (kg)(m/s 2 )(m) = kg!m 2 /s 2 = 1 joule g = acceleration gravity (9.8 m/s 2 ), and h is object s height. A single Fritos chip snack has a content o 5.0 Cal. How many joules is this? E = 5 Cal 1000 cal 1 Cal 4.18J 1 cal = J What is the kinetic (in kilojoules) o a 2300 lb car moving at 55 mi/h? Know how to play with these conversions 1 calorie (cal) = 4.18J 1000 calorie (cal) = 1 Cal (ood calorie) 1 British Thermal Unit (BTU) = 1055 Joules KE = 1 / 2 mv 2 =? 1 Joule = 1 kg"m 2 /s 2 1 mile = km = 1609 m 1 lb = kg Thermodynamics uses a specialized vocabulary that we need to get used to. Energy is the capacity to do work and exchange heat-- it can take on many many orms. Work = Energy = Force x Distance. Internal Energy is the kinetic o molecules in motion (translational, rotational, and vibrational). Heat is thermal transerred between two objects at dierent temperatures. Heat lows only when there is a dierence in temperature. In 1843, James Joule ound that heat and work can be interconverted and are maniestations o the same thing,. This may seem to be trivial but it s a actually a signiicant observation & discovery! Joule ound 1 cal heat = J work amount o heat required to raise the temperature o 1g o water by 1 C W = Force x distance Chemical Energy is stored in a chemical bond. The 1st Law o Thermodynamics that is neither created nor destroyed in any process, it is just transormed to other orms o across two boundaries. The total o a and its surroundings remains constant.!e +!E surroundings = 0!E = -!E surroundings Thermodynamic Vernacular & Deinitions A is that part o the universe that exchanges matter and with its surroundings across a boundary. The surrounding that part o the universe which is not the. A is deined when its variable o states (independent variables) are deined. A process is an event in which a physical or chemical change causes a state property o the to change. When a process occurs matter, heat or work must cross a boundary between the and surroundings. A path is a speciic way to carry out a process rom initial state through intermediate states and inal state. Work and heat are both path unctions.

3 Thermodynamic Formalism: A has a bulk property o matter that we call internal. There are 3-classes o s depending on how each exchanges matter and with its surroundings. Reactants and products are the ; everything else is the surroundings. We track the low into and out o the. Loss o rom the to the surroundings has a negative sign. surroundings - out in + Open System Matter Matter exchanged Heat or work exchanged Closed System Matter not exchanged Heat or work exchanged Isolated System No matter exchanged No heat or work exchanged Adibatic Increase o the rom the surroundings has a positive sign. q q The internal, E, o a is a bulk property o matter that on a microscopic level represents all within the. On a microscopic scale the internal is: Translational kinetic. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons. Nuclear Electrochemical The internal o a can only change by exchange o heat (q) and mechanical work (w) with the surroundings. Add Energy +q = heat added to the +w = work done on the Internal!ESys = q + w Surroundings Internal Energy E System Rotation, Vibrations, Electronic, translational o molecules Surroundings Heat Mechanical Work Lose Energy -q = heat lost rom the -w = work done by the We can not measure the absolute internal o a...we only measure changes in internal ---we call this a change in state. Measurements are relative to something. What s the tallest mountain in the world? Change in Internal Energy Heat!ESYS = Einal - Einitial = q + w Work Eproducts - Ereactants = q + w 1. Mount Everest is king i measured rom sea level (8,848 meters) 2. Mauna Kea is when measured rom the depths o the Paciic Ocean loor. 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 arthest rom the earth's core. change in state rom initial conditions beore chemical reaction to inal condition ater a chemical reaction.

4 Chemical reaction causes a change in internal that is accounted or completely by measuring the exchange o heat (q) and mechanical work (w) with the surroundings. A can do work on the surroundings or the surroundings can do mechanical work on the.!esys = q + w = q - P!V w = P V surroundings +/-w surroundings height Initial State +/-q Final State w = F orce Distance F orce Area w = height Area w = P (Area height) w = P V The internal, U can be changed only by exchanging heat (q) or work (w) with its surroundings. Surroundings in + out - +q = heat -q = heat lost added rom the +w = work done on the System!E Surroundings!ESys = q + w Internal Heat Mechanical work -w = work done by the It s a state unction! Determining A Change in Internal Energy When gasoline burns in a car engine, the heat released causes the products CO 2 and H 2 O to expand, which pushes the pistons outward. Excess heat is removed by the car s cooling. I the expanding gases do 451 J o work on the pistons and the loses 325 J to the surroundings as heat, calculate the change in (!E) in J, kj, and kcal. SOLUTION: kj -776J 103 J q = J and w = J!E = q + w = -325 J + (-451 J) = -776 J kcal = kJ k 4.18k = kcal Announcements Hour Exam II: Date: Sept 5 Time: 6:00-7:30PM (40 mult. choice 2 Long problems Room: WATCH BLOG Chapter 4-6 Tips: 1) Look at learning objectives at the end o each Chapter. 2) Know the homework problems and the type o problems done in class. 3) Note the skipped material posted on blog 4) Practice the Silberberg quizzes in the links o blog Chemists don t use internal to monitor changes in the lab. Instead, we deine and use a related quantity called enthalpy.!esys = q + w = q - P!V CASE 1 At constant pressure:!e = q P - P!V q P =!E + P!V CASE 2 At constant volume:!v=0!e = q V - P!V!E = q V!H = q P =!E + P!V

5 An Exothermic reaction releases heat to the surroundings,!h < 0. 2H 2 (g) + O 2 (g) H 2 O (g) heat + 2HgO (s) heat + H 2 O (s) 2H 2 O (l) + heat H 2 O (l) + heat 2Hg (l) + O 2 (g) H 2 O (l)!h = <0 heat written as a product (given o) An Endothermic reaction absorbs heat rom the surroundings into the,!h > 0.!H = >0 heat written as a reactant (consumed) There are 3-Cases When!H =!E!H = qp =!E + P!V I. In s 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 o moles o gas does not change rom reactants to products; no work is done. N2(g) + O2(g) ==> 2NO(g)!V = 0 and!h =!E 3. Reactions in which the number o moles o gas does change but q is >>> P!V. Calculate the work (in kilojoules) done during a reaction in which the volume expands rom 12.0 L to 14.5 L against an external pressure o 5.0 atm. Watch the conversion o PV units to joules. P = 5.0 atm!v = ( ) L = 2.5 L w = P!V w = -(5.0 atm)(2.5 L) = L"atm (-12.5 L"atm)(101 J/L"atm) = -1.3 # 10 3 J = -1.3 kj Note PV has units o! J = 1 L atm (a) (b) A sample o nitrogen gas expands in volume rom 1.6 L to 5.4 L at constant temperature. What is the work done in joules i the gas expands (a) against a vacuum and (b) against a constant pressure o 3.7 atm? w = -P!V!V = 5.4 L 1.6 L = 3.8 LP = 0 atm W = -0 atm x 3.8 L = 0 L atm = 0 joules!v = 5.4 L 1.6 L = 3.8 LP = 3.7 atm w = -3.7 atm x 3.8 L = L atm w = L atm x J 1L atm = J Imagine a chemical reaction that results in a change in both volume and temperature as shown below. A state unction is any math unction that depends only on the initial and inal states and not the path taken to between the two states. All are state unctions! h PE = mgh Final State!E = E inal - E initial!p = P inal - P initial 1) Has any work been done? I so, is its sign positive or negative? 2) Has there been an enthalpy change? I so, what is the sign o "H? Is the reaction exothermic or endothermic?!v = V inal - V initial!t = T inal - T initial!h = H inal - H initial 0 Initial State The potential o hiker 1 and hiker 2 is the same even though they took very dierent paths to get to the mountain top. PE only depends on the initial and inal heights only.

6 Why Do We Care About State Functions? A change in a state unction tells us that the unction has one unique value between any two arbitrary states. This act is why thermodynamics is so powerul and useul. A thermochemical equation shows the enthalpy change (!H) o a chemical reaction and also the reactions stoichimetric actors. C3H8(g) + 5O 2 (g) #$ 3CO 2 (g) + 4H 2 O(l)!H = kj Final State Arbitray State Initial State mgh2!h h = 1 1)!H <0 = REACTION IS EXOTHERMIC--gives o heat 2) 2043 kj are released and can be written as product C3H8(g) + 5O 2 (g) #$ 3CO 2 (g) + 4H 2 O(l) kj 3) more useul stoichiometric actors or calculations 1 mol C3H8(g) = kj 5O 2 (g) = kj Stoichiometeric actors 3CO 2 (g) = kj 4H 2 O(l) = kj Using the Heat o Reaction (!H rxn ) to Find Amounts The major source o aluminum in the world is bauxite which is mostly aluminum oxide. Its thermal decomposition can be represented by: Using the thermochemical equation calculate how much heat is evolved (or what is the enthalpy) when 266 g o white phosphorus (P 4 ) is combusted in air? The molar mass o P4 is g/mol. Al 2 O 3 (s) 2Al(s) + 3/2O 2 (g)!h rxn = 1676 kj I aluminum is produced this way, how many grams o aluminum can orm when x 10 3 kj o heat is is used to decompose the oxide? P 4 (s) + 5O 2 (g) P 4 O 10 (s)!hrxn = kj/mol!h = H (products) H (reactants) SOLUTION: g Al = kj 2 mol Al 1676 kj g Al 1 mol Al = 32.2 g Al H rxn = 266. g P 4 1 mol P kj = 6470 kj g P 4 1 mol P 4 Rules For Manipulating Thermochemical Equations. 1. The stoichiometric coeicients always reer to the number o moles o a substance---it s extensive! Look at the H 2 O (s) H 2 O (l)!h = 6.01 kj coeicients! 2. I you reverse a reaction, the sign o!h changes H 2 O (l) H 2 O (s)!h = kj 3. I you multiply both sides o the equation by a actor n, then!h must change by the same actor n. 2H 2 O (s) 2H 2 O (l)!h = 2 x 6.01 = 12.0 kj 4. The physical states o all reactants and products must be speciied in thermochemical equations. H 2 O (s) H 2 O (l)!h = 6.01 kj H 2 O (l) H 2 O (g)!h = 44.0 kj Thermochemistry is concerned primarily with calculating!hrxn (the heat o reaction). 1 Multiple Thermochemical Equations To Rearrange-- Use Rules! 2-Ways or Problem Types For Hess Law Enthalpy Calculations 2 Use!H data 1) standard state 2) ormation equation 3) Calculation o!h rxn!h rxn = % ni!hi (products) - % mi!hj (reactants)

7 Hess s Law: The enthalpy change or any reaction is equal to the sum o the enthalpy changes or any individual step in the reaction. CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2!H1 = kj CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O!H2 = kj CH4 (g) + 2O2 ==> CO2 + 2H2O!H3 =? kj Because enthalpy is a state unction, its value only depends on the inal and initial state and not the path. This gives us a powerul way to predict enthalpy o many chemical reactions. H 3 =? CH4 (g) + 2O2 CO2 + 2H2O H 1 = kj CO(g) + 2H2O + 1/2 O2(g) H 2 = kj Rules For Manipulating Thermochemical Equations. 1. The stoichiometric coeicients always reer to the number o moles o a substance---it s extensive! Look at the H 2 O (s) H 2 O (l)!h = 6.01 kj coeicients! 2. I you reverse a reaction, the sign o!h changes H 2 O (l) H 2 O (s)!h = kj 3. I you multiply both sides o the equation by a actor n, then!h must change by the same actor n. 2H 2 O (s) 2H 2 O (l)!h = 2 x 6.01 = 12.0 kj 4. The physical states o all reactants and products must be speciied in thermochemical equations. H 2 O (s) H 2 O (l)!h = 6.01 kj H 2 O (l) H 2 O (g)!h = 44.0 kj Energy level diagrams are useul pictorials o the thermodynamic transitions that take place during a chemical reaction. Enthalpy, H CH4 (g) + 2O2 H 3 = -890 kj CO2 + 2H2O H 1 = kj CO(g) + 2H2O + 1/2 O2(g) H 2 = kj CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2!H1 = kj CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O!H2 = kj CH4 (g) + 2O2 ==> CO2 + 2H2O!H3 = -890 kj Two gaseous pollutants that orm in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmul gases through the ollowing equation: CO(g) + NO(g) CO 2 (g) + 1/2N 2 (g)!h =? Given the ollowing inormation, calculate the unknown!h: CO(g) + 1/2O 2 (g) N 2 (g) + O 2 (g) Multiply Equation B by 1/2 and reverse it. CO(g) + NO(g) CO(g) + 1/2O 2 (g) CO 2 (g)!h A = kj NO(g) 1/2N 2 (g) + 1/2O 2 (g)!h B = kj CO 2 (g) + 1/2N 2 (g) CO 2 (g)!h A = kj 2NO(g)!H B = kj!h rxn = kj Calculate the enthalpy o the ollowing reaction and draw and level diagram that summarize the thermodynamic enthalpies involved in this reaction. NO 2 (g)! NO(g) + 1/2O 2 (g)!h = kj 1/2N 2 (g) + O 2 (g)! NO2(g)!H = kj "N 2 (g) + O 2 (g)! NO + 1/2O 2 (g)!h = kj Use the properties o thermochemical equations with Hess s Law to calculate the enthalpy o the ollowing reaction using the reactions and their associated enthalpies taken rom a handbook below. FeCl 2 (s) + 1/2 Cl 2 (g) FeCl 3 (s)!h =? 1. Fe(s) + Cl 2 (g) FeCl 2 (s)!h = kj/mol 2. Fe(s) + 3/2 Cl 2 (g) FeCl 3 (s)!h = kj/mol

8 Thermochemistry is concerned primarily with calculating!hrxn (the heat o reaction). 1 Multiple Thermochemical Equations To Rearrange-- Use Rules! 2-Ways or Problem Types For Hess Law Enthalpy Calculations 2 Use!H data 1) standard state 2) ormation equation 3) Calculation o!h rxn!h rxn = % ni!hi (products) - % mi!hj (reactants) Physicists and chemists deine a standard thermodynamic standard state and denote with a superscript degree sign,!h Thermodynamic Standard State Means: 1 atmosphere pressure 25 C = K 1 Molar concentration or solutions!hrxn Conditions o P, T, [] are given in problem and stated!h rxn The superscript indicates thermodynamic standard state: P = 1 atm, T = 298, [1M] The Standard Enthalpy o Formation,!H, is the enthalpy change associated with the ormation o 1 mole o product rom its naturally occurring elements under standard state conditions. Examples: H 2 (g) + O 2 (g) #$ H 2 O(l )!H = kj/mol 3C(s) + 4H 2 (g) #$ C 3 H 8 (g)!h = kj/mol Ag(s) + 1/2Cl 2 (g) #$ AgCl(s)!H = kj/mol Note: ractional coeicients allowed Note: 1 mol product! Are amilar molecular gases as O 2, N2, F2, H2 etc as at 25 C Elemental Carbon exists as solid graphite C(graph) at 25 C. Elemental Sulur exits as S8 as a solid at 25 C Water is H 2 O(l ) in its standard state (not ice or water vapor). It is not possible to measure the absolute enthalpy. We deine an arbitrary scale with the standard enthalpy o ormation (!H ) as a reerence point. 1) The standard enthalpy o ormation,!h o any element in its most stable orm is zero.!h 0 (O 2 ) = 0!H 0 (O 3 ) = 142 kj/mol!h 0 (C, graphite) = 0!H 0 (C, diamond) = 1.90 kj/mol!h 0 (S8, rhombic) = 0 kj/mol The standard enthalpy o ormation,!h o many compounds is tabulated in many handbooks. This gives us predictive capability and is someone useul in the real world. 1) Writing Formation Equations Write balanced equations or the ormation o 1 mol o the ollowing compounds rom their elements in their standard states and include!h 0. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen sulide, H2S, a gas at standard conditions. PLAN: ---Use the table o heats o ormation or values. ---Remember 1 mole o product is ormed and ractions are allowed as coeicients.

9 PLAN: Writing Formation Equations Write balanced equations or the ormation o 1 mol o the ollowing compounds rom their elements in their standard states and include!h 0. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. Use the table o heats o ormation or values. METHOD 2: The enthalpy o any reaction,!h rxn can be obtained by using!h and the ollowing equation:!h rxn = " ni!hi (products) - " mi!hj (reactants) where % means the sum o ni is the respective stoich coeicient or i th product mi is the respective stoich coeicient or each i th reactant SOLUTION: (a) Ag(s) + 1/2Cl 2 (g) AgCl(s) (b) Ca(s) + C(graphite) + 3/2O 2 (g) (c) O 2 (g) + 1/2N2(g) NO2(g)!H 0 = kj CaCO!H 0 3 (s) = kj!h 0 = 33.9 kj Example: Suppose aa + bb cc + dd!h rx =?!H rx =!H (Products)!H (Reactants)!H rx = [c!h (C) + d!h (D)] [a!h (A) + b!h (B)] Calculate the heat o decomposition or the ollowing process using standard enthalpies o ormation ound in Appendix. What kind o reaction is this? Is given o or needed to make this reaction proceed? Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including ertilizer, dyes, and explosives. The irst step in the industrial production process is the oxidation o ammonia: CaCO 3 (s) CaO (s) + CO 2 (g)!h 0 rxn =? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g)!H rxn = % ni!hi (products) - % mi!hi (reactants)!h 0 rxn = [!H 0 (CaO) +!H 0 (CO 2 )] - [!H 0 (CaCO 3 )]!H 0 rxn = [ ] [ ] = 179 kj Reaction is decomposition and endothermic. Energy must be absorbed to proceed. PLAN: Calculate!H 0 rxn rom!h 0 values. Look up the!h 0 values in Appendix and use Hess s Law to ind!h rxn.!h 0 NH 3 (g) = kj/mol!h 0 O 2 (g) = 0!H 0 H 2 O(g) = kj/mol!h 0 NO(g) = 90.4 kj/mol Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including ertilizer, dyes, and explosives. The irst step in the industrial production process is the oxidation o ammonia: 4NH 3 (g) + 5O 2 (g) Calculate!H 0 rxn rom!h 0 values. 4NO(g) + 6H 2 O(g) Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole o benzene combusted? The standard enthalpy o ormation o benzene is kj/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l)!h 0 rxn = % n!h 0 (products) - % m!h 0 (reactants) SOLUTION:!H rxn = % m!h 0 (products) - % n!h 0 (reactants)!h rxn = [4(!H 0 NO(g) + 6(!H 0 H 2 O(g)] - [4(!H 0 NH 3 (g) + 5(!H 0 O 2 (g)] = (4 mol)(90.4 kj/mol) + (6 mol)( kj/mol) - [(4 mol)(-46.3 kj/mol) + (5 mol)(0 kj/mol)]!h rxn = -904 kj!h 0 rxn = [ 12!H 0 (CO 6!H 0 2 ) + (H 2 O) ] - [ 2!H 0 (C 6 H 6 )]!H 0 rxn = [ 12x x ] [ 2x49.04 ] = kj kj 2 mol = kj/mol C 6 H 6

10 Enthalpies (!H) o chemical reactions can be determined using the physics o heat transer a lab method called calorimetry. Physics o heat transer: the heat absorbed by a substance depends on the material s speciic heat capacity, mass and temperature change. n q i = 0 i=1 q = C!T = s m!t No heat enters Observed Physics: The amount o heat, q, transerred rom an object at higher temperature to an object at lower temperature is proportional to the dierence in temperature o the two objects! Hot Thigh Block A Heat Transer Block B Cold Tlow Heat transer only occurs when there is a temperature dierence between two bodies Heat is transerred occurs through conduction, convection or radiation. Principle 1: The amount o heat, q, transerred rom an object at higher temperature to an object at lower temperature is proportional to the dierence in temperature o the two objects! Thermometer T1 T1 q!t q = C!T C = heat capacity Thermometer T2 The heat capacity (C) o a substance is the amount o heat (q) required to raise the temperature o a quantity o the substance by one degree Celsius or Kelvin (units o J/ C or J/K or cal/ C). Heat transerred = q & (T inal - T initial ) q = C!T material dependent constant heat capacity temperature dierence between objects Object 1 Object 2 Heat Q C has units o Energy per degree T (J/ C, J/K, cal/ C, cal/k) C = q!t The heat capacity, C, o any material is ound to be proportional to mass o a substance under question. C m C m s The speciic heat (s) o a substance is the amount o heat (q) required to raise the temperature o one gram o the substance by one degree Celsius. (How thermally sensitive a substance is to the addition o!) mass speciic heat Units o s: J/g C, J/g K q = C!T C m s (1) (2) s is called the speciic heat (s) o a substance. It is the amount o heat (q) required to raise the temperature o one gram o the substance by one degree Celsius. Substituting (2) into 1 gives: Heat (q) absorbed or released: q = s m!t

11 The molar heat capacity o a substance is the amount o heat (q) required to raise the temperature o one mole o the substance by one degree Celsius. 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 o kg o iron and 1000 kg o water rom 25 to 75 C? Note that the speciic heat capacity o iron is 0.45 J/g C Heat (q) absorbed or released: q = s m!t Units: J/mol C or J/mol K Units: moles Units: C or K q Fe = kg 1000 g 1 kg (75 25 ) 0.450J C g = J q H2 O = kg 1000 g 1 kg (75 25 ) 4.18J C g = J Determine the speciic heat capacity o lead when grams o lead pellets at C are added to 50.0 ml o water held at 22.0 C. The inal temperature o the insulated calorimeter is 28.8 C. Assume the container and thermometer do not absorb heat g C 2. Add pellets to 50.0 ml 22.0 C 3. What s Up? 28.8 C sp. heat? 1. Use the law o conservation o pertains to ALL OF THE PARTICIPANTS! qpb + qwater + qcontainer = 0 n q i = 0 i=1 qpb = - qwater mpb cpb!tpb = -mwat cwat!twat 1. Conservation o is the key undamental physics! clead = mwat cwat!twat / mlead!tlead q water = mc#t = (50.0 g)(4.184 J/g C)( ) C = 1.40 x10 3 J q lead = mc!t = (150.0 g)(c)( ) C = x10 3 J c lead = 0.13 Jg -1 C -1 A 237 g piece o molybdenum, initially at o C, is dropped into an insulated container with 244 g o water at 10.0 o C. When the comes to thermal equilibrium, the temperature o the is 15.3 o C. What is the speciic heat o molybdenum? n q i = 0 i=1 qmb + qwater = 0 -qmb = -smb mmb!t = swtr mwtr!t = qwater!