6.4-6.6

Quantifying heat and work One way transfer of heat- from hotter object to cooler- no additional net transfer of heat @ thermal equilibrium Q (heat) and ΔT are directly proportional, with C (heat capacity) as the constant of proportionality Cs= Specific heat capacity of a substance = the amount of heat required to raise 1 gram of the substance 1 degree Celsius (units are J/g * degree C), molar heat capacity raises 1 mol 1 degree C Heat capacity depends on the amount of a substance Q= mc ΔT, where q is heat (J), m is mass (g), Cs is specific heat capacity in (J/g * degree C), and T is change in temperature (Degrees Celsius)

Practice Question How much heat is absorbed by a penny as it warms from -8 degrees Celsius to 37.0 degrees Celsius? The penny’s mass is 3.10 grams. q= mc ΔT = 3.10g * (.3845 J/g * degree C) *( (37 degree C- (-8 degree C)) = 53.6 Joules

Thermal Energy Transfer If 2 substances are thermally isolated from everything else, the heat lost by one equals the heat gained by the other One substance as the system, the other as the surroundings (like a block submerged in water, the block is the system, the water is the surroundings) qsys = -qsurr

Practice Problem A 32.5 g cube of aluminum at 45.8 degrees Celsius is submerged in 105.3 g of water at 15.4 degrees C. What is the final temperature of both substances at thermal equilibrium? (Hint- final temperatures same) qAl = -qH2o mAl * Cs, Al * ΔTAl = -mH2o * Cs H2o * ΔT H2o 32.5 g * .902 J/g*C* ΔTAl = -105.3 g * 4.18 J/g *C * ΔTH2o 29.315 * ΔTAl = -440.15 * ΔTH2o ΔTAl = -15.014 * ΔTH2O Tf - TiAl = -15.014 (Tf- TiH2O) Tf= (-15.014*Tf) + 15.014TiH2O + TiAl Tf+ 15.014Tf = 15.014TiH2O + TiAl 16.014Tf= 15.014TiH2O + TiAl Tf= (15.014* 15.4 C) + 45.8 C/ 16.014 = 17.3 C

Pressure- Volume Work Work = Force acting through a distance Pressure-volume work occurs when a force is caused by a volume change against an external pressure (like in a combustion engine) w= -P ΔV (units are L*atm), this needs to be converted to Joules (1 L*atm = 101.3 J) The work caused by an expansion of volume is the negative of the pressure that the volume expands against multiplied by the change in volume

Practice Problem If you inflate a balloon from a volume of .100 L to 1.85 L against an external pressure of 1 atm, how much work is done in Joules? w= -P ΔV = -1 atm (1.75 L) =-1.75 L* atm -1.75 L * atm * (101.3 J/ 1 L*atm) = -177 J

Measuring ΔE for Rxns: Constant Volume Calorimetry ΔE = q+w If reaction is carried at constant volume, w=0, so ΔErxn= q (heat evolved or given off) Calorimetry= measuring thermal energy exchanged between the reaction and the surroundings The magnitude of the temperature change in surroundings depends on the magnitude of ΔE and the heat capacity Bomb calorimeter measures Delta E for combustion reactions qcal= Ccal * Delta T, where Ccal is heat capacity of entire calorimeter assembly, and qcal is the Heat absorbed by entire calorimeter qcal= -qrxn (because if no heat escapes the calorimeter, the heat gained by calorimeter equals the heat lost by the reaction) qrxn= qv= Delta Erxn (qv is heat @ constant volume) To find Delta Erxn per mole of a reactant, divide by the number of moles that actually reacted

Practice Question When 1.010 g of sucrose (C12H22O11) undergoes combustion, the temp rises from 24.92 C to 28.33 C. Find Delta Erxn in kJ/mol sucrose. The heat capacity of the bomb calorimeter is 4.90 kJ/ Degrees C. (qcal → qrxn→ Delta Erxn→ kJ/mol) Delta T = 3.41 Degrees C qcal = Ccal * Delta T = 4.90 kJ/ C * 3.41 Degrees C= 16.7 kJ qcal=-qrxn, so qrxn = -16.7 kJ Delta Erxn = qrxn/ mol sucrose -16.7 kJ/ (1.010 g sucrose * (1 mol sucrose/ 342.3 g sucrose) = -5.66 * 10^3 kJ/mol sucrose, answer negative because the combustion reaction gives off energy

Enthalpy Energy can evolve as both heat and work (when it’s open to the atmosphere) Enthalpy is a way to just express the energy (work is irrelevant) Enthalpy (H)= E (internal energy) + PV ΔH = ΔE + PΔV ΔE is a measure of all of the energy (heat and work), while ΔH is a measure of only the heat exchanged For chemical reactions with not a lot of exchange of work, ΔH and ΔE are almost the same in value

Reactions A positive ΔH indicates that heat flows into the system= Endothermic reaction= absorbs heat from its surroundings A negative ΔH indicates that heat flows out = exothermic reaction= gives off heat Summary: ΔH is the amount of heat absorbed or evolved in a reaction under constant pressure

Practice Exothermic or Endothermic? Sweat Evaporating Water freezing Endothermic (heat needed to change from liquid to gas) Water freezing Exothermic (the freezer is removing heat from the system) Wood Burning Exothermic- releases heat

Stoichiometry ΔHrxn is the heat of reaction/ enthalpy of reaction, and it is the enthalpy change The amount of heat generated or absorbed depends on the amounts of reactants The magnitude of the enthalpy of reaction is for the stoichiometric amounts of reactants and products for the reaction as written

Practice Problem A gas tank in a BBQ contains 13.2 kg of propane (C3H8). Calculate the heat in kJ associated with the complete combustion of all the propane in the tank. C3H8 (g) + 5O2(g) → 3CO2(g) + 4H2O (g), and Delta H = -2044 kJ 13.2 kg C3H8 * (1000g/1kg) * (1 mol C3H8/ 44.09 g C3H8) * (-2044 kJ/1 mol C3H8) -6.12 x 10^5 kJ