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Tuesday February 20, 2024   Day 30
The First Law of Thermodynamics!!!

Textbook Readings:

6.1: Light the Furnace:
       The Nature of Energy and Its Transformations

6.2: The Nature and Types of Energy

6.3: The First Law of Thermodynamics
Course Lecture

6.1  pdf  Video Heat Transfers
Energy & Chemistry: Crash Course Chemistry

 Internal Energy

Objectives

1. State the First Law of Thermodynamics in non-technical terms.

2. Identify those things that gain and lose heat in energy conversion situations.

3. Apply the First Law of Thermodynamics in energy conversion situations and determine missing
    heat energy values.

Homework Problems

30.1. The First Law of Thermodynamics is nature's way of saying the energy "bank account" must balance
    at the end of the month. 

    In other words:  Energy can't be lost or destroyed but can only change form. 

    Consider the following over-simplified budgeting situation:

         Your income                           = $2000/month. 

         Your costs:      Rent                = $850/month
                                   Car payment = $150/month
                                   Food                = $300/month
                                   Credit Card   = $500/month
                                   Savings Acct  =         $ ?
                                  
          Now, use the numbers above to determine the amount of money that can be put into a savings
          account at the end of each month.

30.2.  In "calorimetry experiments", a chemical reaction releases heat energy inside a device known as
     a "calorimeter."   The calorimeter is a well insulated, water-filled cup equipped with a lid,
     stirrer, and thermometer.  Most heat is absorbed by the calorimeter and its components.
     However, the device isn't perfect and some heat energy leaks into the "surroundings" (everything
     that isn't the calorimeter ... e.g. tabletop, surrounding air, etc).

     Use the following information to determine how much heat energy is released into the surroundings.
     
           Heat released by chemical reaction           =    2000 kJoules

           Heat absorbed by the surrounding water  =  850 kJoules
           Heat absorbed by the thermometer            =   150 kJoules
           Heat absorbed by the stirrer                       =   300 kJoules
           Heat absorbed by the calorimeter cup       =   500 kJoules
           Heat absorbed by the "surroundings"       =      ?  kJoules
              
30.3.    In a calorimetry experment, the following heat values were obtained.  How much heat energy
       was produced by the chemical reaction assuming that heat lost to the surroundings is
       negligable (i.e. zero ... often a safe assumption)?

   
           Heat absorbed by the surrounding water  =  1250 kJoules
           Heat absorbed by the thermometer            =    50 kJoules
           Heat absorbed by the stirrer                       =   130 kJoules
           Heat absorbed by the calorimeter cup       =   325 kJoules

30.4.    Hot water at 85oC is poured into a ceramic coffee cup filled with cool water at 25oC.
        a.   What loses heat energy?
        b.   What gains heat energy?
        c.   Click and drag to see part "c" of this question below....

               If the hot water loses 10,000 Joules of heat energy and the cold water gains 6300 Joules
               of heat energy, how much heat energy was gained by the ceramic coffee cup assuming
               no energy is lost to the surroundings?
       
30.5.   Automotive engineers often claim that today's engines are approximately 35% efficient.
      Actually, an internal combustion engine is 100% efficient in converting the energy of the
      gasoline into other forms (heat, sound and the motion of the automobile).

      What is the engineer really saying when they claim 35% efficiency for the engine?


30.6.    When conventional automobiles brake and stop, their kinetic energy is converted into
       heat energy by the brake pads and rotors.  This heat energy is released into the surroundings.

       "Regenerative braking" is different technique minimizes energy loss and boosts the range
       of electric automobiles.  It is not available on engine driven cars.

        What are the key components of regenerative braking and how does the system work to improve
         the range and efficiency of the automobile?
        (You won't find this in the text book readings ...have at it!  :))

Click and drag the region below for correct answers
                   
30.1  $200 can go into the savings account monthly.
30.2   200 kJ  of heat energy is lost to the surroundings.
30.3   1755 kJ of heat energy is produced by the chemical reaction
30.4  a. The hot water loses heat energy and its temperature drops.
         b. The cold water gains heat energy and its temperature increases.
          ALSO...since the ceramic cup initially was in contact with the cold water, it gains heat energy too.
          c.  3700 J of heat energy is gained by the ceramic coffee cup

30.5 Automotive engineers are saying that 35% of the gasoline's chemical potential energy is converted
    into the motion of the automobile.  The remaining 65% of the gasoline's energy is converted to
    heat that is lost to the surroundings.  Note, that some of this waste heat does come in handy during
    the Minnesota winters as it is used to heat up the interior of the car and keep us warm.

30.6 Regenerative braking occurs when the car's drive motors are instead used as generators.
     This happens automatically when the driver presses on the brake pedal.
     In this mode, regenerative braking slows the automobile and converts the car's kinetic energy
     into electricity that charges an on-board battery that is  later used to get the car moving again.
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Wednesday February 21, 2024   Day 31
Heat and Work

Textbook Readings:

6.4: Quantifying Heat and Work
Course Lecture

6.2  pdf  Video   Heat and Work
Heat and Temperature

 		The First Law of Thermodynamics:
 Internal Energy, Heat, and Work

Objectives

1.  Describe how work (w) and heat (q) affect the internal energy change that takes place for
     chemical reactions.

2.  Identify chemical reactions that do significant amounts of work.

3.  Give examples of what's known as a "state function."

4.  Provide all relavent details of "endothermic" and "exothermic" reactions.

5.  Calculate work done given pressure and volume information

6.  Calculate change in energy given work and heat information.

Homework Problems

31.1. When work is done by a chemical reaction, it  produces a gas that pushes  against
    the atmosphere as the gas expands. 
    Which of the following chemical reactions does significant work when it occurs in an open container?

     a.  4 HNO3(aq) + Cu(s)   →   Cu(NO3)2(aq) + 2 H2O(l) + 2NO2(g)
     b.  Cu(OH)2(s)   →   CuO(s) + H2O(l)
     c.  CuO(s) + H2SO4(aq)   →   CuSO4(aq) + H2O(l)
     d.  CaCO3(s)    +   2HCl(aq)    →    CaCl2(aq)    +    H2O(l)    +    CO2(g)

31.2. The internal energy of a system is affected by the heat it releases and the work that it does.
    For example, as you go throughout your day, you move & work and you release heat energy.
    Both of these lower the internal energy of your body.
    How do you re-plenish your body's internal energy?

31.3.  What are the meanings of the terms "exothermic" and "endothermic"  ?

31.4.  What is a "State Function"  ?

31.5.   The work done by a chemical system is defined as:   wsys =   -Pext ΔVsys
      a.  Is  wsys  positive or negative if the system expands  (i.e. Vfinal > Vinitial)?
      b.  How much work in kJ  is done when a piston expands from a volume of 13.27 L to 76.55 L
            against a pressure of 14.89 atm?     w = - P Δ V            Useful Info:   101.33 J = 1 L atm

31.6. . What is the energy change in kJ  if a chemical system releases  32.146 kJ of heat into the
           surroundings  while at the same time the system expands from 1.465 L to 3.687 L
           against a  pressure of 3.64  atmospheres?

Click and drag the region below for correct answers

31.1  Only reactions a and d have a gas as a product.  In these two cases, the gas produced will push back
    the atmosphere and the reaction will do work.

31.2 To compensate for the heat your  body releases and the work you do, you must consume energy
    food, which is to say your internal energy is replenished by your consumption of chemical
     potential energy.
   
31.3 Exothermic refers to a reaction that releases heat energy into the surroundings.  Heat is considered
    a product and is sometimes explicitly written that way.  Exothermic reactions have negative
    ΔH  values. Will discuss ΔH tomorrow.

    Endothermic refers to a reaction that takes heat energy from the surroundings.  Heat is considered
    a reactant and is sometimes explicitly written that way.  Endothermic reactions have positive
    ΔH  values.  Will discuss ΔH tomorrow.

31.4 A "state function" is a quantity that doesn't depend on anything but initial and final values.
    For example, your bank account balance is a state function.  At the end of the month, all that
    matters as you assess your finances is your initial and final account balances.  How you got there,
    (i.e. your spending) doesn't matter... just how the inital and final amounts compare.

31.5  a. Δ V = Vfinal  -   Vinitial     Since the system expands... Vi < Vf    and thus  Δ V  is positive.
         However, since  wsys =   -Pext ΔVsys  ,  work is negative since both   Pext  and  ΔVsys are positive.

      b.   - 942.2 L atm =   - 95.5 kJ
           
31.6 The system is losing heat energy and doing work on the surroundings.  Both result in the
      system losing energy.  The result is a 32.966 kJ loss in system energy.            
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Thursday February 22, 2024   Day 32
Enthalpy: Heat at Constant Pressure

Textbook Readings:

6.6: Enthalpy: The Heat Evolved in a
       Chemical Reaction at Constant Pressure
Course Lectures:

6.3  pdf  Video     State Functions and Enthalpy
6.4  pdf  Video     Enthalpy Changes
6.5  pdf  Video     Enthalpy In Chemical Reactions
Crash Course Chemistry: Enthalpy of Reaction

 		Enthalpy of Reaction

Objectives

1. Describe the significance of enthalpy and how & why it doesn't depend on chemical work.

2. Identify exothermic and endothermic reactions via their ΔH values.

3. Explain the significance of enthalpy's units:  kJ per mole.

3. Given a reaction's ΔH, calculate the heat released or consumed for any reactant
     mole or gram value.

Homework Problems

32.1.  Enthalpy changes (ΔH), refer to changes in heat energy when work (wsys =   -Pext ΔVsys)
     is small or zero and can be ignored. Thus, heat (q) is all that matters.

     What do each of the following ΔH's refer to?
     a.  ΔHvap           b.  ΔHfus            c. ΔHrxn           d. ΔHcombustion   

32.2.   What is the sign of ΔHrxn if the reaction is exothermic?   ...endothermic?

32.3.    The ΔHcombustion for methane (CH4) is -802.3 kJ/mol. 

       Notice that ΔH values are given on a "per mol"  basis.   Thus, enthalpy values are
       considered "extensive" properties in that they depend on the amount of substance.

       a.  How much heat energy would be released by the combustion of 5.00 grams of methane
       b.  How does the system's energy change during the combustion process?
       c.  How does the surroundings energy change as a result of this combustion?

32.4.    Usually, we'll write ΔH values to the right of reactions like this...

                       H2O(s)     →    H2O(l)          ΔHfus =  +  6.02 kJ/mol

       a.  Is this an exo or endothermic reaction?
       b.  How much heat energy would be required to melt a moderate iceburg (1.00 x 109 kg) for
            drinking water?

       c.  Use dimensional analysis to convert the energy value in part b into $.

            Useful information:  3600. kJ  =  $ 0.1409     (electrical energy cost average value for MN)

       d.   It's been suggested that melting iceburgs might be a source of drinking water in arid areas.
             Is this a feasable idea?

32.5.    Given:          H2O(l)     →    H2O(g)          ΔHvap =  +  40.8 kJ/mol

       a.   How is this reaction different from the reaction in #4?
       b.   Does it require more heat energy to melt ice or vaporize liquid water on a per mol basis?
       c.   Physically, what explains the large difference between ΔHvap  and ΔHfus ?
       d.   In our labs, we use distilled water made by using electricity to boil tap water which is
             then  re-condensed.  During the process, impurities are left behind (i.e. the distilled
             water is mostly mineral-free).

              Our largest distilled water reservoir has a capacity of  110. liters.  Convert this value to grams
              and moles using density and molar mass values.  Then, determine how much it costs to make
              110. L of distilled water.

Click and drag the region below for correct answers

32.1.   a.  ΔHvap     refers to the change in enthalpy and is the heat energy required to vaporize one mole
                          of liquid substance.  This will  always be a positive value as the process requires heat
                          energy and is endothermic.

           b.  ΔHfus     refers to the change in enthalpy and is the heat energy required to melt one mole
                               of solid substance.  This will  always be a positive value as the process requires heat
                               energy and is endothermic.

            c. ΔHrxn     This is a general reference to enthalpy changes for any reaction    

            d. ΔHcomb   This is the enthalpy change for a combustion (burning) reaction.  Combustion reactions
                                are exothermic and thus ΔHcomb  is always a negative value.

32.2.    ΔHrxn < 0   (negative)   for exothermic reactions and
            ΔHrxn > 0   (positive)    for endothermic reactions.

32.3.    a. -250. kJ    
            b. The system's change is -250. kJ as it releases heat energy into the surroundings
            c.  The surroundings change is +250. kJ as they absorb the heat produced by the reaction.

32.4. a.  Endothermic    b. 3.3416 x 1011 kJ     c. 13.1 million dollars   
         d.   There's a lot to consider here.  Electricity is expensive and there may be less expensive (free?)
               energy sources available like solar that could be used to melt the ice.  Also, energy will be
               consumed towing the ice burg from it's original location to areas where the water is needed.

32.5.  a.  This reaction isn't "melting ice" but rather vaporizing liquid water to produce steam.

          b.   It takes approximately 6.8 X more energy to vaporize one mole of water than it takes to melt
                 one mole of ice.

          c.   When you melt ice, you're providing the molecules with enough energy to break free of the
                 crystal lattice.  The liquid molecules are still close together but free to move.

                  When you vaporize water, you are pulling water molecules apart from one another and
                  separating them over great distances (that's a gas).  Overcoming the attractive forces
                  between water moleules in this way requires a lot of energy.

          d.    $9.75 required to make 110 L of distilled water.
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Friday February 23, 2024   Day 33
Constant Pressure Calorimetry: The Coffee Cup Calorimater

Textbook Readings

6.7: Constant Pressure Calorimetry:
        Measuring ΔH for Chemical Reactions
Course Lectures

6.6  pdf  Video   Calorimetry
6.7  pdf  Video   Calorimetry: Mixing Solutions

Objectives

1. Determine heat lost and gained using
     q = mcΔT   
   
2.  Determine heat gained by calorimeter
     assembly and calculate a calorimeter
     constant. 

3. Determine total heat gained in a calorimetry
    experiment.

4.  Calculate the ΔHrxn  given heat and mole
      information.

 		 Heat Capacity, Specific Heat, and Calorimetry

Homework Problems

33.1   The heat absorbed or released by a substance is determined by the relationship (know this) :
q   =   m  ✕  c  ✕  ΔT

q = heat energy lost or gained
m =  mass (g)
ΔT = Tf - Ti
c = specific heat ...depends on substance

          a. How much heat (kJ) is absorbed when 500. grams of water is warmed
               from 25.0 oC to 35.0 oC   (Specific heat for water is 4.184 J/goC)

          b.  How much heat (kJ) is absorbed when 500. grams of copper is warmed
                from 25.0 oC to 35.0 oC   (Specific heat for copper  is 0.385 J/goC)
   
          c.  Based on your answers to questions 1 and 2, which material (copper
                or water) requires more energy per gram for the same temperature change?
                Which material would undergo a larger temperature change with the addition
                 of the same amount of heat energy?
33.2    A 75.0 gram piece of copper is heated to 47.00 oC.  The hot copper is then placed
           in a styrofoam cup calorimeter filled with 500.0 mL of water initially at 24.95 oC.
           After a brief time, the copper and water reach the same temperature:  25.20 oC.

           a. Why must the copper and the water have the same final temperature?

           b.  Use q =  mc ΔT  to calculate the heat lost by the copper.

           c.  Use q =  mc ΔT Calculate the heat gained by the water.

           d.  How does the heat lost by the copper compare to the heat gained by the water?

33.3   In 33.2 we determined that something, other than the water, was absorbing heat energy.
          In our calorimeter experiments we'll assume the combination of the calorimeter cup,
          lid, thermometer and stir bar,  are also responsible for absorbing heat energy.

          The calorimeter constant, Ccal, accounts for all parts of the calorimeter and the heat that
          they absorb.   The equation that describes the heat gained by the calorimeter is:

qcal   =   Ccal  ✕  ΔT

            where ....                   qcal is the heat absorbed by the calorimeter

                                       Ccal is the "calorimeter constant" which is independent of mass
                                      (Ccal lumps together contributions from all parts of the calorimeter)

                             ΔT is the temperature change of the calorimeter (same as water's ΔT)

            Use the equation above to determine the calorimeter constant, Ccal , for problem 33.2.

33.4    A student determine mixes 100.0 g of  water at 58.5 °C with 100.0 g of water at 22.8 °C in a
           calorimeter cup.   The final temperature of of all water in the calorimeter is 39.7 °C.
           Calculate the heat lost by the hot water, heat gained by the cold water, heat difference and
           calorimeter constant (J/°C) for this experiment.  

33.5    100. mL solution of 1.0 M H2SO4 (25.0°C) is mixed with 100. mL of 2.0 M LiOH (25.0°C)
           in a calorimeter.  The following exothermic reaction takes place:

                          H2SO4(aq)     +     2 LiOH(aq)     →     2 H2O(aq)     +     Li2SO4(aq)
 
           The final temperature at the end of the reaction is 26.5 °C
           The calorimeter constant is 10.3 J/°C.

           a. Use the calorimeter constant to determine the heat gained by the apparatus.

           b. Assume the final solution to be enough like water that we can use  D = 1.00 g/mL
                and c = 4.184 J/goC.   Calculate the heat gained by the final solution.
                                                      (Remember, volumes add)

           c.  Determine the total heat produced by the reaction by summing up the above
                 heat contributions.

           d.  Calculate the initial moles of H2SO4   and determine the ΔHrxn by dividing the
                total heat by the moles of H2SO4.


Click and drag below for answers:

33.1   a. 20.92 kJ       b.  1.9250 kJ
          c.   Water has a higher specfic heat and requires a lot more heat energy than copper for
                 the same temperature change.
                 For equal masses of copper and water, copper's temperature will change a lot more than
                 water's temperature for the same amount of heat energy added or removed.

35.2.   a.  Heat flows as long as the water and copper have different temperatures.  After they
                 reach the same temperature, heat flow ceases and the temperatures stabilize.
           b.  qCu     =  - 629.5  Joules   (negative because the copper loses heat to water)            
           c.   qH2O = + 523.0  Joules  (positive  because the water gains heat from copper) 
           d. The heat lost by the copper should equal the heat gained by the water as long as
                heat energy doesn't escape into the surroundings.  However, in this case, the copper
                releases more heat energy than the water absorbs.  Something else must be absorbing heat!

33.3   The calorimeter absorbs 106.5 Joules of heat energy and undergoes a 0.25oC temperature change.
          So ...   Ccal = 426 J/oC

33.4   Ccal = 794.96 J / 16.9oC =      47.04 J/oC

33.5   a. 15.45 Joules
          b.  Total solution volume = 200. mL  =   200. grams
                qsolution =  1255.2 J
          c.  qtot = 1270.65 J
          d.  molesH2SO4 =    0.100 moles       ΔHrxn  = 1.27 x 104 J/mol    

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