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					          Section 7.2—Calorimetry &
                      Heat Capacity



Why do some things get hot more quickly than others?
Temperature

   Temperature – proportional to the
   average kinetic energy of the molecules
            Energy due to motion
            (Related to how fast the
            molecules are moving)




        As temperature           Molecules move
        increases                faster
Heat & Enthalpy

   Heat (q)– The flow of energy from
   higher temperature particles to lower
   temperature particles

   Enthalpy (H)– Takes into account the
   internal energy of the sample along with
   pressure and volume
   Under constant pressure (lab-top conditions), heat and enthalpy
   are the same…we’ll use the term “enthalpy”
Energy Units
  The most common energy units are Joules (J) and
  calories (cal)

                           Energy Equivalents

                             4.18 J   =   1.00 cal

                            1000 J    =   1 kJ

                          1000 cal    =   1 Cal (food calorie)
                                          1 kcal
  These equivalents can be used in dimensional analysis to convert units
Heat Capacity
 Heat Capacity – The amount of energy that
 can be absorbed before any mass of a
 substance has increased its temperature by
 one degree. (This changes for one substance depending on how much)

 Specific Heat Capacity (Cp) – The
 amount of energy that can be absorbed
 before 1 g of a substance’s temperature
 has increased by 1°C (This stays the same for one
 substance because it is for a set amount.)

            Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C
Heat Capacity
       High Heat Capacity                          Low Heat Capacity
   Takes a large amount of                   Small amount of energy can
   energy to noticeably                      noticeably change
   change temp                               temperature
   Heats up slowly                           Heats up quickly
   Cools down slowly                         Cools down quickly
   Maintains temp better with                Quickly readjusts to new
   small condition changes                   conditions
  •A pool takes a long time to warm up and remains fairly warm over night. (high)
  •The air warms quickly on a sunny day, but cools quickly at night. (low) This is why
  a desert has extremes of hot and cold due to a lack of water which can maintain the
  temp.
  •A cast-iron pan stays hot for a long time after removing from oven. (high)
  •Aluminum foil can be grabbed by your hand from a hot oven because it cools so
  quickly (low)
What things affect temperature change?

 Heat Capacity of substance
  The higher the heat capacity, the slower the
   temperature change
 Mass of sample
  The larger the mass, the more molecules there are to
   absorb energy, so the slower the temperature change
                                    Specific heat capacity of
                                    substance

                  H  m  C p   T

 Energy added     Mass of sample          Change in temperature
 or removed
Positive & Negative T
 Change in temperature (T) is always T2 – T1
  (final temperature – initial temperature)
  If temperature increases, T will be positive
      A substance goes from 15°C to 25°C.
      25°C - 15°C = 10°C
      This is an increase of 10°C
  If temperature decreases, T will be negative
      A substance goes from 50°C to 35°C
      35°C – 50°C = -15°C
      This is a decrease of 15°C
Positive & Negative H

Energy must be put in for temperature to
 increase
  A “+” T will have a “+” H
Energy must be removed for temperature
 to decrease
  A “-” T will have a “-” H
Let’s Practice #1

            Example:
    How many joules must be
   removed from 25 g of water
      at 75.0°C to drop the
    temperature to 30.0°? Cp
       water = 4.18 J/g°C
Let’s Practice #1

              Example:                       H  m  C p   T
      How many joules must be
    removed from 25.0 g of water       H = change in energy
        at 75.0°C to drop the          m = mass
      temperature to 30.0°? Cp         Cp = heat capacity
         water = 4.18 J/g°C            T = change in temperature (T2 - T1)



                 
H  25 .0 g   4.18           
                        J    30 .0  75 .0 C        
                        g C


                                        H = - 4702.5 J
                                       H = - 4.70 x 103 J
Example

              Example:
    If 285 J is added to 45 g of
     water at 25°C, what is the
  final temperature? Cp water
            = 4.18 J/g°C
Example

                Example:                                    H  m  C p   T
      If 285 J is added to 45 g of
       water at 25°C, what is the                     H = change in energy
    final temperature? Cp water                       m = mass
              = 4.18 J/g°C                            Cp = heat capacity
                                                      T = change in temperature (T2 - T1)



                            
                                  
 285 J  45 g   4.18 J    T2  25  C                        285 J
                                                                                      
                                                                                      T2  25  C   
                         g C
                                                            45 g   4.18 J g C 
                                                                                  
                                                                                  


                      285 J
                                       25  C  T2
             45 g   4.18 J g C 
                                   
                                                           T2 = 26.515°C
                                   
                                                           T2 = 27°C
Let’s Practice #2

            Example:
   If the specific heat capacity
  of aluminum is 0.900 J/g°C,
  what is the final temperature
  if 437 J is added to a 30.0 g
         sample at 15.0°C
Let’s Practice #2

                Example:                                       H  m  C p   T
       If the specific heat capacity
      of aluminum is 0.900 J/g°C,                        H = change in energy
      what is the final temperature                      m = mass
      if 437 J is added to a 30.0 g                      Cp = heat capacity
             sample at 15.0°C                            T = change in temperature (T2 - T1)



                               
                                      
437 J  30 .0 g   0.900 J    T2  15 .0 C                         437 J
                                                                                             
                                                                                             T2  15 .0 C   
                            g C                              30 .0 g   0.900 J g C 
                                                                                         
                                                                                         


                         437 J
                                           15 .0 C  T2
             30 .0 g   0.900 J g C 
                                       
                                                                 T2 = 31.2°C
                                       
Calorimetry
Conservation of Energy

   1st Law of Thermodynamics – Energy
   cannot be created nor destroyed in
   physical or chemical changes

   This is also referred to as the Law of Conservation of Energy


   If energy cannot be created nor destroyed, then energy lost by the
   system must be gained by the surroundings and vice versa
Calorimetry
   Calorimetry – Uses the energy change
   measured in the surroundings to find
   energy change of the system
   Because of the Law of Conservation of Energy,
   The energy lost/gained by the surroundings is equal to but opposite
   of the energy lost/gained by the system.

                       Hsurroundings = - Hsystem

              (m×Cp×T)surroundings = - (m×Cp×T)system

   Don’t forget the “-” sign on one side
   Make sure to keep all information about surroundings together and all
   information about system together—you can’t mix and match!
Two objects at different temperatures


    Thermal Equilibrium – Two objects at
    different temperatures placed together
    will come to the same temperature

 Why is this important?

 So you know that T2 for the system is the same as T2 for the surroundings!
An example of Calorimetry
                      Example:
    A 23.8 g piece of unknown metal is heated to
     100.0°C and is placed in 50.0 g of water at
     24°C water. If the final temperature of the
     water is 32.5°,what is the heat capacity of
                      the metal?
  An example of Calorimetry
                                       Example:
                A 23.8 g piece of unknown metal is heated to 100.0°C
                and is placed in 50.0 g of water at 24.0°C water. If the
                final temperature of the water is 32.5°,what is the heat
                                 capacity of the metal?
Metal:            Water:
M = 23.8 g        M = 50.0 g
T1 = 100.0 °C     T1 = 24.0 °C                                       H metal  H water
T2 = 32.5 °C      T2 = 32.5 °C
Cp = ?            Cp = 4.18 J/g°C
                                                   m  C   p    T metal  m  C p  T water

                                                                 
                                                                                                             
                                                                                                              
                           23 .8 g  C p  32 .5 C  100 .0 C   50 .0 g  4.18 J   32 .5 C  24 .0 C 
                                                                                    gC                       
                                              
                                                                                        
                                              50 .0 g  4.18 J   32 .5 C  24 .0 C           
                                        Cp                                             
                                                                 gC
                                                                     
                                                                      
                                                      23 .8 g  32 .5 C  100 .0 C
                                                                                
                                                                                            
                                                                                            Cp = 1.04 J/g°C
Let’s Practice #3
                       Example:
    A 10.0 g of aluminum (specific heat capacity
       is 0.900 J/g°C) at 95.0°C is placed in a
    container of 100.0 g of water (specific heat
    capacity is 4.18 J/g°C) at 25.0°. What’s the
                  final temperature?
Let’s Practice #3
                             Example:
          A 10.0 g of aluminum (specific heat capacity
             is 0.900 J/g°C) at 95.0°C is placed in a
          container of 100.0 g of water (specific heat
           capacity is 4.18 J/g°C) at 25.0°C. What’s
                      the final temperature?

                                                H metal  H water
   Metal:
m = 10.0 g
                                 m  Cp  T metal  m  Cp  T water
                                              T2  95 .0 C   100 .0 g  4.18 J   T2  25 .0 C 
T1 = 95.0°C                                                                                              
                 10 .0 g  0.900 J
T2 = ?                               g C                                           gC                   
Cp = 0.900 J/g°C
                         9.0  T2  95  C   418  T2  25 .0 C 
   Water:
m = 100.0 g                                                                             11305
T1 = 25.0°C
                         9.0T2  855  418 T2  10450                             T2 
                                                                                        427.0
T2 = ?
Cp = 4.18 J/g°C          427 .0T2  11305
                                                                                     T = 26.5 °C
                                                                                      2

				
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posted:3/13/2012
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