; Chemistry You Need to Know
Documents
User Generated
Resources
Learning Center
Your Federal Quarterly Tax Payments are due April 15th

# Chemistry You Need to Know

VIEWS: 3 PAGES: 23

• pg 1
```									          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

```
To top