Transcript Chemistry You Need to Know

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
increases
Molecules move
faster
Heat & Enthalpy
Heat (q)– The flow of energy from
higher temperature particles to lower
temperature particles
Heat and Enthalpy are the same as long as you have
constant pressure conditions. In this class q=H=Heat
Enthalpy (H)– heat changes under
constant pressure conditions.
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
required to get a sample to move up by 1°C.
(This changes for one substance depending on how much)
Specific Heat Capacity (Cp) – The
amount of energy required to get 1 g of
a substance to increase 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
energy to noticeably
change temp
Small amount of energy can
noticeably change
temperature
Heats up slowly
Cools down slowly
Heats up quickly
Cools down quickly
Maintains temp better with
small condition changes
Quickly readjusts to new
conditions
•Do you think Air has a high or low heat capacity? Why or Why Not?
Ans: LOW – The Desert can go from 32 °F at night and 113 °F at day.
•Do you think the ocean has a high or low heat capacity? Why or Why Not?
Ans: HIGH - The heat capacity of ocean water is about four times that of air.
•Why can you grab Aluminum foil out of the oven but not an aluminum cookie sheet?
Ans: Both have the same specific heat (about 0.9J/g°C) but the cookie pan is a much greater mass.
How can we calculate the amount of energy
we put into a sample or take out?
Q  H  m  C p  T
 This shows that how much energy something has absorbed or
released is dependent upon three things
 The Specific Heat Capacity of the substance (
 The Mass of the substance
 The temperature change of the substance (T2-T1)
Specific heat capacity of
substance
H  m  C p  T
Energy added
or removed
Mass of sample
Change in temperature
How can we calculate the amount of energy
we put into a sample or take out?
Q  H  m  C p  T
 This shows that how much energy something has absorbed or
released is dependent upon three things
 The Specific Heat Capacity of the substance (
 The Mass of the substance
 The temperature change of the substance (∆T = T2 - T1)
Specific heat capacity of
substance
H  m  C p  T
Energy added
or removed
Mass of sample
Change in temperature
Let’s Practice #1
Example:
How many joules must be
removed from 25.0 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
H  m  C p  T
Example:
How many joules must be
removed from 25.0 g of water
at 75.0°C to drop the
temperature to 30.0°? Cp
water = 4.18 J/g°C
H = change in energy
m = mass
Cp = heat capacity
T = change in temperature (T2 - T1)




J
H  25.0 g   4.18

30
.
0

75
.
0
C
 
g C


H = - 4702.5 J BUT SIG FIGS WOULD BE
H = - 4.70 x 103 J
Let’s Practice #1
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 #1
H  m  C p  T
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



437J  30.0 g   0.900 J    T2  15.0 C
g C

H = change in energy
m = mass
Cp = heat capacity
T = change in temperature (T2 - T1)

437J
 15.0 C  T2
30.0 g   0.900 J g C 



437J
 T2  15.0 C
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
Qsystem = - Qsurrounding
Calorimetry
Calorimetry – the science of heat flow
between the system and the surroundings
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!
Here’s a little tip to remember with
calorimeters!
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?
NOTE TO THE TEACHER! SHOW THE
STUDENTS HOW TO DO THIS PROBLEM
ON THE BOARD BEFORE THIS REVEAL!!
IT WILL HELP THEM TREMENDOUSLY!
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
T2 = 32.5 °C
T2 = 32.5 °C
Cp = ?
Cp = 4.18 J/g°C
H metal  H water
m C
p
 T metal  m  Cp  T water


23.8g  C p  32.5 oC 100.0 oC  50.0g  4.18 J o )  32.5 oC  24.0 oC
gC







  50.0 g  4.18 J   32.5 C  24.0 C 
gC

Cp  


23.8 g  32.5 C  100.0 C


Cp = 1.10 J/g°C
Let’s Practice #2
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 #2
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 Cp  T metal  m Cp  T water
m = 10.0 g


T1 = 95.0°C
10.0 g  0.900 J
 T2  95.0 C   100.0 g  4.18 J   T2  25.0 C 
g C
gC
T2 = ?


Cp = 0.900 J/g°C
9.0  T2  95 C   418 T2  25.0 C 
Water:
11305
m = 100.0 g
9.0T2  855  418T2  10450
T2 
T1 = 25.0°C
427 .0
T2 = ?
427.0T2  11305
Cp = 4.18 J/g°C
T = 26.5 °C
2