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THERMODYMANICS
Thermodynamics is the
study of the motion of heat
energy as it is transferred
from the system to the
surrounding or from the
surrounding to the system.
The transfer of heat could be due to
a physical change or a chemical
change.
There are three laws of chemical
thermodynamics.
CHEMICAL
THERMODYMANICS
The first law of thermodynamics:
Energy and matter can be neither created nor destroyed;
only transformed from one form to another. The energy
and matter of the universe is constant.
The second law of thermodynamics:
In any spontaneous process there is always an increase in
the entropy of the universe. The entropy is increasing.
The third law of thermodynamics:
The entropy of a perfect crystal at 0 K is zero.
There is no molecular motion at absolute 0 K.
HEAT
The energy that flows into or out of a
system because of a difference in
temperature between the thermodynamic
system and its surrounding.
Symbolized by "q".

When heat is evolved by a system, energy
is lost and "q” is negative (-).

When heat is absorbed by the system, the
energy is added and "q" is positive (+).
HEAT FLOW
Heat can flow in one of two directions:
Exothermic
To give off heat; energy is lost from the
system: (-q)
Endothermic
To absorb heat; energy is added to the
system: (+q)
If the heat transfer involves a chemical reaction then q is called:
HEAT OF REACTION
The heat energy (DH; enthalpy) required to
return a system to the given temperature
at the completion of the reaction.
q = DH
at constant pressure
The heat of reaction can be specific to a reaction like:
HEAT OF COMBUSTION
The quantity of heat energy given off when
a specified amount of substance burns in
oxygen.
UNITS: kJ/mol (kilojoules per mole) or kcal/mol (kilocalories per mole)
HEAT CAPACITY &
SPECIFIC HEAT
HEAT CAPACITY: The quantity of heat
needed to raise the temperature of a
substance one degree Celsius (or one
Kelvin).
q = Cp DT
SPECIFIC HEAT: The quantity of heat
required to raise the temperature of one
gram of a substance by one degree
Celsius (or one Kelvin).
q = s x m x DT
Both Cp & s are chemical specific constants found in the textbook or CRC
Handbook.
UNITS for HEAT ENERGY
Heat energy is usually measured in
either Joules, given by the unit (J), and
kilojoules (kJ) or in calories, written
shorthand as (cal), and kilocalories
(kcal).
1 cal = 4.184 J
NOTE: This conversion correlates to the specific heat of water
which is 1 cal/g oC or 4.184 J/g oC.
SPECIFIC HEAT

Determine the energy (in kJ) required to
raise the temperature of 100.0 g of water
from 20.0 oC to 85.0 oC?
m = 100.0 g
q = m x s x DT
DT = Tf -Ti = 85.0 - 20.0 oC = 65.0 oC
s (H2O) = 4.184 J/ g - oC
q = (100.0 g) x (4.184 J/g-oC) x (65.0oC)
q = 27196 J (1 kJ / 1000J) = 27.2 kJ

Determine the specific heat of an
unknown metal that required 2.56 kcal of
heat to raise the temperature of 150.00 g
from 15.0 oC to 200.0 oC?
S = 0.0923 cal /g -oC
LAW OF CONSERVATION OF
ENERGY

The law of conservation of energy (the
first law of thermodynamics), when
related to heat transfer between two
objects, can be stated as:
The heat lost by the hot object = the heat
gained by the cold object
-qhot = qcold
-mh x sh x DTh = mc x sc x DTc
where DT = Tfinal - Tinitial
LAW OF CONSERVATION OF ENERGY
 Assuming no heat is lost, what mass of cold
water at 0.00oC is needed to cool 100.0 g of
water at 97.6oC to 12.0 oC?
-mh x sh x DTh = mc x sc x DTc
- (100.0g) (1 cal/goC) (12.0-97.6oC) = m (1 cal/goC) (12.0 - 0.0 oC)
8560 cal = m (12.0 cal/g)
m = 8560 cal / (12.0 cal/g)
m = 713 g

Calculate the specific heat of an unknown metal if a 92.00 g
piece at 100.0oC is dropped into 175.0 mL of water at 17.8 oC.
The final temperature of the mixture was 39.4oC.
s (metal) = 0.678 cal/g oC
PRACTICE PROBLEM #7
1. Iron metal has a specific heat of 0.449 J/goC. How much heat is
transferred to a 5.00 g piece of iron, initially at 20.0 oC, when it is
placed in a beaker of boiling water at 1 atm?
180. J
2. How many calories of energy are given off to lower the temperature of
100.0 g of iron from 150.0 oC to 35.0 oC?
1.23 x 103 cal
3. If 3.47 kJ were absorbed by 75.0 g H2O at 20.0 oC, what would be the
final temperature of the water?
31.1 oC
4. A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g
of lead shots (at 100 oC) was added to the calorimeter and the final
temperature of the mixture was 34.4 oC. What is the specific heat of
lead?
1.28 J/g oC
5. A 17.9 g sample of unknown metal was heated to 48.31 oC. It was then
added to 28.05 g of water in an insulted cup. The water temperature
rose from 21.04 oC to 23.98oC. What is the specific heat of the metal
in J/goC?
0.792 J/goC
GROUP STUDY PROBLEM #7

_____1. A 250.0 g metal bar requires 5.866 kJ to change its
temperature from 22.0oC to 100.0oC. What is the specific heat
of the metal in J/goC?

_____2. How many joules are required to lower the temperature
of 100.0 g of iron from 75.0 oC to 25.0 oC?

_____ 3. If 40.0 kJ were absorbed by 500.0 g H2O at 10.0 oC,
what would be the final temperature of the water?

_____ 4. A 250 g of water at 376.3 oC is mixed with 350.0 mL of
water at 5.0 oC. Calculate the final temperature of the mixture.

_____5. A 400 g piece of gold at 500.0oC is dropped into 15.0 L
of water at 22.0oC. The specific heat of gold is 0.131 J/goC or
0.0312 cal/goC. Calculate the final temperature of the mixture
assuming no heat is lost to the surroundings.