Thermodynamics
1. Define temperature and state the units in which it is
measured.
2. Define heat and state its units.
3. Perform specific heat problems.
Temperature
 Temperature is the measure of the average kinetic
energy of the particles of a substance.
 Thermometers are used to measure temperatures.
 Units: °C and K
(and °F of course)
Heat (q)
 Heat is the flow of energy from a substance of
higher temperature to a substance of lower
temperature, when they are placed in thermal
contact with each other.
HOT 
 Units of Heat:
COLD
1000 J = 1 kJ
 Calorimeters: Measure heat flow and direction
Venn Diagram
Temperature
°C , K , °F
Heat
J, kJ, cal
Venn Diagram
Temperature
°C , K , °F
Heat
J, kJ, cal
Energy
Other Energy Units
1 J = 1 kg · m2 / s2
4.184 J = 1 cal
1000 cal = 1 kcal = 1Cal
(Nutritional Calorie)
q = m cp ΔT
 q = Heat ( units are J and kJ)
 m = Mass (units are g or kg)
 ΔT = Change of Temperature (Tf – Ti)
for ΔT: °C is the same size as a K
 cp = Specific Heat (units are J / g · °C )
Specific Heat
 The amount of heat (energy) required to raise the
temperature of 1 gram of a substance by 1°C.
Water
Aluminum
Iron
Silver
4.184 J / g °C
0.897 J / g °C
0.449 J / g °C
0.235 J / g °C
Calorimeter
 A device that uses a thermometer to measure the
change of temperature of water caused by a physical
change or a chemical change.
 Foam Cup Calorimeters (styrofoam)
 Bomb Calorimeters (steel)
Foam Cup Calorimeter
These calorimeters are
used for mixing
aqueous solutions and
then measuring the
temperature change.
 Exothermic: If the
temperature goes up.
 Endothermic: If the
temperature goes down.
Bomb Calorimeter
Bomb Calorimeters
 Bomb calorimeters are used for determining the
heat lost or gained by combustion reactions and
others that release a lot of heat.
 Exothermic: If the temperature of the water
increases (rises).
 Endothermic: If the temperature of the water
decreases (drops).
Calculations
q
mass
 How much heat is required to heat 53.8 g of water
from 20.0°C to 45.0 °C ?
 Waters Cp = 4.184 J / g °C
Temperature
q = mcp∆T
q = (53.8 g) (4.184 J /g °C)(45.0 – 20.0°C)
= 5627.48 J or 5.63 kJ
Specific Heat
of Unknown Substance
mass
 An unknown substance with a mass of 0.506 kg required
q
0.567 kJ of heat to raise its temperature 8.0°C.
 What is the unknown substances specific heat?
 Identify the substance.
q = mcp∆T
0.567 kJ
567 J
0.506 kg
= (506 g) (cp)(8.0°C)
∆T
Specific Heat
of Unknown Substance
 An unknown substance with a mass of 0.506 kg required
0.567 kJ of heat to raise its temperature 8.0°C.
 What is the substances specific heat?
 Identify the substance.
0.567 kJ
567 J
0.506 kg
= (506 g) (8.0°C) (cp)
567 J
= cp = 0.140 J / (g °C )
(506 g) ( 8.0·C)
Mercury
Heat Lost = Heat Gained
 When one substance loses heat during a chemical or
physical change, another substance must gain that lost
heat.
 When two aqueous solutions of chemicals are being
mixed together, the reaction will either give off heat to
the water or absorb heat from the water.
When the Water in a Calorimeter Changes
Temperature
 Does the reaction release heat into the water or absorb heat
from the water, when the temperature rises?
****If it releases heat –
exothermic – Temperature rises!
If it absorbs heat from the water endothermic – Temperature decreases!
Heat Lost = Heat Gained
Diagram
25.0 g Metal at 95°C = ti
Final temperature at 37°C = tf
50.0 g Water at 25°C = ti
Heat Lost = Heat Gained
Diagram
25.0 g Metal at 95°C = ti
Heat Lost
Final temperature at 37°C = tf
Heat Gained
50.0 g Water at 25°C = ti
The formula
q=mcp∆T
25.0 g Metal at 95°C = Ti
Final temperature at 37°C = Tf
50.0 g Water at 25°C = Ti
-(Heat (q) of Metal) = (Heat (q) of Water)
-q(metal) = qwater
-m (metal)cp (metal)ΔT(metal) = m(water) cp (water) ΔT(water)
cp (metal) = - m(water) cp (water) ΔT (water)
m (metal) ΔT (metal)
cp (metal) = - (50.0 g)(4.184 J/g0C)(370C-250C)
(25.0 g)(370C-950C)
cp =
1.73 J / g °C
Changing a
solid to a
liquid
Changing liquid
into gas
Heating gas
Heating liquid
Heating solid
Hess’s Law and Physical Changes
Heat Curve for water
ΔH = q 2 = Mole(∆Hfus)
Be Sure that all
the q’s are in
the same units!
Joules (J) or
kilojoules (kJ)
ΔH =q 4 = Mole(∆Hvap)
ΔH = q 5 = m cp ( of gas) ΔT
ΔH = q 3 = m cp ( of liquid ) ΔT
qt=q1+q2+q3+ q4+q5
ΔH = q 1 = m cp ( of solid) ΔT
Example:
A 32.5 g popsicle is in a freezer at
-20.0 oC. It is removed from the freezer
and left on the counter. It melts and
when mom comes home she finds a
puddle of flavored water at 25.0 oC.
How much energy did it take
end
start
Example:
 A 32.5 g popsicle is in a freezer at -20.0 oC. It is removed from the
freezer and left on the counter. It melts and when mom comes
home she finds a puddle of flavored water at 25.0 oC. How much
energy did it take to do this?
 Steps:
1-Heat Solid -20 oC to 0 oC
q1=mcp∆T = 32.5 g(2.06J/g oC)(20 oC) = 1339 J =
2 – Melt Solid
q2 = mole∆Hfus = ____32.5 g (6.01 KJ/mole) =
18.015 g/mole
3 – Heat the liquid from 0 oC to 25 oC
q3=mcp∆T = 32.5 g(4.184J/g oC)(25 oC) = 3399.5 J =
Then add q1 + q2 + q3 = ∆H =
1.34 kJ
= q1
10.8 kJ
= q2
3.40 kJ
= q 3+
15.5 kJ