So… about Thermal Energy
What’s up with Temperature vs Heat?
Temperature is related to the average
kinetic energy of the particles in a
You know the SI unit for temp is Kelvin
K = C + 273 (10C = 283K)
C = K – 273 (10K = -263C)
Thermal Energy is
the total of all the
kinetic and potential
energy of all the
particles in a
Thermal energy relationships
As temperature increases, so does
thermal energy (because the kinetic
energy of the particles increased).
If the temperature stays the same,
the thermal energy in a more massive
substance is higher (because it is a
total measure of energy).
Cup gets cooler while
hand gets warmer
The flow of thermal
energy from one
object to another.
Heat always flows
from warmer to
Ice gets warmer
while hand gets
Specific Heat (c, sometimes s, but
Things heat up or cool down at
Land heats up and cools down faster than
water, and aren’t we lucky for that!?
Specific heat is the amount of heat
required to raise the temperature of 1 kg
(but in Chem we use g) of a material by
one degree (C or K, they’re the same
C water = 4184 J / kg C (“holds” its heat)
C sand = 664 J / kg C (less E to change)
This is why land heats up quickly
during the day and cools quickly at
night and why water takes longer.
Why does water have such a
high specific heat?
Water molecules form strong bonds with each
other water molecule; (including H-bonds!)so it
takes more heat energy to break the bonds.
Metals have weak bonds (remember the “sea of
e-) and do not need as much energy to break
WHERE’S THE MATH, MATE?!
Q = m x T x Cp
Q = change in thermal energy
m = mass of substance
T = change in temperature (Tf – Ti)
Cp = specific heat of substance
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many
joules of heat energy are lost by the Al?
heat gain/lose = q = (c)(mass)(∆T)
where ∆T = Tfinal - Tinitial
q = (0.897 J/g•K)(25.0 g)(37 - 310)K
q = - 6120 J
Notice that the negative sign on q signals
heat “lost by” or transferred OUT of Al.
Heat can be Transferred even if
there is No Change in State
q transferred = (c)(mass)(∆T)
Or… Heat Transfer can cause a
Change of State
Changes of state involve energy (at constant T)
Ice + 333 J/g (heat of fusion) -----> Liquid water
Is there an equation? Of course!
q = (heat of fusion)(mass)
Heat Transfer and Changes of
Liquid (l) Vapor (g)
Requires energy (heat).
Why do you…
cool down after
use water to put out a
Remember this – it’s the Heating/Cooling
Curve for Water! Woa!
Note that T is
constant as a phase
So, let’s look at this equation again…
q = (heat of fusion)(mass)
(There’s also q = (heat of vaporization)(mass), by the
way, for when we are talking about vaporization)
WHY DO I NEED THIS WHEN I HAVE
q transferred = (c)(mass)(∆T)
Well, when a phase changes THERE IS
NO change in temperature… but there is
definitely a change in energy!
So… if I want the total heat to
take ice and turn it to steam I
need 3 steps…
1) To melt the ice I need to
multiply the heat of fusion with
the mass…q = (heat of fusion)
2) Then, there is moving the
temperature from 0 C to 100C… for this
there is a change in temperature so we
can use… q transferred = (c)(mass)
3) But wait, that just takes us to 100 C,
what about vaporizing the molecules?
Well, then we need q = (heat of
Add ‘em all up and there it is!
Now, lucky for us, just like there are
tables for specific heats, there are also
tables for heats of fusion and heats of
At least we don’t have to worry about
Heat & Changes of State
What quantity of heat is required to
melt 500. g of ice and heat the water
to steam at 100 oC?
Heat of fusion of ice = 333 J/g
Specific heat of water = 4.2 J/g•K
Heat of vaporization = 2260 J/g
And now… More! Heat & Changes of
How much heat is required to melt 500. g of ice and
heat the water to steam at 100 oC?
To melt ice
q = (500. g)(333 J/g) = 1.67 x 105 J
To raise water from 0 oC to 100 oC
q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J
To evaporate water at 100 oC
q = (500. g)(2260 J/g) = 1.13 x 106 J
4. Total heat energy = 1.51 x 106 J = 1510 kJ
Aka… How we Measure Heats of
In a Constant
Volume or “Bomb”
burn a combustible
From the heat
change , we
evolved in a
reaction to get ∆E
BOOM! Combustible material ignited at
constant volume! This heats up the
“bomb”, which heats up the water
First, some heat from reaction
warms the water, which we
know the mass of and “c” for…
qwater = (c)(water mass)(∆T)
THEN, some heat from reaction
warms “bomb,” which has a
known specific heat for the entire
apparatus (typically), so we don’t
need the mass…
qbomb = (heat capacity, J/K)(∆T)
Total heat evolved = qtotal = qwater + qbomb
The Mathy bit… Measuring Heats of
Calculate heat of combustion of octane.
C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O
You could burn 1.00 g of octane… or you could just note
Temp rises from 25.00 to 33.20 oC
Calorimeter contains 1200 g water
Heat capacity of bomb = 837 J/K
Step 1 Calc. heat transferred from reaction to water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2 Calc. heat transferred from reaction to bomb.
q = (bomb heat capacity)(∆T)
= (837 J/K)(8.20 K) = 6860 J
Step 3 Total heat evolved
41,170 J + 6860 J = 48,030 J
Heat of combustion of 1.00 g of octane = 48.0 kJ
One More (ok, LAST) Example… With a
If I burn 0.315 moles of hexane (C6H14) in a bomb calorimeter containing 5.65
liters of water, what’s the molar heat of combustion of hexane is the water
temperature rises 55.40 C? The heat capacity of water is 4.184 J/g0C.
∆H = mCp∆T
∆H = (5,650 grams H2O)(4.184 J/g0C)(55.40 C)
∆H = 1310 kJ
Do you think we’re done?
NOPE! We were asked for the MOLAR
heat of combustion! Tricky, tricky…
Molar Heat of Combustion
The amount of energy released in burning completely one
mole of substance.
Heat of combustion
The amount of heat released per unit mass or unit volume of
a substance when the substance is completely burned.
We also have heat of fusion vs MOLAR heat of fusion and
heat of vaporization vs MOLAR heat of vaporization, so be
What we calculated is the amount of energy generated
when 0.315 moles of hexane is burned, which is close
but not quite what we were asked for...
To find the molar heat of combustion,
we need to multiply this by (1
mole/0.315 moles) = 3.17. As a
result, the molar heat of combustion
of hexane is 4150 kJ/mol.