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So… about Thermal Energy
What’s up with Temperature vs Heat?
Temperature is related to the average kinetic energy of the particles in a
substance.
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 substance.
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).
Heat
The flow of thermal energy from one object to another.
Heat always flows from warmer to cooler objects.
Ice gets warmer while
hand gets cooler
Cup gets cooler while hand gets
warmer
Specific Heat (c, sometimes s, but usually c)
Things heat up or cool down at different rates.
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 size).
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
them.
water metal
WHERE’S THE MATH, MATE?!
Q = m x T x CpQ = change in thermal energy
m = mass of substance
T = change in temperature (Tf – Ti)
Cp = specific heat of substance
Specific Heat CapacitySpecific Heat Capacity
If 25.0 g of Al cool from 310 If 25.0 g of Al cool from 310 ooC to 37 C to 37 ooC, how many C, how many joules of heat energy are lost by the Al?joules of heat energy are lost by the Al?
heat gain/lose = q = (c)(mass)(∆T)
where ∆T = Twhere ∆T = Tfinalfinal - T - Tinitialinitial
q = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = (0.897 J/g•K)(25.0 g)(37 - 310)K
q = - 6120 Jq = - 6120 J Notice that the negative sign on q Notice that the negative sign on q signals heat “lost by” or transferred signals heat “lost by” or transferred
OUT of Al.OUT of Al.
Heat can be Transferred even if Heat can be Transferred even if there is No Change in Statethere is No Change in State
q transferred = (c)(mass)(∆T)
Or… Heat Transfer can cause a Or… Heat Transfer can cause a Change of StateChange of State
Changes of state involve energy Changes of state involve energy (at constant T)(at constant T)Ice + 333 J/g (heat of fusion) -----> Liquid waterIce + 333 J/g (heat of fusion) -----> Liquid waterIs there an equation? Of course!Is there an equation? Of course!
q = (heat of fusion)(mass)q = (heat of fusion)(mass)
Heat Transfer and Changes of StateHeat Transfer and Changes of State
Liquid (l) Liquid (l) Vapor (g) Vapor (g)
Requires energy (heat).Requires energy (heat).Why do you…Why do you…cool down after cool down after
swimming ???swimming ???use water to put out a use water to put out a
fire???fire???
Remember this – it’s the Heating/Cooling Curve for Water! Woa!
Note that T is Note that T is constant as a constant as a
phase changesphase changes
Evaporate waterEvaporate water
So, let’s look at this equation again…So, let’s look at this equation again…q = (heat of fusion)(mass)q = (heat of fusion)(mass)
(There’s also q = (heat of vaporization)(mass), by (There’s also q = (heat of vaporization)(mass), by the way, for when we are talking about the way, for when we are talking about
vaporization)vaporization)
WHY DO I NEED THIS WHEN I HAVEWHY DO I NEED THIS WHEN I HAVEq transferred = (c)(mass)(∆T)
HUH???HUH???Well, when a phase changes THERE IS Well, when a phase changes THERE IS NO change in temperature… but there NO change in temperature… but there
is definitely a change in energy!is definitely a change in energy!
So… if I want the total heat to So… if I want the total heat to take ice and turn it to steam I take ice and turn it to steam I need 3 steps…need 3 steps…1) To melt the ice I need to 1) To melt the ice I need to multiply the heat of fusion with multiply the heat of fusion with the mass…the mass…q = (heat of fusion)q = (heat of fusion)(mass)(mass)
2) Then, there is moving the 2) Then, there is moving the temperature from 0 C to 100C… for temperature from 0 C to 100C… for this there is a change in temperature this there is a change in temperature so we can use… so we can use… q transferred = (c)(mass)(∆T)
3) But wait, that just takes us to 100 3) But wait, that just takes us to 100 C, what about vaporizing the C, what about vaporizing the molecules? Well, then we need molecules? Well, then we need q = q = (heat of vaporization)(mass)…(heat of vaporization)(mass)…
Add ‘em all up and there it is!Add ‘em all up and there it is!
Now, lucky for us, just like there are Now, lucky for us, just like there are tables for specific heats, there are tables for specific heats, there are also tables for heats of fusion and also tables for heats of fusion and
heats of vaporization. heats of vaporization.
Whew, Whew, At least we don’t have to worry about At least we don’t have to worry about
that! that!
Heat & Changes of Heat & Changes of StateStateWhat quantity of heat is required to melt What quantity of heat is required to melt
500. g of 500. g of iceice and heat the water to and heat the water to steamsteam at 100 at 100 ooC?C?
Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g
+333 J/g+333 J/g
+2260 +2260 J/gJ/g
And now… More! Heat & Changes And now… More! Heat & Changes of Stateof State
How much heat is required to melt 500. g of ice and heat the How much heat is required to melt 500. g of ice and heat the water to steam at 100 water to steam at 100 ooC?C?
1. 1. To melt iceTo melt ice q = (500. g)(333 J/g) = 1.67 x 10q = (500. g)(333 J/g) = 1.67 x 1055 J J
2.2. To raise water from 0 To raise water from 0 ooC to 100 C to 100 ooCC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 10q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 1055 J J
3.3. To evaporate water at 100 To evaporate water at 100 ooCC q = (500. g)(2260 J/g) = 1.13 x 10q = (500. g)(2260 J/g) = 1.13 x 1066 J J
4. 4. Total heat energy = 1.51 x 10Total heat energy = 1.51 x 1066 J = 1510 kJ J = 1510 kJ
CALORIMETRYCALORIMETRYAka… How we Measure Heats of
Reaction
In a Constant Volume or “Bomb” Calorimeter, we burn a combustible sample.From the heat change , we measure heat evolved in a reaction to get ∆E
for reaction.
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
BOOM! Combustible material ignited BOOM! Combustible material ignited at constant volume! This heats up the at constant volume! This heats up the “bomb”, which heats up the water “bomb”, which heats up the water surrounding it…surrounding it…
The Mathy bit… Measuring Heats of The Mathy bit… Measuring Heats of Reaction using…Reaction using…
CALORIMETRYCALORIMETRY
Calculate heat of combustion of octane.Calculate heat of combustion of octane.
CC88HH1818 + 25/2 O + 25/2 O22 --> 8 CO --> 8 CO22 + 9 H + 9 H22OO
You could burn 1.00 g of octane… or you could just note You could burn 1.00 g of octane… or you could just note that…that…
Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC
Calorimeter contains 1200 g waterCalorimeter contains 1200 g water
Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K
Step 1Step 1 Calc. heat transferred from reaction to water. Calc. heat transferred from reaction to water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2Step 2 Calc. heat transferred from reaction to bomb. Calc. heat transferred from reaction to bomb.
q = (bomb heat capacity)(∆T)q = (bomb heat capacity)(∆T)
= (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 J
Step 3Step 3 Total heat evolved Total heat evolved
41,170 J + 6860 J = 48,030 J41,170 J + 6860 J = 48,030 J
Heat of combustion of 1.00 g of octane = - 48.0 kJHeat of combustion of 1.00 g of octane = - 48.0 kJ
VIOLA!VIOLA!
One More (ok, LAST) Example… With a twist…
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 = mCpT
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 CombustionThe amount of energy released in burning completely one
mole of substance.Vs.
Heat of combustionThe 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
watchful…
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.