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.