๐Ÿฆ– Bellaziraptor

Topic 2

Heat, specific heat & calorimetry

How much energy does it take to change something’s temperature โ€” and what happens when the temperature refuses to change at all? Two equations, Q = mcΔT and Q = mL, and the whole skill is knowing which one you are in.

On this page

  1. Heat vs. temperature vs. internal energy
  2. Specific heat
  3. Latent heat and phase change
  4. The heating curve
  5. Calorimetry
  6. How heat travels
  7. Worked examples

Heat vs. temperature vs. internal energy

These three get used interchangeably in everyday speech and never in physics.

QuantitySymbolWhat it is
TemperatureT A measure of average kinetic energy per molecule. Intensive โ€” does not depend on how much stuff you have.
Internal energyU The total microscopic energy in the sample. Extensive โ€” twice the water, twice the internal energy at the same T.
HeatQ Energy in transit because of a temperature difference. A process, not a property.
Say it right

An object never “contains heat.” It contains internal energy. Heat is what crosses the boundary. A bathtub at 30 °C is at a lower temperature than a cup of coffee at 80 °C but holds far more internal energy โ€” and that is not a contradiction.

Specific heat

Specific heat c is the energy needed to raise one kilogram by one kelvin:

Q = mc ΔT Q in J, m in kg, c in J/(kg·K), ΔT = TfTi. Positive Q = energy in, temperature rises.

A big c means the substance is hard to heat up and slow to cool down. Water’s is enormous โ€” the highest of any common substance โ€” which is why coastal climates are mild, why radiators and engine coolant use water, and why the ocean is a planetary thermal flywheel.

Substancec (J/kg·K)Note
Water (liquid)4186The benchmark. Memorize this one.
Ethanol2440
Ice2100Half of liquid water โ€” same molecules, different phase
Steam2010
Air1005At constant pressure
Aluminum900
Glass840
Iron / steel450
Copper385
Silver235
Mercury140
Lead128Heats fast, cools fast
Metals cluster low; water sits an order of magnitude above them.
Conceptual question they love

Equal masses of water and copper absorb the same energy. Which ends up hotter? Copper โ€” its c is ~11× smaller, so the same Q buys ~11× the temperature rise. Same energy in, very different thermometer reading.

Latent heat and phase change

During melting or boiling, energy pours in and the temperature does not move. The energy is going into breaking intermolecular bonds โ€” raising potential energy, not kinetic energy. Since temperature only tracks kinetic energy, the thermometer sits still.

Q = mL Lf = latent heat of fusion (melting/freezing); Lv = latent heat of vaporization (boiling/condensing). Units J/kg. No ΔT appears โ€” there is no temperature change.
SubstanceMelt pt.Lf (J/kg)Boil pt.Lv (J/kg)
Water0 °C3.34×105100 °C2.26×106
Ethanol−114 °C1.04×10578 °C8.54×105
Lead327 °C2.45×1041750 °C8.70×105
Nitrogen−210 °C2.55×104−196 °C2.00×105

Notice Lv for water is about 6.8× its Lf. Melting ice only has to loosen the crystal; boiling has to rip molecules entirely free of each other. That ratio explains two everyday facts: a pot takes forever to boil dry after it reaches 100 °C, and steam at 100 °C burns far worse than water at 100 °C โ€” condensing on your skin dumps 2.26×106 J/kg before the temperature even starts to drop.

The heating curve

Put 1 kg of ice at −20 °C on a burner of constant power and plot temperature against energy added. You get the single most useful diagram in this topic:

0 °C 100 °C T Q ice melting water boiling steam Q = mLf Q = mLv Q = mcΔT
Shape only โ€” the plateau widths are not to scale. Drawn honestly, the boiling plateau would swallow 73% of the horizontal axis (see the table below). Sloped segment ⇒ Q = mcΔT; flat segment ⇒ Q = mL.

How to read it:

Stage (1 kg)EquationEnergyShare
Ice, −20 → 0 °CmciceΔT = (1)(2100)(20)42 kJ1.4%
Melting at 0 °CmLf = (1)(334 000)334 kJ10.8%
Water, 0 → 100 °CmcwaterΔT = (1)(4186)(100)419 kJ13.5%
Boiling at 100 °CmLv = (1)(2 260 000)2260 kJ73.0%
Steam, 100 → 120 °CmcsteamΔT = (1)(2010)(20)40 kJ1.3%
Total3095 kJ100%
Boiling costs more than every other stage combined, roughly three times over.

Calorimetry

Put hot and cold things in an insulated container. No energy leaves, no work is done, so all the heat lost by some objects is gained by the others:

ΣQ = 0    ⇒    Qhot + Qcold = 0 Write every term as mc(TfTi) with signs included, then solve. The hot object’s Q comes out negative automatically โ€” do not also insert a minus sign by hand.

Method

  1. Write Q for every object, each as mc(TfTi), all with the same unknown Tf.
  2. Add any phase-change terms as separate mL pieces (positive if the substance absorbs energy: melting or boiling).
  3. Set the sum to zero and solve.
  4. Sanity check: Tf must land between the starting temperatures. If it does not, you have a sign error โ€” or a phase change you did not account for.
The classic trap

When ice is involved, check whether it all melts before you solve for Tf. Compare the energy needed to melt every gram of ice against the energy available from cooling everything else to 0 °C. If there is not enough, the answer is simply Tf = 0 °C with some ice left over โ€” and any algebra assuming otherwise returns a nonsense negative temperature.

How heat travels

ModeMechanismNeeds matter?Example
ConductionMolecule-to-molecule collisions; in metals, mostly free electronsYesSpoon handle heating in soup
ConvectionBulk motion of a heated fluid โ€” warm fluid is less dense and risesYes (fluid)Boiling water, sea breeze
RadiationElectromagnetic wavesNoSunlight; heat felt from a fire across a room

For steady conduction through a slab:

P = kA ΔTL P = rate of energy transfer (W), k = thermal conductivity (W/m·K), A = cross-sectional area, L = thickness, ΔT = temperature difference across it.

Read the structure: transfer is faster through a wide, thin, conductive slab with a big temperature difference, and slower through a narrow, thick, insulating one. Copper’s k is about 400 W/(m·K); fiberglass insulation is about 0.04 โ€” a factor of 104. Most insulation works by trapping air (k ≈ 0.026) in pockets small enough to stop convection.

Why metal feels colder than wood at the same temperature

Both are at room temperature; your skin is not. Metal conducts energy away from your finger far faster, so your nerves report a rapid heat loss and you call it “cold.” You are sensing rate of transfer, not temperature. This is a favorite free-response prompt.

Worked examples

1. Identify a metal from a calorimetry run

A 150 g metal block is heated to 95.0 °C and dropped into 200 g of water at 21.0 °C in a calorimeter of negligible heat capacity. The mixture settles at 25.8 °C. What is the metal?

Step 1 โ€” heat gained by the water.

Qw = (0.200)(4186)(25.8 − 21.0) = +4.02×103 J

Step 2 โ€” the metal supplied it. ΣQ = 0, so Qm = −4.02×103 J.

−4.02×103 = (0.150)c(25.8 − 95.0) = −10.38 c c = 4.02×103 / 10.38 = 387 J/(kg·K)

That matches copper (385). Note the sign took care of itself: ΔT for the metal is negative because it cooled.

2. Ice into water โ€” does it all melt?

50 g of ice at 0 °C is added to 300 g of water at 25 °C. Find the final temperature.

Step 1 โ€” test the phase change first.

  • Energy to melt all the ice: mLf = (0.050)(3.34×105) = 16 700 J
  • Energy available from cooling the water to 0 °C: (0.300)(4186)(25) = 31 400 J

31 400 > 16 700, so all the ice melts and there is energy left over. Good โ€” the final temperature will be above 0 °C.

Step 2 โ€” spend the surplus warming the combined 350 g of water.

ΔT = 31 400 − 16 700(0.350)(4186) = 14 7001465 = 10.0 K Tf = 10.0 °C

Sanity check: between 0 and 25 °C. โœ“   Had the ice been 200 g instead, melting it all would have cost 66 800 J โ€” more than the 31 400 J available โ€” and the answer would have been 0 °C with ice remaining.

3. Reading specific heat off a graph

A 0.500 kg sample is heated at a constant 200 W. Its temperature rises from 20 °C to 60 °C in 90 s, with no phase change. Find c.

Q = Pt = (200)(90) = 1.80×104 J c = Q / (mΔT) = 1.80×104 / [(0.500)(40)] = 900 J/(kg·K) โ€” aluminum.

On a T-vs-Q graph this is the reciprocal of the slope divided by mass: c = 1/(m · slope). Steep line, small specific heat.

4. Conceptual: two burns

Why is a burn from 100 °C steam worse than one from 100 °C water, when both are at the same temperature?

Because of the phase change. Steam condensing on skin releases Lv = 2.26×106 J/kg before its temperature drops at all. Only after condensing does it start cooling from 100 °C like ordinary water. Per gram, condensation alone delivers about 2260 J โ€” roughly the same as cooling liquid water from 100 °C all the way down to 46 °C. The steam burn simply carries far more energy.

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