Heat capacity and calorimetry: how much heat it takes, and how transitions show up

Heat capacity is how much heat it takes to warm something by one kelvin

Intuition

Think of filling a bucket with water. A big bucket needs more water to rise by the same amount than a small one. Heat capacity is like the width of the bucket: how much heat it takes to raise the temperature by one degree. Water is a wide bucket. Gold is a thin one.

Phase transitions often involve transfers of heat. Calorimetry is the experimental technique for measuring heat capacity and heat flow.

Heat capacity, $C$, is the heat needed to raise an object’s temperature by one kelvin. Its units are J K$^{-1}$.

$C = \frac{dQ}{dT}$

$C$ is extensive: the bigger the object, the more heat it needs. So we usually quote the specific heat capacity (per gram, or per unit mass) or the molar heat capacity (per mole).

$C_s = \frac{C}{m} \qquad C_m = \frac{C}{n}$

The lecture’s table of specific heat capacities, in J/(kg K): water 4186, ice 2090, steam 2010, aluminium 900, iron 448, glass 837, copper 387, mercury 138, gold 129. Water has a high heat capacity.

Calorimeters measure heat from a heater, a hot object or a process

There are two ways to find $C$.

1. Known heat in. Put in a known amount of heat with an electrical heater, then measure the temperature change.
2. Hot object in. Heat the object to a known temperature, then drop it into a calibrated calorimeter. The water in the calorimeter warms up, and the heat released by the object is found from that.

Calorimetry can also measure the heat given out or taken in by a chemical or physical process, at either constant pressure or constant volume. The constant-pressure version in the slide is an insulated beaker of water with a thermometer, where a salt dissolves. The constant-volume version is a bomb calorimeter: a steel bomb with ignition wires and a sample cup, sitting in a water bath.

A heating curve. Slide 11 plots temperature against heat added for water. The solid warms from -40 °C to 0 °C and the plot reaches 1.5 kJ. Then there is a plateau at 0 °C, the melting plateau, up to 7.5 kJ. This is $\Delta H_{\text{fus}}$. Then the liquid warms to 100 °C, reached at 15 kJ. Another plateau at 100 °C runs to 56 kJ. This is $\Delta H_{\text{vap}}$. After that the gas warms. These kJ values are read from the plot, so use them to read the shape rather than as exam numbers.

First-order transitions have latent heat, second-order transitions do not

Intuition

Melting ice in a drink: the ice keeps absorbing heat but stays at 0 degrees until it is all gone. That hidden heat is the latent heat, and it is the sign of a first-order transition. A second-order transition changes the material’s behaviour without any such hidden heat.

• First-order phase transitions involve a latent heat: energy is released or absorbed during the change of state. Examples: boiling and freezing.
• Second-order phase transitions do not involve a latent heat. Examples: the ferromagnetic and superconducting transitions.

Either kind can be seen calorimetrically. The Ehrenfest classification says which derivative of $G$ (or $\mu$) jumps.

Compare

First order

The first derivative of $G$ is discontinuous. So $S$ and $V$ jump. $C_p$ shows a sharp spike. Latent heat. Boiling, freezing.

Second order

The second derivative of $G$ is discontinuous. $S$ and $V$ are continuous (they may show a kink). $C_p$ shows a step. No latent heat. Ferromagnetic and superconducting transitions.

The relations behind this are

$S = -\left(\frac{\partial G}{\partial T}\right)_p \qquad V = \left(\frac{\partial G}{\partial p}\right)_T$

and heat capacity is a second derivative of $G$:

$C_p = \left(\frac{dQ}{dT}\right)_p = T\left(\frac{dS}{dT}\right)_p = -T\left(\frac{d^2G}{dT^2}\right)_p$

Optional aside

Slide 14 prints $S = (\partial G/\partial T)_p$ with no minus sign, and the $C_p$ line with $+T(d^2G/dT^2)_p$. The standard relation is $S = -(\partial G/\partial T)_p$. It comes from $dG = V\,dp - S\,dT$, and Lecture 22 uses it as $d\mu = V_m\,dp - S_m\,dT$, so $(\partial\mu/\partial T)_p = -S_m$. This page uses the standard sign. With it, $C_p = T(dS/dT)_p = -T(d^2G/dT^2)_p$. The idea that $C_p$ is a second derivative of $G$ does not depend on the sign.

Exam detail

First order: latent heat, $S$ and $V$ discontinuous, examples boiling and freezing. Second order: no latent heat, $S$ and $V$ continuous, examples ferromagnetic and superconducting transitions. $C_p$ is a second derivative of $G$.

Differential scanning calorimetry reads heat capacity against temperature

DSC is a more precise calorimetric technique, and it works at the mg scale. It scans across a range of temperature, and it is differential because it compares a sample with a reference, such as an empty pan. A computer controls electric heaters so that the sample and reference stay at the same temperature while the temperature is raised over time. It watches which side needs more heater power, meaning which side is absorbing more heat.

A DSC scan shows heat flow against temperature. At a constant scan rate, heat flow is proportional to heat capacity.

$\frac{\text{Heat flow}}{T\ \text{scan rate}} = \frac{dQ/dt}{dT/dt} = C$

Slide 16 shows a thermogram of PEEK, with endotherm up and exotherm down. The values below are read from the plot and are approximate. There is a small step near 145 °C, the glass transition. There is a sharp dip near 170 °C, crystallisation, which gives away heat. There is a large peak near 340 °C, melting, which absorbs heat.

The lecture lists the advantages of DSC over conventional calorimetry: it scans the temperature, and it can be done on a small scale, as little as 1 mg.

Try it: worked example

A DSC thermogram shows a large upward peak (endotherm up). Is the sample absorbing or releasing heat, and what kind of transition could it be?

Show answer and steps

AnswerAbsorbing heat. In the lecture's PEEK scan the large peak is melting. A dip would be heat given away, like crystallisation.

1. The slide’s convention: endotherm up, exotherm down.
2. An upward peak means the sample needs more heater power than the reference to keep up. It is absorbing heat.
3. In the PEEK scan, melting is the large peak, near 340 °C (read from the plot).
4. Crystallisation gives away heat, so it appears as a dip, near 170 °C in that scan.
5. The glass transition is a small step, because it changes the heat capacity without a latent heat peak.

Pitfall

Do not mix up the direction on a DSC plot. In the lecture’s thermogram, endotherms (absorbing heat, such as melting) point up and exotherms (giving away heat, such as crystallisation) point down. Always check the axis label before you read a peak.

Recall

Why does a first-order transition show as a spike in C_p, while a second-order one shows as a step?

A first-order transition jumps in the first derivative of $G$ (latent heat absorbed at one temperature), so the heat capacity, a second derivative, spikes. A second-order transition jumps only in the second derivative, so $C_p$ steps.