Latent heat and thermal expansion: worked exercises

In Year 2 of high school, thermal physics is almost always taught between November and December: temperature and scales, expansion, specific heat, latent heat, how heat travels. The test arrives in four to eight weeks, in December, and it matters above all for students at a technical school (Italian istituto tecnico, technology track), though the syllabus is the same in many Year 2 classes. Now is the right time to prepare, because this block uses vectors much less than dynamics does, and it rewards students who are tidy with numbers.

Thermal physics gets lost for unglamorous reasons: mixed-up units (degrees Celsius and kelvin, joules and kilojoules), the plateau on the graph forgotten, and an equilibrium equation written with the wrong signs. In my lessons I find these three almost every time. I have already covered specific heat and thermal equilibrium in specific heat and thermal equilibrium: here we concentrate on expansion and changes of state, and I repeat only what is needed about $c$.

Temperature scales

Celsius and kelvin have the same step: an interval of $1{}^{\circ }\text{C}$ is an interval of $1\text{K}$. What changes is the zero:

So $0{}^{\circ }\text{C}$ corresponds to $273.15\text{K}$. In formulas that contain a change $\Delta T$ you can safely use degrees Celsius, because the difference is identical. Conversion is only needed when a formula uses absolute temperature, as in the gas laws you will meet later.

Thermal expansion

Almost all solids get longer when they are heated. For a long, thin body (a bar, a rail, a cable) the linear expansion is

where $L_0$ is the initial length and $\lambda$ is the coefficient of linear expansion, which depends on the material and is measured in ${\text{K}}^{-1}$. For steel $\lambda \approx 12\cdot 10^{-6}{\text{K}}^{-1}$.

For volumes there is the analogous formula

with $\beta \approx 3\lambda$ for an isotropic solid. In Year 2 you mostly use the linear form; the volume form appears in problems about containers and liquids.

One thing to remember: $\Delta L$ is tiny compared with $L_0$. If you get an expansion of metres on a bar a few metres long, you have got a power of ten wrong.

Latent heat and changes of state

When a substance changes state, the heat supplied does not raise its temperature: it is used to break (or form) the bonds between the molecules. The amount of heat needed is

where $L$ is the latent heat of the change, in $\text{J/kg}$ or $\text{kJ/kg}$. For water at atmospheric pressure:

Solidification and condensation give out the same amount of heat, with the opposite sign.

The heating curve

If you supply heat to a block of ice at $-10{}^{\circ }\text{C}$ and plot temperature against the heat absorbed, the graph has five sections:

The rises are sloping sections, the plateaus are horizontal sections. The vaporisation plateau is much longer than the fusion one, because it takes about seven times more energy.

For the calculation you need two specific heats: $c_{ice}=2.1\text{kJ/(kg}\cdot \text{K)}$ and $c_{water}=4.186\text{kJ/(kg}\cdot \text{K)}$. For the rest, $Q=mc\Delta T$ is the specific heat formula you already know.

Exercise 1: the steel rail

A section of steel rail is $25\text{m}$ long at $-5{}^{\circ }\text{C}$. In summer it reaches $35{}^{\circ }\text{C}$. By how much does it lengthen? Use $\lambda =12\cdot 10^{-6}{\text{K}}^{-1}$.

Step 1: the change in temperature. $\Delta T=35-(-5)=40\text{K}$. Watch the sign: subtracting a negative number is the same as adding.

Step 2: the formula.

The rail lengthens by $12\text{mm}$. That is why a small gap is left between two sections of rail: without it, the metal would buckle in the sun. The result is small but sensible: $12\text{mm}$ over $25\text{m}$ is about $0.05\%$.

If you had instead converted the temperatures to kelvin, $268.15\text{K}$ and $308.15\text{K}$, the difference would still have been $40\text{K}$: same result, longer calculations.

Your school’s physics test might also ask for the volume expansion of a container; as far as method goes it is identical, with $\beta =3\lambda$.

Exercise 2: from ice to vapour

How much energy does it take to turn $200\text{g}$ of ice at $-10{}^{\circ }\text{C}$ into vapour at $100{}^{\circ }\text{C}$?

Step 0: units. $m=200\text{g}=0.200\text{kg}$. We work in kJ, with $c$ in $\text{kJ/(kg}\cdot \text{K)}$ and $L$ in $\text{kJ/kg}$: everything is consistent.

Section 1, ice from $-10$ to $0{}^{\circ }\text{C}$:

Section 2, fusion:

Section 3, water from $0$ to $100{}^{\circ }\text{C}$:

Section 4, vaporisation:

Total:

The figure that stands out: vaporisation alone accounts for $452/606.72\approx 75\%$ of the total. Heating the water is almost a detail compared with the change of state. Whoever forgets a plateau, usually the fusion one, is off by almost 70 kJ.

To practise on the first-two-years syllabus there are the test practice papers, where the correct answer is free and the guided step-by-step solution is unlocked with the first lesson. An honest note: for Year 2 thermal physics the dedicated practice paper is calibrated on the liceo scientifico (science-track high school) syllabus, and a technical school covers some of the same topics but with less emphasis on certain parts. Use it as training on the common content (specific heat, latent heat, scales), not as an exact copy of your test.

Exercise 3: water and ice in the same container

In an insulated container there are $300\text{g}$ of water at $40{}^{\circ }\text{C}$. You add $100\text{g}$ of ice at $0{}^{\circ }\text{C}$. What is the final temperature?

First step: does all the ice melt? Compare two quantities.

• Heat the water can give up as it cools to $0{}^{\circ }\text{C}$: $Q_{given,max}=0.300\cdot 4.186\cdot 40=50.232\text{kJ}$.
• Heat needed to melt all the ice: $Q_{fus}=0.100\cdot 334=33.4\text{kJ}$.

Since $50.232>33.4$, the ice melts completely and there is heat left over to warm the whole mass. The final temperature will therefore be above zero.

Second step: the equilibrium equation. Heat given up by the hot water = heat absorbed by the ice (fusion, then warming the melted water from $0$ to $T_f$):

Expand the products: $1.2558(40-T_f)=50.232-1.2558T_f$. The equation becomes

The temperature lies between $0$ and $40$, and it is plausible: the ice has cooled the water by about thirty degrees.

Check. Given up by the hot water: $1.2558\cdot (40-10.05)\approx 37.6\text{kJ}$. Absorbed: $33.4+0.4186\cdot 10.05\approx 37.6\text{kJ}$. The two values match.

If you had found $Q_{given,max}<Q_{fus}$ instead, the ice would not all melt: the final temperature would be exactly $0{}^{\circ }\text{C}$, with a mixture of water and ice, and the question would become “how much ice is left”.

The mistakes that cost marks

1. The forgotten plateau. Going from ice to vapour there are two changes of state: one is never optional. Draw the five-section graph before writing the calculations.
2. Celsius and kelvin mixed up. Fine for changes, never for absolute temperatures. In the formulas of this post $\Delta T$ is always a difference.
3. J versus kJ. If $L$ is in $\text{kJ/kg}$ and $c$ in $\text{J/(kg}\cdot \text{K)}$, the sum makes no sense. Convert everything at the start, and check that the result has the right order of magnitude: hundreds of kJ for a bottle of water, not thousands.
4. Grams instead of kilograms. With $L$ and $c$ in units per kilogram, the mass goes in kg. It sounds trivial, but it happens.
5. Signs in the equilibrium. Write “heat given up = heat absorbed” with both sides positive, and build $\Delta T$ in each term as the higher temperature minus the lower temperature.
6. Pressure. The values of $L$ given here hold at atmospheric pressure. If the problem mentions different pressures, check the units ($\text{Pa}$, $\text{atm}$, $\text{bar}$) before substituting into the formula.

How to prepare

Redo the three exercises without looking at the solutions, changing the masses and temperatures. Then try inverting the problem: given the total heat, find the mass or the final temperature. If dynamics is still an unresolved knot from the first term, review dynamics and the inclined plane in Year 2, and to understand pressure, which sometimes appears alongside thermal physics, there is fluid statics, Stevin and Archimedes. The general overview of the subject is in Physics.

If you want a check before the test

If the formulas make sense but the problems do not, what is usually missing is a method for reading the text: what changes state, what only heats up, where the plateau is. Write to me from the contact page and we will do a free introductory call: send me a problem that does not add up and I will tell you where it gets stuck.