Specific heat and thermal equilibrium: worked exercises

End of September 2026: in Year 2 at the liceo scientifico (science-track Italian high school), fluids are coming to a close — pressure, Stevin’s law, Pascal, Archimedes. The next block is thermal physics, which starts between the end of October and the beginning of November, and the test typically falls in December, before the holidays: about nine to ten weeks from now. If you are reading this post the evening of the first lesson on heat, you are in the right place — it is written for exactly that moment, before the teacher gets to the second or third exercise and the class starts to lose the thread.

The block is called thermal physics but its core is just one idea: heat is energy being transferred, not a substance that “sits inside” bodies. All the formulas in these pages follow from it. Thermal expansion, which often opens the chapter in the textbook, I touch only in passing here: the real topic, the one that weighs in the test, is specific heat and thermal equilibrium.

Temperature and heat are not the same thing

The first hurdle, even before the formulas, is conceptual: temperature measures how “hot” a body is in a precise sense — it is linked to the average agitation of the particles it is made of. Heat is instead energy being transferred from a hotter body to a colder one, until the temperatures become equal. A body does not “contain” heat: it has a temperature, and it exchanges heat when it is in contact with a body at a different temperature — just as in mechanics position is a state and work is a transfer.

Temperature scales: Celsius and Kelvin

The Celsius scale (${}^{\circ }\text{C}$) is the everyday one, with zero at the melting point of ice and 100 at the boiling point of water (at atmospheric pressure). The Kelvin scale (K, without the degree symbol) has the same size of degree as Celsius, but with its zero moved to absolute zero, the lowest possible temperature:

In problems you almost always use the approximation $273$ instead of $273.15$. A point that often causes confusion: when you convert a temperature difference (a $\Delta T$), the $+273$ cancels in the subtraction, so $\Delta T$ in kelvin has exactly the same numerical value as $\Delta t$ in degrees Celsius — it only matters when you convert an absolute temperature, not a change in it. The Fahrenheit scale, if your textbook mentions it, is just a side note: you do not need to know how to convert it for the test.

Heat as energy: joules and calories

Heat is measured, like any energy, in joules (J). There is also a historical unit, the calorie (cal), defined as the energy needed to raise the temperature of 1 gram of water by $1{}^{\circ }\text{C}$. The conversion between the two is fixed:

In problems where the starting unit is the calorie, it is best to convert to joules straight away and work in a single system of units to the end, so you do not mix the two halfway through the calculation.

The fundamental law of calorimetry

The amount of heat $Q$ needed to change the temperature of a body of mass $m$ by an amount $\Delta T$ depends on the material, through a constant called the specific heat $c$:

Specific heat is the amount of heat needed to raise the temperature of 1 kg of that material by $1{}^{\circ }\text{C}$, and it is measured in $\text{J}/(\text{kg}\cdot {}^{\circ }\text{C})$ (or the equivalent in kelvin, since it is a $\Delta T$). It is a property of the material, not of the quantity of substance: a litre of water and a drop of water have the same specific heat, but very different amounts of heat are needed to warm them by the same amount, because the mass $m$ is different.

Some standard values that the textbook often expects you to know by heart or gives in a table in the text of the problem:

Notice how high the specific heat of water is compared with metals: it takes much more energy to warm water than a piece of metal of the same mass. That is why, in exercises with several bodies, it is almost always the water that “wins” in the sense of changing temperature less.

Heat capacity

Sometimes the problem gives you directly the heat capacity $C$ of an object (not of a material), defined as $C=mc$: the amount of heat needed to raise by $1{}^{\circ }\text{C}$ the temperature of that specific object, regardless of the material the mass is made of. In that case the formula simply becomes

with no need to know the mass and the specific heat separately.

Thermal equilibrium: $Q$ given up = $Q$ absorbed

When two bodies at different temperatures are put in contact (in an isolated system, which exchanges no heat with the outside), heat flows from the hotter body to the colder one until they reach the same equilibrium temperature $T_{eq}$. The principle is very simple to state and just as treacherous to apply with the right signs:

The heat given up by the hot body (which cools down, so its $\Delta T$ is negative) is, in absolute value, equal to the heat absorbed by the cold body (which warms up, $\Delta T$ positive). Writing the equation with the correct signs from the start, for two bodies 1 and 2:

Here you do not need to decide in advance who gives up heat and who absorbs it: the formula takes care of the signs by itself, because one of the two terms $(T_{eq}-T)$ will come out negative (the body that cools down) and the other positive (the one that warms up), and the sum is zero by the definition of equilibrium. From this equation you get $T_{eq}$ by isolating it algebraically.

The calorimeter and the water equivalent

In the laboratory, thermal equilibrium is observed inside a calorimeter, a thermally insulated container. But its walls also absorb a little heat: to account for this without knowing the specific heat of the calorimeter’s material, we introduce the water equivalent $m_{eq}$, the mass of water that would absorb the same heat for the same $\Delta T$. In the equilibrium equation the calorimeter is treated as an extra mass $m_{eq}$ of water, added to that of the liquid it contains.

Latent heat and changes of state

When a body changes state (from solid to liquid, from liquid to gas) at constant temperature, the heat exchanged does not change the temperature but breaks (or forms) the bonds between the molecules. This heat is called latent heat $\lambda$, and it is proportional only to the mass that changes state, not to any $\Delta T$ (because during the change of state the temperature stays fixed):

The two values that come up most often in problems:

• latent heat of fusion of ice: ${\lambda }_f=3.34\times 10^5\text{J/kg}$
• latent heat of vaporisation of water: ${\lambda }_v=2.26\times 10^6\text{J/kg}$

The heating curve

If you heat ice from a very low temperature until it becomes vapour, the temperature-heat graph is not a single straight line: it alternates sections in which the temperature rises (ice, then liquid water, then vapour warming up — $Q=mc\Delta T$ applies) with flat sections, at constant temperature, in which the change of state takes place ($Q=m\lambda$ applies: $0{}^{\circ }\text{C}$ during melting, $100{}^{\circ }\text{C}$ during boiling, at atmospheric pressure). Recognising which section you are in — whether you are warming a phase or changing it — is the first step in choosing the right formula.

The post does not go into thermal expansion, a topic that usually comes before this one in the syllabus: it is a chapter of its own, with its own formulas ($\Delta L=L_0\alpha \Delta T$) and its own logic.

Typical mistakes

Three worked exercises

1. A hot metal dropped into water. A copper block of mass $m_1=0.2\text{kg}$ at temperature $T_1=90{}^{\circ }\text{C}$ is immersed in $m_2=0.5\text{kg}$ of water at $T_2=20{}^{\circ }\text{C}$, in an insulated container. Find the equilibrium temperature.

I use $Q_{\text{given up}}=Q_{\text{absorbed}}$, with $c_{\text{copper}}=385\text{J}/(\text{kg}\cdot {}^{\circ }\text{C})$ and $c_{\text{water}}=4186\text{J}/(\text{kg}\cdot {}^{\circ }\text{C})$:

The final temperature is very close to the water’s starting $20{}^{\circ }\text{C}$: this is the direct consequence of the specific heat of water, much higher than that of copper — even starting much hotter, the small mass of copper cannot shift the temperature of the water, five times larger in mass, by very much.

2. The calorimeter, to find an unknown specific heat. In a calorimeter with water equivalent $m_{eq}=0.02\text{kg}$, containing $m_a=0.3\text{kg}$ of water at $T_a=18{}^{\circ }\text{C}$, you immerse a block of mass $m_x=0.15\text{kg}$ of an unknown material, heated to $T_x=100{}^{\circ }\text{C}$. The measured equilibrium temperature is $T_{eq}=23{}^{\circ }\text{C}$. Find the specific heat of the material.

The heat given up by the unknown block is absorbed both by the water and by the calorimeter (treated as extra water mass):

A plausible value for a metal alloy, between that of aluminium and that of water: it is a good final check to verify that the number you found is the right order of magnitude for a solid.

3. Ice in water, checking whether it all melts. In an insulated container there are $m_a=0.4\text{kg}$ of water at $T_a=25{}^{\circ }\text{C}$. You add $m_g=0.05\text{kg}$ of ice at $0{}^{\circ }\text{C}$. Does the ice melt completely? Find the final temperature.

First of all I check whether the heat the water can give up as it cools to $0{}^{\circ }\text{C}$ is enough to melt all the ice.

Maximum heat available from the water (cooling from $25{}^{\circ }\text{C}$ to $0{}^{\circ }\text{C}$):

Heat needed to melt all the ice:

Since $Q_{\text{available}}=41860\text{J}>Q_{\text{fusion}}=16700\text{J}$, all the ice melts, and there is heat left over to warm the melted water (at $0{}^{\circ }\text{C}$) together with the rest of the water up to a common final temperature $T_{eq}$ above zero. I write the full balance, with the whole final mass ($0.45\text{kg}$ of water) warming from $0{}^{\circ }\text{C}$ to $T_{eq}$, balanced by the residual heat given up by the hot water cooling from $25{}^{\circ }\text{C}$:

The “does it all melt?” check before writing the final balance is the step that almost everyone skips, and it is exactly the step the problem wants to see: if the available heat had been less than the heat of fusion, the answer would have been “the ice does not all melt, the final temperature is $0{}^{\circ }\text{C}$”, with no further calculation.

Test yourself with the dedicated practice test

On the site you will find the thermal physics test practice paper, calibrated exactly on this Year 2 scientifico syllabus: temperature scales, specific heat, thermal equilibrium, latent heat. As always, the correct answer is free, and the guided step-by-step solution is unlocked with the first lesson.

If you want to review the rest of the physics syllabus too, you will find everything on the Physics page. If you are just now finishing the previous block, the companion post is Fluid statics in Year 2: pressure, Stevin, Archimedes. And for a wider review of the Year 2 programme, equations and inequalities included, there is Year 2: the whole maths programme in one formula sheet.

If you want a hand setting up this block before the December test, the first introductory call is free and lasts half an hour.