Calorimetry and latent heat

Drop a hot metal block into a bucket of cool water and, within a minute, both settle at some in-between temperature. Nothing was created or destroyed — the heat the metal lost is exactly the heat the water gained. That one bookkeeping rule, the principle of calorimetry, is powerful enough to predict the final state of almost any mixing or heating problem, from a coffee cooling to a glacier melting. But not all heat shows up as a temperature change. Melt an ice cube and its thermometer reading refuses to budge past 0°C until every crystal has turned to liquid — the heat is going somewhere else entirely: prying molecules apart, not speeding them up. This hidden energy is latent heat, and it is why phase changes dominate so many real heat budgets even though they happen at a single fixed temperature. This article builds the two governing equations, $Q = mc\Delta T$ and $Q = mL$ , shows how they combine in the principle of calorimetry, and works through a full multi-stage ice-to-steam calculation — the classic exam problem that tests whether you can tell sensible heat from latent heat at a glance.

The core insight

In an isolated system heat lost by hotter bodies equals heat gained by colder ones, with temperature changes costing $Q = mc\Delta T$ and phase changes costing $Q = mL$ at constant temperature.

The core idea

Specific heat capacity. Raising the temperature of a mass $m$ by $\Delta T$ without changing its phase costs heat proportional to both:

Q = mc\,\Delta T

where: $Q$ is heat transferred (J), $m$ is mass (kg), $c$ is the specific heat capacity (J kg $^{-1}$ K $^{-1}$ — heat needed to raise 1 kg by 1 K), and $\Delta T$ is the temperature change (K or °C, since the size of a degree is the same on both scales). Water's $c_w = 4186$ J kg $^{-1}$ K $^{-1}$ is unusually large, which is why large water bodies resist temperature swings.

Principle of calorimetry. In an insulated system where no heat escapes to the surroundings, energy conservation demands

\text{heat lost by hotter body(ies)} = \text{heat gained by colder body(ies)}

This single equality, applied with $Q = mc\Delta T$ on each side, is enough to solve any mixing problem — find the shared final temperature so that both sides balance.

Latent heat. A substance changing state (solid to liquid, liquid to gas) at its fixed melting or boiling point absorbs or releases heat with no change in temperature, because the energy goes into breaking (or forming) intermolecular bonds rather than speeding up molecules:

Q = mL

where: $L$ is the specific latent heat (J kg $^{-1}$ ) — the heat needed to change the state of 1 kg. For water, latent heat of fusion $L_f \approx 3.34\times10^5$ J/kg (melting at 0°C) and latent heat of vaporisation $L_v \approx 2.26\times10^6$ J/kg (boiling at 100°C, atmospheric pressure). $L_v$ is nearly seven times $L_f$ , because boiling must fully separate molecules into a gas while melting only loosens the rigid lattice. A heating curve for ice-to-steam makes both ideas visible at once: sloped segments where $Q=mc\Delta T$ raises temperature, and flat plateaus at 0°C and 100°C where $Q=mL$ changes phase with the thermometer frozen in place.

Figure

A temperature versus heat added graph for ice warming to steam showing sloped rising segments for sensible heat and flat plateaus at zero and one hundred degrees Celsius for melting and boiling

Derivation of the mixing formula

Derivation AFinal temperature of a two-body mixture from the calorimetry principle

The goal is to turn the principle "heat lost equals heat gained" into an explicit algebraic formula for the common final temperature $\theta$ of two bodies at different starting temperatures, then check it against a real number pair (copper dropped into water) so the algebra and the physical answer can be compared side by side.

Step 1. Set up the two heat quantities. Let a hot body of mass $m_1$ , specific heat $c_1$ , start at temperature $T_1$ , and a cold body of mass $m_2$ , specific heat $c_2$ , start at $T_2$ , with $T_1 > T_2$ . Both end at the same equilibrium temperature $\theta$ , so the hot body's heat loss is $Q_{\text{lost}} = m_1c_1(T_1-\theta)$ and the cold body's heat gain is $Q_{\text{gained}} = m_2c_2(\theta-T_2)$ — both expressions are written as positive quantities because $T_1 > \theta > T_2$ .

Step 2. Apply the calorimetry principle. Setting the two equal, $m_1c_1(T_1-\theta) = m_2c_2(\theta-T_2)$ . Expand both brackets: $m_1c_1T_1 - m_1c_1\theta = m_2c_2\theta - m_2c_2T_2$ .

Step 3. Collect every $\theta$ term on one side. Adding $m_1c_1\theta$ and $m_2c_2T_2$ to both sides gives $m_1c_1T_1 + m_2c_2T_2 = m_2c_2\theta + m_1c_1\theta = (m_1c_1+m_2c_2)\theta$ , so

\theta = \frac{m_1c_1T_1 + m_2c_2T_2}{m_1c_1+m_2c_2}.

This is a weighted average of the two starting temperatures, weighted by each body's heat capacity $mc$ — the body with the larger $mc$ pulls $\theta$ closer to its own starting temperature.

Step 4. Test it against the copper-water numbers: $m_1=0.50$ kg copper at $T_1=200^\circ$ C with $c_1=390$ J kg $^{-1}$ K $^{-1}$ , and $m_2=1.0$ kg water at $T_2=20^\circ$ C with $c_2=4186$ J kg $^{-1}$ K $^{-1}$ . Then $m_1c_1 = 195$ J/K and $m_2c_2 = 4186$ J/K, so

\theta = \frac{195(200) + 4186(20)}{195+4186} = \frac{39000+83720}{4381} = \frac{122720}{4381} = 28.0^\circ\text{C}.

This matches the direct "heat lost = heat gained" solution exactly, and it shows algebraically why $\theta$ lands so close to water's 20°C: water's heat capacity $m_2c_2=4186$ J/K dominates copper's $195$ J/K by more than twenty to one, so the weighted average sits almost entirely on water's side. Check: the formula is dimensionally a weighted average of temperatures (units cancel to °C), and $\theta=28.0^\circ$ C lies strictly between $20^\circ$ C and $200^\circ$ C as physically required.

Worked example

Find the total heat needed to convert 100 g of ice at $-10^\circ$ C completely into steam at $100^\circ$ C. Take $c_{\text{ice}} = 2100$ J kg $^{-1}$ K $^{-1}$ , $c_w = 4186$ J kg $^{-1}$ K $^{-1}$ , $L_f = 3.34\times10^5$ J/kg, $L_v = 2.26\times10^6$ J/kg, $m = 0.100$ kg.

Step 1. Identify the stages. Ice must first warm to 0°C (sensible heat), then melt at 0°C (latent heat), then the resulting water must warm to 100°C (sensible heat), then boil at 100°C (latent heat) — four stages in sequence, each computed separately and then summed.

Step 2. Warm the ice: $Q_1 = mc_{\text{ice}}\Delta T = (0.100)(2100)(10) = 2100$ J. Melt the ice: $Q_2 = mL_f = (0.100)(3.34\times10^5) = 33{,}400$ J.

Step 3. Warm the water: $Q_3 = mc_w\Delta T = (0.100)(4186)(100) = 41{,}860$ J. Boil the water: $Q_4 = mL_v = (0.100)(2.26\times10^6) = 226{,}000$ J.

Step 4. Sum all four stages: $Q = Q_1+Q_2+Q_3+Q_4 = 2100+33{,}400+41{,}860+226{,}000 = 303{,}360$ J $\approx 3.03\times10^5$ J. The two latent-heat stages ( $Q_2+Q_4 = 259{,}400$ J) alone account for over 85% of the total, despite happening at only two fixed temperatures — this is the signature of latent heat dominating a real heat budget.

The total heat required is $\mathbf{3.03\times10^5\ \text{J}}$ (303,360 J).

Competitive stretch

If only $150{,}000$ J were supplied instead, where would the system end up? Stage 1 and stage 2 together need $2100+33{,}400=35{,}500$ J, leaving $114{,}500$ J for stage 3, which needs $41{,}860$ J to fully reach 100°C — so all the ice melts, the water is fully heated to 100°C using $41{,}860$ J, and the remaining $114{,}500-41{,}860=72{,}640$ J goes into partial boiling. Since $Q_4 = mL_v$ needs $226{,}000$ J for complete vaporisation, the mass actually turned to steam is $m' = 72{,}640/2.26\times10^6 \approx 0.0322$ kg — about 32 g of steam, with 68 g still liquid water at 100°C.

Common traps

Applying $Q=mc\Delta T$ across a phase change — no temperature rise occurs while melting or boiling, so that stage must be isolated and handled with $Q=mL$ instead.
Skipping a stage in a multi-step problem (e.g. forgetting to heat the melted ice from 0°C to 100°C before applying $L_v$ ).
Sign confusion — always write "heat lost = heat gained" using positive quantities on both sides, rather than juggling negative $\Delta T$ terms.
Mixing up $L_f$ and $L_v$ , or reusing water's values for a different substance without checking the given data.

Ask the tutor: if a mixing problem or a multi-stage phase-change calculation isn't balancing, ask the AI tutor to walk through your specific numbers stage by stage.

Why is the latent heat of vaporisation so much bigger than the latent heat of fusion?

Melting only has to loosen a solid's rigid lattice enough for molecules to slide past each other while staying close together — most of the intermolecular attraction survives in the liquid. Boiling has to do far more work: it must pull molecules essentially all the way apart, into a gas where they barely interact, against the full strength of the attractive forces. That much larger bond-breaking job is why $L_v$ for water is nearly seven times $L_f$ .

Does the calorimetry principle heat lost equals heat gained always hold exactly?

Only if the system is truly isolated, meaning no heat escapes to or enters from the surroundings. Real calorimeters lose a little heat through their walls and the vessel itself absorbs some heat, which is why careful experiments include a correction for the container's own heat capacity. Textbook problems that say "calorimeter of negligible heat capacity" are simplifying away that correction so you can focus on the two substances being mixed.

If I add up all four stages for the ice to steam problem, do I need to worry about the order I add them in?

No — heat is a scalar quantity being summed, so $Q = Q_1+Q_2+Q_3+Q_4$ gives the same total regardless of the order you add the numbers. What does matter is getting the physical order of stages right when computing each $Q_i$ : you must fully melt the ice before you can start warming liquid water, and fully warm the water to 100°C before boiling can begin, because each stage's starting temperature depends on the previous stage finishing.