ARCHETYPE: Method-of-Mixtures Problems

The instrument never left. Bomb calorimeters burn a food sample in oxygen inside a steel vessel immersed in a stirred water bath, and the calorie count on a cereal packet is the temperature rise of that bath run backwards through $\sum Q_{\text{lost}} = \sum Q_{\text{gained}}$. Differential scanning calorimetry does the same trick with microwatts, tracking the tiny latent heats absorbed as a polymer melts or a protein unfolds — a plateau on a graph, exactly like the one that pins the ice water at zero. The ledger also designs anything built to move heat on purpose. Phase-change materials exploit the latent term deliberately: a wax sealed into a ceiling panel, chosen to melt at room temperature, absorbs a huge slug of heat at almost constant temperature and flattens an office's afternoon peak with no air conditioning at all — the physics of the $6.1$ g of surviving ice, aimed at a building. A laptop's heat pipe runs on $L_v$, the same monster term that made $10$ g of steam worth nearly $30$ K. Every one of them is keeping Laplace's account: joules out on one side, joules in on the other, and the books must balance.

Picture the apparatus, because in an archetype the setup is the physics. A calorimeter is a metal cup — copper, usually — sitting inside an insulating jacket, with a stirrer and a thermometer through the lid. Inside it: some water at a known temperature. Into it: a hot body, or an ice cube, or a jet of steam. The lid goes on, the stirrer turns, and after a short while every part of the contents reads the same temperature. That common end-point is thermal equilibrium, and reaching it is the only thing the apparatus is for. The insulation is the whole reason the problem is solvable. "Isolated" means no heat leaks in from the room and none escapes to it, so the energy inside is a closed account. Whatever the hot contents lose, the cold contents must have received — there is nowhere else for it to go:

where: $\sum Q_{\text{lost}}$ — the total heat given out by everything that ends up colder than it started; $\sum Q_{\text{gained}}$ — the total heat taken in by everything that ends up warmer (or that melts, or boils). This is the principle of calorimetry, and it is nothing more than conservation of energy with the leaks sealed. Read it as a balance sheet: the left column is money out, the right column is money in, and the books must close to the last joule.

Three idealizations turn this cartoon into an exam problem, and a good board answer states them:

The system is isolated. No conduction through the jacket, no convection out of the lid, no radiation to the lab. Real calorimeters leak a little, which is why careful experimenters apply a cooling correction — but for the problem in front of you, the account is closed.

Stirring makes the temperature common. Without it, water near the hot body would be hotter than water at the wall, and "the final temperature" would not be a single number. The stirrer buys us the right to write one symbol, $T_f$.

Everything ends at the same temperature. That is what equilibrium means, and it is what lets you connect two bodies you were handed separately. It is also the assumption that breaks when a phase change is involved: if ice is left over at the end, the final temperature is not free at all — it is nailed to $0\ ^\circ\mathrm{C}$.

Every entry in the ledger is one of exactly two kinds. Nothing else exists at this level, and half of all lost marks come from writing the wrong species.

Sensible heat changes the temperature of a body that stays in the same state:

where: $m$ — mass in kg; $c$ — specific heat capacity in J/kg·K (water $4180$, ice $2100$, steam $2010$, copper $386$, iron $450$); $\Delta T$ — the temperature change, taken as a positive magnitude once the term sits on the correct side of the ledger. "Sensible" is the old word for sensed: this is the heat the thermometer feels. Note the scale of $c_w = 4180$ — water needs about ten times as much heat per kilogram per kelvin as a typical metal, which is why a metal spoon in soup burns your fingers and the soup does not.

Latent heat changes the state of a body at constant temperature:

where: $L$ — the specific latent heat in J/kg: for fusion (melting or freezing) of ice $L_f = 3.34\times10^{5}$; for vaporization (boiling or condensing) of water $L_v = 2.26\times10^{6}$. There is no $\Delta T$ here, and there must not be — the temperature is not changing. "Latent" means hidden: the joules go into breaking the bonds that hold the solid together, not into speeding molecules up, so the thermometer sees nothing. Feel the sizes, because the sizes are what make calorimetry surprising. Divide each latent heat by $c_w$ to get the water-cooling that would buy the same energy: $L_f/c_w = 79.9\ \mathrm{K}$ and $L_v/c_w = 540.7\ \mathrm{K}$. In words: melting one gram of ice costs as much heat as cooling one gram of water by $80\ ^\circ\mathrm{C}$, and condensing one gram of steam releases as much as cooling one gram of water by $540\ ^\circ\mathrm{C}$. The latent terms are giants — they routinely dominate the sum, and they are exactly the terms students forget.

Now assemble the archetype. Take a hot body (mass $m_1$, specific heat $c_1$, temperature $T_1$) and a cold body (mass $m_2$, specific heat $c_2$, temperature $T_2 < T_1$), with no phase change anywhere. Let them come to a common final temperature $T_f$. Heat lost by the hot body equals heat gained by the cold one:

where: each bracket is a positive temperature drop or rise — the hot body falls from $T_1$ to $T_f$, the cold body climbs from $T_2$ to $T_f$. Writing the brackets this way is the sign discipline that saves you: no minus signs to babysit, and every quantity in sight is positive. Solve (4) for $T_f$:

where: the products $m c$ are heat capacities (J/K) — the "thermal inertia" of each body. Read equation (5) physically: the final temperature is the average of the starting temperatures, weighted by heat capacity. The body with the bigger $mc$ drags the answer toward itself, exactly as the bigger mass dominates a centre-of-mass calculation. Two consequences double as instant sanity checks. The final temperature always lies strictly between $T_1$ and $T_2$, because a weighted average of two numbers cannot escape them. An answer of $95\ ^\circ\mathrm{C}$ for a mix of water at $80$ and $20\ ^\circ\mathrm{C}$ is an algebra slip — stop and find it. And if the two bodies are the same substance ($c_1 = c_2$), the $c$ cancels and (5) collapses to a plain mass-weighted average. Equal masses of water at $80$ and $20\ ^\circ\mathrm{C}$ therefore meet at exactly $50\ ^\circ\mathrm{C}$, while $100$ g at $80\ ^\circ\mathrm{C}$ poured into $200$ g at $20\ ^\circ\mathrm{C}$ lands at $40\ ^\circ\mathrm{C}$ — two-thirds of the way toward the cold side, because the cold side carries twice the heat capacity.

Real calorimeters are made of metal, and metal warms up. The copper cup, the stirrer and (strictly) the thermometer bulb all sit at the water's temperature and all absorb a share of the heat. Ignore them and your answer is systematically wrong — in the same direction, every time. The tidy fix is to give the calorimeter an alias. Its heat capacity is $m_{\text{cal}}c_{\text{cal}}$ J/K. Ask: what mass of water would have that same heat capacity? Call it the water equivalent $w$:

where: $w$ — the water equivalent in kg, a fictitious extra slug of water that soaks up heat exactly as the cup does. Once you have $w$, the cup disappears from the problem: just pretend the calorimeter contains $(m_{\text{water}} + w)$ of water and use the master equation unchanged. A $100$ g copper calorimeter has $m_{\text{cal}}c_{\text{cal}} = 0.100 \times 386 = 38.6\ \mathrm{J/K}$, so $w = 38.6/4180 = 9.23\ \mathrm{g}$. A whole $100$ g cup of copper is thermally worth barely nine grams of water — copper's specific heat is roughly a tenth of water's. Small, but not zero. Run the $40\ ^\circ\mathrm{C}$ mixture of Example 1 inside that cup — the cold side becomes $200 + 9.23 = 209.23$ g of water — and the answer drops to $12\,184.6/309.23 = 39.4\ ^\circ\mathrm{C}$. A shift of $0.6$ K is the difference between a right answer and a "close" one, and when a question hands you $m_{\text{cal}}$ and $c_{\text{cal}}$ it is because the examiner intends to charge you for ignoring them.

Four dressings of the same ledger cover essentially every calorimetry question a board has ever set. The equation never changes; only the list of terms does.

Variant 1 — the two-liquid mix. Two masses of liquid at different temperatures; find $T_f$. Straight application of equation (5), and if both are water the specific heat cancels.

Variant 2 — the unknown specific heat. Drop a hot metal block into cool water, measure the final temperature, and solve the ledger for $c_{\text{metal}}$ instead of $T_f$. This is the classic school experiment — Black's original method of mixtures — so the rearranged form is worth memorising:

where: the numerator is the heat the water gained; the denominator is (mass × temperature drop) of the metal. Count the calorimeter and $m_w c_w$ becomes $(m_w + w)c_w$ — a correction worth about $3\%$, as Example 5 shows.

Variant 3 — the ice interrupt. Ice meets warm water. Run the budget test of equation (7) first. If the melt is incomplete, the answer is $T_f = 0\ ^\circ\mathrm{C}$ plus a mass of melted ice; if it completes, carry the leftover heat into an ordinary mix. Ice starting below $0\ ^\circ\mathrm{C}$ needs a sensible term $m_i c_{\text{ice}}\,\Delta T$ first (with $c_{\text{ice}} = 2100$, not $4180$) to reach the melting point before any melting can begin.

Variant 4 — steam condensing. Steam at $100\ ^\circ\mathrm{C}$ blown into cool water gives up two things in order: its latent heat $m_s L_v$ as it condenses to water at $100\ ^\circ\mathrm{C}$, then sensible heat $m_s c_w (100 - T_f)$ as that condensate cools. The ledger reads

where: the $L_v$ term is enormous — $2.26\times10^{6}$ J/kg — which is why a few grams of steam heat a whole beaker startlingly far. The mirror-image test applies here too: if the steam offers more than the water needs to reach $100\ ^\circ\mathrm{C}$, not all of it condenses and the mixture is pinned at $100\ ^\circ\mathrm{C}$ with steam left over. Same interrupt logic, other end of the thermometer.

Every one of these can be read straight off equation (5) or the budget test, and each is a five-second check on an answer you are unsure of. Note where the final temperature is allowed to land: a pure mix lands strictly between the two starts, and only an unfinished phase change can pin it to a plateau — $0\ ^\circ\mathrm{C}$ with ice left over, $100\ ^\circ\mathrm{C}$ with steam left over.

1 · Board level — the two-liquid mix (CBSE/ICSE pattern, 3 marks)

$100$ g of water at $80\ ^\circ\mathrm{C}$ is poured into $200$ g of water at $20\ ^\circ\mathrm{C}$ in a vessel of negligible heat capacity. Find the final temperature of the mixture. Take $c_w = 4180$ J/kg·K.

Step 1. Check for phase changes. Both bodies are liquid water and any $T_f$ between $20$ and $80\ ^\circ\mathrm{C}$ keeps them liquid — no melting, no boiling, so the ledger carries sensible terms only. Saying this in one line is what earns the method mark.

Step 2. Name the sides — hot water falling from $80$ to $T_f$, cold water rising from $20$ to $T_f$:

Step 3. The specific heat is common to both sides, so cancel it:

Step 4. Collect and solve: $12\,000 = 300\,T_f$, so $T_f = \boxed{40\ ^\circ\mathrm{C}}$.

Step 5. Check the books. Lost $= 0.100 \times 4180 \times 40 = 16\,720$ J; gained $= 0.200 \times 4180 \times 20 = 16\,720$ J ✓. And $40\ ^\circ\mathrm{C}$ lies between $20$ and $80$ ✓, nearer the cold side because the cold side has twice the heat capacity — the hot water fell $40$ K while the cold water rose only $20$ K.

2 · Board level — the unknown specific heat of a metal (the method of mixtures)

A $200$ g block of metal is heated to $100\ ^\circ\mathrm{C}$ and dropped into $300$ g of water at $20\ ^\circ\mathrm{C}$. The mixture settles at $25\ ^\circ\mathrm{C}$. Neglecting the calorimeter, find the specific heat capacity of the metal.

Step 1. Direction of flow: the metal is hot, so it loses; the water is cold, so it gains. Nothing melts or boils, so there are no latent terms.

Step 2. Heat gained by the water, which rises by $25 - 20 = 5$ K:

Step 3. Heat lost by the metal, which falls by $100 - 25 = 75$ K. That drop is the number students most often get wrong — it runs to the final temperature, not to the water's starting temperature.

Step 4. Close the books and solve:

Step 5. Sanity: one tenth of water's $4180$, squarely in the metal range (copper $386$, iron $450$). A value larger than water's would mean you divided by the wrong temperature drop. The experiment's design says the same thing — a big fall on the metal side ($75$ K) against a small rise on the water side ($5$ K) is exactly what a small $c$ looks like.

3 · Competitive level — the ice interrupt (JEE-Main/NEET pattern, ~3 minutes)

$300$ g of water at $25\ ^\circ\mathrm{C}$ is poured onto $100$ g of ice at $0\ ^\circ\mathrm{C}$ in an insulated flask. Find the final temperature and the mass of ice that melts. Take $L_f = 3.34\times10^{5}$ J/kg, $c_w = 4180$ J/kg·K.

Step 1. Do not write a linear equation yet. Run the budget test. Heat available from the water, cooling all the way to the plateau at $0\ ^\circ\mathrm{C}$:

Step 2. Heat demanded to melt the whole block:

Step 3. Compare: $31\,350 < 33\,400$. The water runs out of energy $2050$ J short of a complete melt. So the melt is incomplete — ice survives, and while ice survives, the mixture cannot leave the melting point:

Step 4. Mass melted, from equation (7): the $31\,350$ J that were available bought

leaving $100 - 93.9 = \boxed{6.1\ \mathrm{g}}$ of ice floating in $393.9$ g of water, the whole flask sitting at exactly $0\ ^\circ\mathrm{C}$.

Step 5. The counterfactual, so the boundary is unforgettable. Had the ice been only $50$ g, the demand would have been $0.050 \times 3.34\times10^{5} = 16\,700$ J — comfortably inside the budget. All of it melts, $31\,350 - 16\,700 = 14\,650$ J is left over, and that surplus warms the combined $350$ g of water: $T_f = 14\,650/(0.350 \times 4180) = 10.0\ ^\circ\mathrm{C}$. Same water, half the ice, and a completely different kind of answer. Timing note: the budget test is two multiplications — under thirty seconds — and it decides whether the rest of the question takes one line or five. In an MCQ, "$0\ ^\circ\mathrm{C}$ with ice left over" is usually one of the options precisely because most candidates never test for it. Threshold worth memorising: with $300$ g of water you would need a start of $26.6\ ^\circ\mathrm{C}$ to melt all $100$ g of the ice.

4 · Competitive level — steam condensing (JEE/NEET pattern, ~3 minutes)

$10$ g of steam at $100\ ^\circ\mathrm{C}$ is passed into $200$ g of water at $20\ ^\circ\mathrm{C}$ in an insulated container of negligible heat capacity. Find the final temperature. Take $L_v = 2.26\times10^{6}$ J/kg, $c_w = 4180$ J/kg·K.

Step 1. Completion test first, at the upper plateau this time. The most the steam can ever deliver is its latent heat plus the cooling of its condensate all the way down to $20\ ^\circ\mathrm{C}$:

The heat needed to take the water all the way up to $100\ ^\circ\mathrm{C}$ is $0.200 \times 4180 \times 80 = 66\,880$ J. Since $25\,944 < 66\,880$, the water never reaches $100\ ^\circ\mathrm{C}$: all the steam condenses, and $T_f$ lands strictly between $20$ and $100\ ^\circ\mathrm{C}$.

Step 2. Now write the ledger, equation (9). Steam side loses twice — condensing, then cooling:

Step 3. Expand: $22\,600 + 41.8(100 - T_f) = 836(T_f - 20)$, i.e. $22\,600 + 4180 - 41.8\,T_f = 836\,T_f - 16\,720$.

Step 4. Collect: $43\,500 = 877.8\,T_f$, so

Step 5. Feel the size of the latent term. Ten grams — one twentieth of the water's mass — dragged the mixture up by nearly $30$ K. Replace the steam with $10$ g of liquid water at $100\ ^\circ\mathrm{C}$ and equation (5) gives only $23.8\ ^\circ\mathrm{C}$: the phase change alone is worth an extra $25.7$ K. That is why steam burns are so much worse than boiling-water burns — a gram of steam landing on $37\ ^\circ\mathrm{C}$ skin dumps $2523$ J, against $263$ J for a gram of boiling water: a factor of $9.6$. Timing note: candidates lose this question by dropping the $m_s c_w(100 - T_f)$ term — forgetting that the condensate is still at $100\ ^\circ\mathrm{C}$ and has more heat to give. Budget ninety seconds: completion test, ledger, solve. If the numbers ever give $T_f > 100\ ^\circ\mathrm{C}$, do not "round down" — it means some steam never condensed and the true answer is exactly $100\ ^\circ\mathrm{C}$.

5 · Competitive level — JEE-Advanced stretch: cold ice, and the calorimeter correction (~5 minutes)

Part (a): repeat Example 3 with the ice starting at $-10\ ^\circ\mathrm{C}$ instead of $0\ ^\circ\mathrm{C}$ ($c_{\text{ice}} = 2100$ J/kg·K). Part (b): repeat Example 2, but now the $300$ g of water sits in a $100$ g copper calorimeter ($c_{\text{Cu}} = 386$ J/kg·K) that starts at $20\ ^\circ\mathrm{C}$ with it. By what percentage was the metal's specific heat underestimated?

Step 1. Part (a): the ice cannot melt until it reaches its melting point, so a sensible term now comes first. Warming $100$ g of ice from $-10$ to $0\ ^\circ\mathrm{C}$ costs

Use $c_{\text{ice}} = 2100$, not $c_w$ — ice is not water, and this is a favourite one-mark theft.

Step 2. Re-run the budget. The water still offers $31\,350$ J, but $2100$ J of it is eaten before any melting begins, leaving $29\,250$ J for the melt. Demand is still $33\,400$ J, so the melt is still incomplete:

with $12.4$ g of ice left — about $6.3$ g more ice saved than in Example 3, bought by the ice's $10$ K head start.

Step 3. Part (b): the calorimeter joins the cold side, because it starts at $20\ ^\circ\mathrm{C}$ with the water and warms with it. The gaining heat capacity is now

Step 4. Redo equation (8) with that corrected capacity:

The water-equivalent route must agree, and does: $(0.300 + 0.00923) \times 4180 \times 5 / (0.200 \times 75)$ returns the same $430.9$.

Step 5. The verdict. Ignoring the cup gave $418$ instead of $430.9$ — an underestimate of $3\%$, and always an underestimate, because the heat the cup absorbed was silently credited to nobody. A systematic error, not a random one: repeating the experiment a hundred times would never fix it. Timing note: part (a) is Example 3 with one extra line at the front — recognise it and it costs ninety seconds. The trap is arithmetic drift: students subtract the $2100$ J from the wrong budget, or add it to the demand side. Keep the sides honest — the ice warming is heat GAINED, so it comes out of the available pot before the melt begins.

Skipping the budget test — the number-one mark killer. Any question containing ice or steam must begin with the completion test, before a single equation for $T_f$ is written. Compare the heat available with the heat a complete phase change demands. Get this wrong and everything downstream is wasted ink, however neat the algebra. (CBSE and ICSE reward the test explicitly; JEE and NEET simply put "$0\ ^\circ\mathrm{C}$" in the options and wait.)

Sign discipline on the ledger. Write each term as a positive quantity and put it on the side it belongs to: bodies that cool go on the LOST side with $(T_{\text{start}} - T_f)$; bodies that warm, melt or boil go on the GAINED side with $(T_f - T_{\text{start}})$ or $mL$. Mixing $\Delta T = T_f - T_{\text{start}}$ into a "lost" term flips a sign and silently corrupts the answer. If you prefer the $\sum Q = 0$ convention, use it consistently — but never blend the two.

Latent heat is $mL$, never $mL\,\Delta T$ — and the wrong $c$ is just as fatal. There is no temperature change during a phase transition, so there is no $\Delta T$ to multiply by; conversely, a solid that merely sits at $0\ ^\circ\mathrm{C}$ without melting gets no latent term at all. Ask of every term: is this body changing temperature, or changing state? It cannot do both at once. And use the right specific heat when it is changing temperature — ice $2100$, water $4180$, steam $2010$ J/kg·K. Writing $4180$ for ice roughly doubles that term and quietly poisons the melt budget. (For $\Delta T$ itself, Celsius and kelvin are interchangeable: a change of $1\ ^\circ\mathrm{C}$ is a change of $1$ K.)

Forgetting the calorimeter. If the question gives you the cup's mass and specific heat, it wants them used — via $(m_w + w)c_w$ with $w = m_{\text{cal}}c_{\text{cal}}/c_w$, or by adding $m_{\text{cal}}c_{\text{cal}}$ straight to the gaining heat capacity. Both routes give the same number. Ignoring the cup is wrong by a few percent in a predictable direction: measured specific heats come out too low, as Example 5 shows ($418$ instead of $430.9$). CBSE and ISC practicals mark this explicitly.

The metal's temperature drop is to the FINAL temperature. In "hot block into cool water", the drop is $(T_{\text{metal}} - T_f)$, not $(T_{\text{metal}} - T_{\text{water}})$. Both bodies end at $T_f$; that is what equilibrium means. This single slip accounts for more wrong specific heats than any other.

An answer outside the starting temperatures is a bug report. A pure mix must land strictly between the two starts. A negative $T_f$, or one above $100\ ^\circ\mathrm{C}$, is the algebra telling you that a phase change interrupted the sum and pinned the answer to a plateau. Do not round it away. Go back and re-test.

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) · HC Verma, Concepts of Physics Vol. 2, Ch. 25 (Calorimetry) · Halliday, Resnick and Walker, Fundamentals of Physics, Ch. 18 · Zemansky and Dittman, Heat and Thermodynamics, for the experimental technique.

Provenance. Every number here is recomputed in `verify_hs2-calorimetry-mixing-problems.py` from $c_w = 4180$, $c_{\text{ice}} = 2100$, $c_{\text{steam}} = 2010$, $c_{\text{Cu}} = 386$, $L_f = 3.34\times10^{5}$, $L_v = 2.26\times10^{6}$ J/kg. The list a verifier reproduces: the mix at $40\ ^\circ\mathrm{C}$ with $16\,720$ J crossing the ledger; the metal at $418$ J/kg·K, and $430.9$ once the cup is counted; the water equivalent $9.23$ g and the shifted mix at $39.4\ ^\circ\mathrm{C}$; the ice budget $31\,350$ J against $33\,400$ J, giving $93.9$ g melted, $6.1$ g surviving, $T_f = 0\ ^\circ\mathrm{C}$, and a threshold start of $26.6\ ^\circ\mathrm{C}$; the $50$ g branch at $10.0\ ^\circ\mathrm{C}$; the steam mix at $49.6\ ^\circ\mathrm{C}$, against $23.8\ ^\circ\mathrm{C}$ without the latent term; the cold-ice case melting $87.6$ g; and the ratios $L_f/c_w = 79.9$ K, $L_v/c_w = 540.7$ K.

Prerequisites. Specific Heat Capacity · The Principle of Calorimetry · Latent Heat and Change of State. Leads to. Heating Curves and Latent-Heat Numericals — the same ledger drawn as a graph, where the plateaus you tested for become flat stretches you can measure.

The deep dive. Calorimetry takes $c$ as a measured number. Why $c$ has the value it does is a statistical-mechanics question, opened by the undergraduate article "Heat Capacity of Classical Gases", which counts a molecule's degrees of freedom and predicts $c$ from first principles. Not needed for any board exam; irresistible once the bookkeeping feels easy.