Calorimetry: the final temperature when two bodies are mixed

Give it two masses, two specific heats, two starting temperatures and the calorimeter’s own thermal capacity, and it returns the one temperature both bodies end at — with the heat each gave up or took in, the check that the three sums cancel, and what ignoring the vessel would have cost you.

Last updated 11 Nov 2026

Final temperature of a mixture

Two bodies, one calorimeter, one final temperature

Mass of the first body (the one the calorimeter holds)In kilograms: 250 g is 0.25. This is the body already in the vessel, so the calorimeter and stirrer start at ITS temperature and not the other body’s. Curiously, putting both masses in grams leaves the final temperature exactly right, because the mass unit cancels in a ratio of two sums — but every joule on the page is then out by a factor of a thousand.

Specific heat capacity of the first bodyFive values, each with a document behind it. Water is listed three times on purpose: NCERT’s own module prints 4186 in its text and 4.2 × 10³ in one of its own examples, and the measured value is 4181.6 at 25 °C, rising to 4220 at 0 and 4216 at 100. Aluminium, iron, brass, lead and glass are deliberately absent. The figures everyone quotes (900, 450, 380, 130, 840) are in a thousand places and this page could trace none of them to a published source, so it asks you for the one in your question instead of printing them with a confident air. Choose the last option and the box below unlocks.

Specific heat capacity of the first body, typed inLocked while a value is selected above. Pick the last option up there and this becomes the input, which is how aluminium, brass, ethanol or anything else your question names still works. The arithmetic does not care where the number came from — the UNIT does. This field wants J kg⁻¹ K⁻¹: a table in J g⁻¹ K⁻¹ needs ×1000, one in cal g⁻¹ °C⁻¹ needs ×4186. Entering 0.39 for copper instead of 390 is the commonest way a calorimetry answer comes out a thousand times wrong.

Starting temperature of the first body (and of the calorimeter)In degrees Celsius, and it may be negative. Celsius is safe here even though gas-law work demands kelvin: every temperature in this formula sits inside a difference, and a difference of 1 °C is a difference of exactly 1 K by definition. Add 273.15 to every input and the answer shifts by 273.15 and nothing else changes.

Mass of the second body (the one dropped in)In kilograms, same as the first: the hot block, the cold water, the metal out of the boiling tube — whichever body is added rather than already sitting in the vessel.

Specific heat capacity of the second bodyThe same five sourced values as above, for the second body. Choose the last option to type your own.

Specific heat capacity of the second body, typed inLocked unless the last option is chosen above. Wants J kg⁻¹ K⁻¹, not J g⁻¹ K⁻¹.

Starting temperature of the second bodyIn degrees Celsius. Set it equal to the first temperature and the page returns that temperature unchanged — not a fault, and the one case worth trying once. If this page ever returns a temperature OUTSIDE the range between your two starting values, an input is wrong: a mixture cannot end hotter than its hottest part.

Thermal capacity of the calorimeter and stirrerIn joules per kelvin for the whole vessel, not per kilogram. Set it to 0 for the exam question that ignores it, and watch the last two rows. Laboratories often quote it instead as a water equivalent in grams, the mass of water of the same thermal capacity: multiply that mass in grams by 4.186 to get J K⁻¹, so 10 g becomes 41.9.

25.36°CExample

200 g of copper at 100 °C dropped into 250 g of water at 20 °C, in a calorimeter and stirrer of thermal capacity 40 J/K

A weighted mean, not a formula to memorise

T_f = (m₁c₁T₁ + m₂c₂T₂ + CT₁) / (m₁c₁ + m₂c₂ + C) · m₁c₁(T_f−T₁) + m₂c₂(T_f−T₂) + C(T_f−T₁) = 0

m·c: the thermal capacity of a body: the joules needed to raise it one kelvin. It is what the formula weights by, and the thing students skip. Two bodies mix to the average of their starting temperatures weighted by m·c and nothing else — not by mass, not by specific heat, by the product
C: the vessel’s own thermal capacity, in J K⁻¹ for the whole calorimeter and stirrer. It enters at T₁, the temperature of whatever the vessel was already holding, which is why this page asks which body is in it. Set C = 0 and you have the textbook question
Tf: the single temperature both bodies end at. It must lie between T₁ and T₂, because a weighted mean of two numbers cannot fall outside them. If yours does, a unit is wrong, and the usual culprit is a specific heat in J g⁻¹ K⁻¹
ΔT: a temperature DIFFERENCE, and the reason Celsius is allowed here. The degree Celsius is defined equal in magnitude to the kelvin, so a drop of 80 °C is a drop of exactly 80 K. Gas-law pages need absolute temperature because they use T itself in a ratio; this page never does
heat lost = heat gained: the form most textbooks lead with, and the same equation with the signs taken out by hand. The version above keeps them in — every body gets m·c·(T_f−T_start), positive if it warmed — and the terms sum to zero. Harder to get backwards, and it extends to three bodies or ten, where “lost = gained” needs you to know in advance which is which

Worked example

200 g of copper at 100 °C dropped into 250 g of water at 20 °C, in a calorimeter and stirrer of thermal capacity 40 J/K
THE THERMAL CAPACITIES FIRST, because they are what the formula weights by. Water: 0.250 × 4186 = 1046.5 J K⁻¹. Copper: 0.200 × 390 = 78.0 J K⁻¹, on NCERT's 0.39 J g⁻¹ K⁻¹. Vessel: 40, given. Total 1164.5. Note what has already happened: the copper is four fifths the mass of the water and carries one thirteenth of its thermal capacity, so the answer will sit close to 20 °C and nowhere near the middle
THE WEIGHTED SUM. 1046.5 × 20 + 78.0 × 100 + 40 × 20 = 20 930 + 7800 + 800 = 29 530. The vessel is multiplied by 20 and not by 100, because it was sitting in the water at 20 °C before the copper arrived
DIVIDE. T_f = 29 530 / 1164.5 = 25.36 °C. It lies between 20 and 100 as a weighted mean must, 5.36 K above the water and 74.64 K below the copper — a 6.7 to 93.3 split, which is exactly the inverse ratio of the two thermal capacities
CHECK IT WITH THE HEAT ROWS. The copper gives up 78.0 × (25.36 − 100) = −5822 J; the water takes in 1046.5 × 5.36 = +5608 J and the vessel 40 × 5.36 = +214 J. The three sum to zero to the last decimal the page prints. That row is on the page rather than in a footnote because it is the only self-check available, and it catches a swapped temperature, a wrong sign and a mass in the wrong unit all at once
AND WHAT THE VESSEL WAS WORTH. Drop the 40 J K⁻¹ and the answer becomes 25.55 °C — only 0.19 K away, but 3.4 per cent of the 5.5 K rise, which is the quantity the experiment actually measures. Run it the other way round, deducing copper's specific heat from an observed 25.36 °C, and ignoring the vessel makes you credit all the heat to the water alone: you would report 376 J kg⁻¹ K⁻¹ for copper instead of 390, which is 3.7 per cent low and low every time

Why the vessel cannot usually be ignored: one mixture, four calorimeters

The mixture is this page’s own default — 200 g of copper at 100 °C into 250 g of water at 20 °C — with only the vessel changed. Read the last two columns together, because they are the same number twice: a vessel worth 3.43 per cent of the mixture’s thermal capacity takes 3.43 per cent off the temperature RISE, and 0.191 K off a 5.55 K rise is 3.43 per cent of it. That is not a coincidence. For a fixed quantity of heat the rise goes as one over the total thermal capacity, so the fractional error in the rise is simply the vessel’s fractional share — which means the shift is small in degrees and not small as a fraction of the thing being measured. It is also one-directional: ignoring the vessel always makes the predicted rise too large, every repeat, so it does not average out. That is the whole difference between a systematic and a random error.
Vessel	Thermal capacity	Water equivalent	Share of the total	Final temperature	Shift from ignoring it
Ignored, as the exam question says	0 J/K	0 g	0.00%	25.55 °C	—
A light aluminium can	20 J/K	4.78 g	1.75%	25.45 °C	0.097 K
A copper calorimeter and stirrer	40 J/K	9.56 g	3.43%	25.36 °C	0.191 K
A heavy brass vessel, thick stirrer	100 J/K	23.89 g	8.17%	25.10 °C	0.453 K

Heat lost equals heat gained, with the signs left in and the vessel left in too

The formula here is a weighted mean, and saying so out loud is the fastest route through most calorimetry questions. Two bodies brought together end at the average of their starting temperatures, weighted by thermal capacity m·c. Nothing else is in it. That is why 200 g of copper at 100 °C barely warms 250 g of water: copper’s specific heat is about a tenth of water’s, so its thermal capacity is 78 J K⁻¹ against the water’s 1046, and the weighted mean sits within six degrees of where the water started. A student who expects 60 °C has weighted by mass. A student who expects 25 has weighted by m·c.

This page keeps the signs in rather than taking them out. Most textbooks write “heat lost by the hot body = heat gained by the cold body”, which makes you decide in advance which is which and then write two expressions with the signs already arranged. The version here gives every body the same term, m·c·(T_f − T_start), positive if it warmed and negative if it cooled, and sets the sum to zero. Same physics, impossible to get backwards, and it extends without thought to the three-body case — water, vessel and sample — which is what every real experiment is.

Where this goes wrong, almost always, is the unit on the specific heat. Tables print copper as 0.39, as 390 and as 0.093, and all three are right in their own units: J g⁻¹ K⁻¹, J kg⁻¹ K⁻¹ and cal g⁻¹ °C⁻¹. This page wants the middle one. A value a thousand times too small makes the copper’s thermal capacity negligible, and the answer then comes out at almost exactly the water’s starting temperature — plausible, and wrong, which is the dangerous kind. It is why both thermal capacities are printed as rows, where one of them being absurd is visible.

Two assumptions sit underneath the answer and they are not equally safe. The first is that no heat leaks to the room, and nothing here can check that; it is controlled by lagging, by starting as far below room temperature as you finish above it, or by a cooling correction. The second is that neither body changes state, and that one the page does check. If the final temperature lands at or below 0 °C or at or above 100 °C a note appears, because water reaching a phase boundary stops changing temperature and starts changing state: freezing a kilogram releases 333 000 J, as much as cooling it eighty degrees, so a mixture that crosses zero does not continue down the line this page drew. It stops dead and sits there while ice forms. The latent-heat page solves that properly, including the common case where the mixture ends part-way through melting.

Frequently asked questions

Why can I use Celsius here when gas-law questions insist on kelvin?

Because every temperature in this formula appears inside a DIFFERENCE, and a difference is the same number in both scales. The SI Brochure defines the degree Celsius as equal in magnitude to the kelvin, with t/°C = T/K − 273.15 — an offset and no scale factor — so a rise of 5.36 °C is a rise of 5.36 K exactly. A gas law uses the absolute temperature itself, in a ratio such as V₂/V₁ = T₂/T₁, and a ratio of Celsius numbers is meaningless: 100 °C is not five times 20 °C, it is 1.27 times it. The test of whether a formula needs an absolute scale is whether adding 273.15 to every input leaves it unchanged.

Why is the final temperature so close to the water’s starting temperature?

Because thermal capacity, not mass, sets the weighting, and water’s specific heat is far larger than any common solid’s. With 250 g of water at 4186 against 200 g of copper at 390, the water’s m·c is 1046 J K⁻¹ and the copper’s is 78 — thirteen to one — so the 80 K gap splits in the inverse ratio: the water rises about 5 K while the copper falls about 75. For a mixture that ends near the middle you need to match the two m·c products, which for water against copper means about eleven times as much copper by mass.

What is the water equivalent of a calorimeter, and how does it relate to its thermal capacity?

The same quantity in different clothing. The water equivalent is the mass of water with the same thermal capacity, so it is the thermal capacity divided by water’s specific heat: 40 J K⁻¹ ÷ 4.186 = 9.56 g. Its attraction is that you can add those grams to the water already in the vessel and use one specific heat for both; the J K⁻¹ form’s attraction is that it does not pretend the can is made of water. Note that the conversion carries your choice of water constant with it: 9.56 g on 4186 and 9.52 g on 4200, so a water equivalent quoted to three figures is quoting that choice as much as the vessel.

The answer came out above both starting temperatures. What did I do wrong?

A weighted mean of two numbers cannot lie outside them, so an input is at fault and the candidates are few: a negative mass, a negative specific heat, or a thermal capacity typed into a mass field. The commonest is a specific heat with the wrong sign of exponent — 3.9e-1 where 3.9e2 was meant. Check the two thermal-capacity rows: if one reads a fraction of a joule per kelvin for a 200 g block of metal, that is your answer.

Related calculators

References

NCERT / CIET, Physics-02 (Keph_201104): Thermal Properties of Matter, Unit 7 Module 15, “Change of State” (chapter 11, class XI), 2019, on the Central Institute of Educational Technology e-content server (read 11 November 2026). The source of every material constant this batch prints without hedging: ice 2100 J kg⁻¹ K⁻¹, water 4186 — written 4.2 × 10³ in a different worked example in the same module — copper 0.39 J g⁻¹ K⁻¹, latent heat of fusion 3.33 × 10⁵ J kg⁻¹ in the text against 3.35 × 10⁵ and 335 J g⁻¹ in its own examples, and latent heat of vaporisation 22.6 × 10⁵ in the text against 2.256 × 10⁶ in an example. Those are not misprints to tidy away: one official module gives the same quantity two ways, 0.6 per cent apart on fusion and 0.18 per cent apart on vaporisation. That is why this batch takes the constant from the reader instead of choosing one for them.
The International System of Units (SI), 9th edition, BIPM, published May 2019; the version read for this batch is 4.01, revised June 2026 (read 11 November 2026). The source of the kelvin’s CURRENT definition: the kelvin is defined by fixing the Boltzmann constant at exactly k = 1.380 649 × 10⁻²³ kg m² s⁻² K⁻¹, which the brochure states as an exact relation. The degree Celsius is defined equal in magnitude to the kelvin, with T₀ = 273.15 K and t/°C = T/K − 273.15. Most school textbooks still define the kelvin as 1/273.16 of the triple point of water. The brochure records that as the PREVIOUS definition, and says the triple-point temperature must now be determined experimentally — it stood at 273.16 K with a relative standard uncertainty of 3.7 × 10⁻⁷ at the redefinition, and is a measured quantity from then on.
Specific Heat Capacity of Water: Temperature-Dependent Data and Calculator, The Engineering ToolBox (2004) (read 11 November 2026). The source of the 4181.6 J kg⁻¹ K⁻¹ option here and of the curve behind it: at saturation pressure the isobaric specific heat runs 4.2199 kJ kg⁻¹ K⁻¹ at 0.01 °C, 4.1955 at 10, 4.1844 at 20, 4.1816 at 25, a minimum of 4.1796 at 40, then back up through 4.1851 at 60, 4.1969 at 80 and 4.2157 at 100. Water’s specific heat is a shallow U with its floor near 40 °C, and the 0.96 per cent spread across the liquid range is wider than the gap between the two NCERT figures. The page states that it names no primary source of its own, which is why these figures sit beside the NCERT ones rather than replacing them.
COULD NOT BE READ, and nothing was substituted for any of it. The NCERT textbook pages on ncert.nic.in, and four CIET modules in the same series as the ones cited here — Keph_201101 (temperature and its measurement), Keph_201102 (thermal expansion), Keph_201103 (specific heat capacity and calorimetry) and Keph_201105 (conduction and convection) — were all refused by the server’s robots policy on 11 November 2026. That is why this batch quotes no NCERT table number, and why the coefficients of expansion and the thermal conductivities are attributed to the reference actually read rather than to the textbook table you are more likely to be holding. If your table disagrees, use your table: these pages take the constant from you for exactly that reason.