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Calories to Joules

Calories to Joules

Convert calorimetry heat between calories and joules when the specific-heat table is in cal/g·°C but the answer has to be reported in J or kJ.

Turning a q = m·c·ΔT Answer From Calories Into Joules

Thermochemistry homework has a habit of arriving in one unit and being marked in the other. The specific-heat table at the back of the book is printed in cal/g·°C because that is where calorimetry grew up, but the answer box wants joules, and half the time the question itself asks for kilojoules. Nothing about the physics changes — the mass, the specific heat and the temperature difference are the same three numbers. Only the size of the energy unit moves, so a single multiplication carries the whole result across.

Conversion factor: 1 cal = 4.184 J exactly (the thermochemical calorie). Heating 250 g of water from 20 °C to 80 °C needs q = 250 × 1.000 × 60 = 15 000 cal, and 15 000 × 4.184 = 62 760 J, which you would report as 62.76 kJ.

Three Things Worth Knowing Before the First Substitution

Water Sets the Calorie's Size

The calorie was chosen so that warming one gram of liquid water by one degree Celsius costs exactly one of them. That is why the water row of every table reads 1.000 cal/g·°C and 4.184 J/g·°C — the awkward-looking 4.184 is simply what that same quantity of heat measures in SI.

Specific Heat Changes Unit With the Energy

If you convert q at the end, leave c alone. If you convert c at the start, do not convert q afterwards as well. Doing both is the commonest way a calorimetry answer ends up 4.184 times too large, and a marker spots it immediately.

Phase Changes Sit Outside m·c·ΔT

Melting and boiling happen at constant temperature, so ΔT is zero and the m·c·ΔT term contributes nothing. Those steps use m·L instead: 79.7 cal/g to melt ice (333.5 J/g) and about 540 cal/g to boil water off (2 259 J/g).

Working a Coffee-Cup Problem in One Unit and Reporting in the Other

The tidiest habit is to do the whole arithmetic in calories, where water's specific heat is 1 and most of the multiplication disappears, then convert once at the very end.

1

Finish the calorie arithmetic first

Multiply mass in grams by the tabulated c in cal/g·°C by the temperature change. Keep the sign: a substance that cools has a negative ΔT, and in a heat-lost-equals-heat-gained setup the two q values come out equal and opposite.

2

Type that single total into the calorie field

Enter the magnitude and read the joule figure beside it. Spaces used to group thousands are ignored, and a decimal written with a comma is understood, so a number copied out of a European workbook drops in without editing.

3

Swap when the question hands you joules

Some problems give the heat released in joules and ask you to work back to a specific heat in the older units. Press the swap arrows, or simply start typing on the joule side — each field drives the other, so nothing has to be cleared.

4

Take the digits into your write-up

The copy button above a field puts the plain number on the clipboard with nothing attached. Round it yourself to the significant figures your thermometer and balance actually justify before it goes in the answer box.

Watch the calorimeter itself: a real coffee-cup setup absorbs heat too. When the experiment gives you a calorimeter constant in cal/°C, that term joins the balance before any unit change, and converting it separately afterwards double-counts the same energy.

Specific Heat Capacity in cal/g·°C and J/g·°C

These are the values that appear in most introductory thermochemistry tables, listed side by side so you can see that the two columns are one measurement wearing different units. Every J/g·°C entry is its cal/g·°C neighbour multiplied by 4.184.

Substancec (cal/g·°C)c (J/g·°C)Where it turns up in a problem
Water (liquid)1.0004.184The bath in almost every coffee-cup experiment
Ethanol0.5812.431Alternative solvent in heat-of-solution work
Ice0.5002.092Warming a sample from −20 °C up to 0 °C
Steam0.4481.874The final leg of a full heating-curve question
Air (dry, constant pressure)0.2401.004Ventilation and room-warming estimates
Aluminium0.2150.900The classic hot-metal-into-water determination
Glass0.1200.502Beaker and thermometer corrections
Iron0.1080.452Quenching problems and nail-in-water labs
Copper0.09230.386Calorimeter blocks and conductivity comparisons

Read the first column as an answer to a question: how much heat does one gram need per degree? Water asks for more than nine times what copper asks for, which is why dropping a hot metal block into water barely warms the water while the block plunges tens of degrees — and why the equilibrium temperature in those problems always lands close to the water's starting value.

What the Converter Adds to a Thermochemistry Set

Solve for Either Side of the Equation

Neither box is a fixed output. Whichever unit the question hands you goes in first, and the opposite side re-solves on every keystroke, so a question that changes direction halfway through needs no reset.

A Bare Number for the Lab Report

Copying a field, by the button above it or with Ctrl+C from inside it, yields the digits alone with no unit trailing behind — ready for a results table or a spreadsheet column.

Reach kJ and kcal From the Same Field

Both sides carry a searchable list of all 23 energy units, so the same page handles the kilojoule an exam answer wants and the kilocalorie the enthalpy chapter switches to a fortnight later.

Comma or Dot, Whichever Your Textbook Prints

A value written 0,215 is read exactly like 0.215, and a very small heat — a few millijoules from a thermistor exercise, say — switches to exponent form rather than being flattened to zero.

Calorimetry Questions From the Lab Bench

Why is water's specific heat exactly 1.000 in calories but 4.184 in joules?

Because water is where the calorie came from. Nineteenth-century chemists needed a heat unit and picked the most abundant substance on the bench, defining theirs as whatever it took to warm one gram of it by one degree. The joule was defined from mechanics instead — a newton acting through a metre — with no reference to water at all. When the two systems were tied together the ratio came out at 4.184, and since 1948 the thermochemical calorie has simply been declared equal to 4.184 J. That is why the figure is exact rather than something you would measure and quote with an uncertainty.

Should I convert every term of a coffee-cup problem, or only the final answer?

Only the final answer, and only once. Set the balance up entirely in calories — heat lost by the hot object equals heat gained by the water plus the calorimeter — solve for whatever is asked, then multiply that one result by 4.184. Converting the specific heats first works equally well and lands on the same number, but mixing the two approaches is where marks disappear: a q already in joules that gets multiplied by 4.184 again is out by a factor of more than four, and on paper it still looks like a plausible answer.

There is ice in my calorimeter — where does the 79.7 cal/g belong in the equation?

As its own separate term on the heat-gained side. Ice starting below freezing needs three contributions in sequence: warming the solid up to 0 °C at 0.500 cal/g·°C, melting it at 79.7 cal/g with no temperature change at all, and then warming the meltwater at 1.000 cal/g·°C. Ten grams of ice already sitting at 0 °C absorbs 797 cal — 3 335 J — before the thermometer moves a fraction of a degree. That is exactly why a heating curve has flat shelves in it, and why leaving the fusion term out throws the predicted equilibrium temperature badly off.

My textbook's table and my lecturer's notes disagree slightly — which calorie is meant?

Almost certainly the thermochemical calorie of 4.184 J, which is what this page uses and what chemistry courses have standardised on. Its rival is the International Table calorie of 4.1868 J, a definition that survives in steam tables and older engineering handbooks. The two differ by 0.067 %, so on a three-significant-figure answer they are indistinguishable; the gap only shows up if you are carrying five figures or reconciling a published value to its last digit. Where a source does not say which it means, the thermochemical one is the safe assumption in a chemistry context.

Why does a metal need so little heat per gram compared with water?

Per gram the metals look feeble, but per mole they are remarkably alike. Multiply each specific heat by the molar mass and copper gives about 5.87 cal/mol·K, aluminium 5.80 and iron 6.03 — all clustered near 6 cal/mol·K, which is 25 J/mol·K, the Dulong–Petit result for solid elements. A gram of copper simply holds far fewer and far heavier atoms than a gram of water, so there are fewer particles to share the energy out among. Water then goes further still, because its hydrogen-bond network stores energy in ways a plain metal lattice cannot.

cal
J

Calorimetry Heat Quantities in Joules

1 cal=4.184 J
10 cal=41.84 J
79.7 cal (melt 1 g of ice)=333.5 J
540 cal (boil off 1 g of water)=2 259 J
1 500 cal (25 g water, +60 °C)=6 276 J
15 000 cal (250 g water, +60 °C)=62 760 J

Calorie (cal)

The unit that makes calorimetry arithmetic easy: it is sized so that liquid water takes exactly 1.000 cal per gram per degree Celsius, which is why m·c·ΔT for a water bath collapses to mass times temperature change.

Joule (J)

The SI unit a marked thermochemistry answer is expected in. Since 1948 the calorie has been defined from the joule rather than the reverse, so the 4.184 factor is exact and carries no experimental uncertainty of its own.

Finish the calorimetry sum in calories, then convert the single total once
Press the swap arrows when a problem gives joules and asks for the older units
The copy button hands you digits only, ready to round for the lab report
Search either dropdown for kJ or kcal when the question changes scale
Want to learn more? Read documentation →
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