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The first law of thermodynamics says heat and work are both ways of moving energy. Today it sits in one line of your notes: ΔU = Q − W. In the 1840s, most scientists still thought of heat as a substance that flowed. James Prescott Joule (1818–1889), working near Manchester, spent years measuring how much work it takes to make a given amount of heat. This post explains his experiment, converts his number, and shows what it means for a JEE or NEET problem.
What question was Joule trying to answer?
If heat is a kind of motion, then using up work should always produce heat in a fixed ratio, whatever the method. Joule tested this in many ways. In his paper On the Mechanical Equivalent of Heat, communicated to the Royal Society by Michael Faraday, he reviews his own earlier results. In 1843 he reported that heat is produced when water is forced through narrow tubes. In 1845 and 1847 he used a paddle wheel to churn water, sperm oil and mercury, and got equivalents of 781.5, 782.1 and 787.6 foot-pounds. These results, from different liquids and methods, were already close together.
He also credits Count Rumford's cannon-boring experiments, from which he works out a much higher equivalent of 1034 foot-pounds, and explains the excess: Rumford made no estimate of the heat that went into the wooden box and other parts of the apparatus.
How did the paddle-wheel experiment work?
The scanned text of the 1849 paper describes it. A brass paddle wheel with eight sets of arms turned between four sets of fixed vanes inside a vessel of water. The vanes stopped the water simply spinning with the wheel, so the work went into friction within the water.
The wheel was driven by lead weights on strings. They fell about 63 inches to the laboratory floor. Joule then wound them up and let them fall again, twenty times in a row, measuring the water's temperature before and after. He also measured the temperature of the room at the start, middle and end, to correct for heat leaking in or out.
The temperature rise was tiny, so the thermometers and corrections were the real work. Joule even subtracted the friction of the pulleys and the small amount of energy the weights still carried when they reached the ground.
What number did Joule report?
His value from water was 772.692 foot-pounds, corrected to a vacuum. He judged it the most correct of his results, because of the number of experiments and the large heat capacity of the apparatus, and thought it probably slightly too high, since he could not entirely avoid vibration and a slight sound. He closed the paper with two conclusions. The second reads:
"the quantity of heat capable of increasing the temperature of a pound of water (weighed in vacuo, and taken at between 55° and 60°) by 1° Fahr., requires for its evolution the expenditure of a mechanical force represented by the fall of 772 lbs. through the space of one foot."
The paper is signed from Oak Field, near Manchester, on 4 June 1849.
Worked example: converting 772 foot-pounds to joules per calorie
Use the exact conversion factors from NIST's guide to SI units: 1 foot pound-force = 1.355818 J and 1 pound = 0.4535924 kg. A temperature change of 1 °F equals 5/9 K.
Work: 772 × 1.355818 ≈ 1046.7 J.
Heat in calories: warming 453.59 g of water by 5/9 K needs 453.59 × 5/9 ≈ 252.0 cal, since one calorie warms 1 g of water by 1 K.
Ratio: 1046.7 ÷ 252.0 ≈ 4.15 J per calorie.
Compare: NIST gives the thermochemical calorie as exactly 4.184 J. Joule's value is about 0.7% low. Going the other way, 4.184 J/cal corresponds to about 778 foot-pounds.
Worked example: how warm does the water get?
Question. Two 10 kg weights each fall 1.6 m, twenty times, and all the work goes into 6 kg of water. Take g = 9.8 m/s² and the specific heat of water as 4186 J/(kg K). Find the temperature rise.
Work in one fall of both weights = 2 × 10 × 9.8 × 1.6 = 313.6 J.
Twenty falls = 20 × 313.6 = 6272 J.
ΔT = W ÷ (m c) = 6272 ÷ (6 × 4186) ≈ 0.25 K.
A quarter of a degree after all that winding. That is why Joule's precision mattered. The most common mistake in this kind of problem is mixing calories and joules: convert everything to joules first. Our post on calorimetry and heat transfer problems has more practice.
Why does this matter for your syllabus?
The JEE (Main) 2026 syllabus lists heat, work and internal energy and the first law of thermodynamics in its thermodynamics unit. Joule's result is the reason that law can treat Q and W in the same units. The SI unit of energy, the joule, is named after him.
What can a student take from Joule's way of working?
Hunt for the hidden losses. Joule's corrections, for the room, the pulleys and the weights hitting the floor, were as important as the main measurement. In a problem, ask where else the energy could go.
Check one idea several ways. If two methods give the same answer, you can trust it more. Solve a problem by energy and by forces, and compare.
Respect small numbers. A quarter of a kelvin was the whole signal. Do not round too early.
If the link between work, heat and internal energy is still fuzzy, the Fermi Concept Library covers every concept in physics with its own lesson, so you can revise the first law before attempting mixed problems.
Try it now: redo the second example with 4 kg of water and see how the answer changes.
