Example 1
Reference example: work of {{u:300000:energy:0}} at a power of 2,000 W. Two and a half minutes — that is how long this work takes.
t = 150 s
Enter the work W in joules and the power P in watts. The calculator divides work by power with t = W/P and reports the time in seconds, minutes, hours, and clock format, plus the same energy in kilowatt-hours. 300 kJ at 2,000 W takes 150 s, or 2.5 minutes. The model assumes constant power and does not compute W = P·t or device efficiency.
Enter the work and the power. The time in seconds, minutes, and hours and the t(P) chart will appear here.
For work from power and time, use the W = P·t calculator; for power from work and time, P = W/t; and for instantaneous power from force and speed, P = F·v.
Power is work done per unit time, P = W/t. If the work to be done and the constant power at which it is done are known, the time follows from inverting that definition: t = W/P. Units: a joule divided by a watt gives a second, because 1 W = 1 J/s. The formula is standard in textbooks (Halliday, Resnick, Walker; Young, Freedman) and underlies every “how long will it take” calculation: from a kettle, through an elevator, to charging an electric car.
The calculator works with two numbers: the work W in joules and the power P in watts. The result is their quotient. For 300,000 J and 2,000 W you get 150 s, that is 2.5 minutes. Boiling a liter of water from 20 °C (68 °F) needs about 335 kJ of heat; a 2,000 W kettle delivers it in 167 s, under three minutes. For a phone battery holding 15 Wh (54,000 J) an 18 W charger needs 3,000 s, that is 50 minutes.
What the result panel shows. Besides the time in the most convenient unit, the calculator reports the same value in seconds, minutes, hours, and clock format h:mm:ss, the work in kilowatt-hours (the unit on an electricity bill), the entered work and power, and the time at doubled power — half as long, because the power sits in the denominator. The chart draws the hyperbola t(P) from ½P to 4P for the entered work.
Limits of the model. The formula assumes constant power over the whole time and 100 % efficiency. The calculator does not compute power from force and speed (that is the P = F·v calculator), and it ignores heat losses in a kettle, motor efficiency, and the charging curve of a battery, which slows down near the end. In practice the real time is longer by the reciprocal of the efficiency: at 80 % efficiency, by 25 %.
The unit switch in the header changes the work to ft·lb and shows the power additionally in hp; the values in the text stay in SI units as a reference. Convert kilowatt-hours to joules by multiplying by 3,600,000. Work and power must be greater than zero: at zero power the work would never be done, and at zero work there is nothing to compute.
Examples on this page: 300 kJ at 2,000 W → 150 s; a liter of water in a 2 kW kettle → 167 s; an 800 kg elevator rising 30 m at 12 kW → 19.6 s; a 15 Wh phone battery at 18 W → 50 min; a 60 kWh battery at 11 kW → 5.45 h; a 10 W LED bulb using 1 kWh → 100 h. The “Fill this example” button enters the same numbers into the fields and refreshes the result panel and the chart.
t = W / P
[t] = J / W = J / (J/s) = s; 1 kWh = 3,600,000 J, 1 min = 60 s, 1 h = 3,600 s; W = P·t, P = W/t
Model: time from the definition of average power, P = W/t (the definition of power in Newtonian mechanics; the watt in the SI since 1889). The rearrangements W = P·t and P = W/t follow from the same definition and have their own calculators; this one solves only for t. Order of computation: W/P in SI units, then conversions to minutes, hours, and h:mm:ss format, and W/3,600,000 for kWh.
Average power is the quotient of work and the time in which it was done: P = W/t. The three quantities of this definition can be computed from one another: work from power and time (W = P·t), power from work and time (P = W/t), and — as on this page — time from work and power (t = W/P). Work is understood broadly here as energy transferred: the mechanical work of a force, heat delivered to water, electrical energy stored in a battery.
The formula t = W/P says that time is inversely proportional to power. A device twice as powerful does the same work in half the time; one ten times weaker takes ten times longer. This is a hyperbolic relation, which the chart under the result draws. The unit follows from the definition: a joule divided by a watt, that is by a joule per second, gives a second.
The concept of power as “work over time” took shape in the 18th century together with the steam engine. James Watt, comparing his engines with horses at mill wheels, defined horsepower around 1782 as 33,000 pounds lifted one foot per minute — precisely work divided by time. The question “how long will the engine take to pump the water out of the mine” was the first practical application of t = W/P.
The unit of power was named the watt at the 1882 meeting of the British Association for the Advancement of Science, and in 1889 the International Electrical Congress adopted it. With electrification came the kilowatt-hour: the work done by a power of 1 kW in one hour, that is 3.6 MJ. Electrical energy is billed in it, which is why the result panel also shows the entered work in kWh — so that charging or cooking time can be compared with the bill.
Step 1. From the definition of average power, P = W/t. Step 2. Multiply both sides by t and divide by P: t = W/P. Assumption: the power is constant over the whole interval; when it varies, W is the integral ∫P dt and t must be found from the equation ∫₀ᵗ P(τ) dτ = W. Step 3. For W = 300,000 J and P = 2,000 W: t = 300,000 / 2,000 = 150 s = 2 min 30 s. The same work in kilowatt-hours is 300,000 / 3,600,000 = 0.0833 kWh.
Step 4. When the work is not given directly, it must be computed first. Lifting a mass m to a height h: W = m·g·h, for 800 kg and 30 m — 235,440 J. Heating a mass m of water by ΔT: Q = c·m·ΔT, for 1 kg and 80 K — 4,186·80 = 334,880 J. Accelerating a mass m to speed v: W = ½mv², for 1,500 kg and 27.8 m/s — about 580 kJ. Only such a result is then divided by the power; the neighboring calculators (potential energy, heat, kinetic energy) compute these works separately.
The first assumption is constant power. An electric motor at start-up, a kettle just after switching on, or a charger in its final phase do not run at constant power. Charging a lithium-ion battery proceeds at constant current only to about 80 % of capacity, after which the power drops, so the real time to a full charge is longer than W/P computed from the rated power — usually by 20–40 %.
The second assumption is 100 % efficiency: all of the device’s power goes into useful work. A kettle loses a few percent of heat to the surroundings, an elevator motor has an efficiency of 70–90 %, a phone charger 80–90 %, and a combustion engine only 25–40 %. If you know the efficiency η, divide the calculator’s result by η, or enter the product P·η in the power field. For a 2,000 W kettle at 90 % efficiency the time to boil a liter of water grows from 167 s to 186 s.
The joule is the common unit of all kinds of energy, so the same formula works for mechanical work, heat, and electrical energy. Mechanical work: force times distance, W = F·s, or the change of potential energy m·g·h and kinetic energy ½mv². Heat: Q = c·m·ΔT, where for water c = 4,186 J/(kg·K); boiling a liter of water from 20 °C takes about 335 kJ, and evaporating it another 2.26 MJ. Electrical energy: voltage times charge, U·Q, in practice given in Wh or kWh.
Conversion factors used on this page: 1 kWh = 3.6 MJ, 1 Wh = 3,600 J, 1 kcal = 4,184 J, 1 ft·lb = 1.3558 J. A 4,000 mAh phone battery at 3.8 V holds about 15 Wh = 54 kJ; an electric car battery 60 kWh = 216 MJ; a resting human gives off about 100 W, so 100 kcal (418 kJ) is “burned” in a little over an hour, and at 400 W of effort in seventeen minutes.
The same formula spans an enormous range of scales. A camera flash releases about 10 J in 1 ms, that is at 10 kW. A 2 kW kettle boils a liter of water in under 3 minutes. A 12 kW elevator lifts 800 kg to the tenth floor in 20 s. An 11 kW wall charger fills a 60 kWh battery in 5.5 hours, and a 1.4 kW household outlet in 43 hours. A 10 W LED bulb uses 1 kWh only after 100 hours of light, and a 1 GW power plant produces 1 TWh in 1,000 hours, that is six weeks.
For very long times it is easier to think in hours and days, which is why the result panel shows clock format and picks the unit automatically. For very short ones — milliseconds and microseconds — the instantaneous power can be enormous despite a small energy; that is why pulsed lasers with energies of single joules reach powers of terawatts. The calculator computes in seconds and displays full precision, so both ends of the scale are available.
The definition, derivation, and historical note come from general physics textbooks and metrology sources: D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, chapter on work, energy, and power; H. D. Young, R. A. Freedman, University Physics, chapter “Work and Kinetic Energy,” section “Power”; R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics, vol. I, chapters 4 (Conservation of Energy) and 13 (Work); BIPM, The International System of Units (SI), 9th edition (joule, watt, non-SI units: kWh); NIST SP 811 (conversion factors for kcal, ft·lb, hp); the specific heat of water per the NIST Chemistry WebBook; H. W. Dickinson, James Watt: Craftsman and Engineer, 1936. This page is for educational purposes.
t is the time, W the work, P the power. For 300,000 J and 2,000 W the result is 150 s.
Reference example: work of {{u:300000:energy:0}} at a power of 2,000 W. Two and a half minutes — that is how long this work takes.
t = 150 s
A 2,000 W kettle boils a liter of water from 20 °C: the heat needed is 4,186 J/(kg·K) · 1 kg · 80 K = {{u:334880:energy:0}}. Without heat losses, that takes under three minutes.
t ≈ 167 s
A 12 kW elevator lifts an 800 kg cab to the tenth floor, 30 m up: W = m·g·h = {{u:235440:energy:0}}.
t ≈ 19.6 s
An 18 W charger fills a phone battery holding 15 Wh, that is {{u:54000:energy:0}}. In practice the last 20 % charges more slowly.
t = 50 min
An 11 kW wall charger fills a 60 kWh electric car battery = {{u:216000000:energy:0}}. At constant power, that takes about five and a half hours.
t ≈ 5.45 h
A 1,500 kg car accelerates to 62 mph (100 km/h): ½mv² ≈ {{u:580000:energy:0}} at 80 kW at the wheels. The result is the ideal 0–62 mph time, with no losses.
t ≈ 7.25 s
A cyclist of 80 kg (with bike) climbs 500 m of elevation at 250 W: work against gravity {{u:392400:energy:0}}. The climb takes under half an hour.
t ≈ 26 min
An 800 W microwave heats 300 g of soup by 60 K: Q ≈ {{u:72000:energy:0}}. Heating takes a minute and a half.
t = 90 s
A 1.5 kW pump lifts 50 m³ of water by 3 m into a pool: W = 50,000 kg · 9.81 · 3 m = {{u:1471500:energy:0}}. Pumping takes about sixteen minutes.
t ≈ 16.4 min
An industrial flywheel storing 274 MJ is spun up by a 100 kW motor. Charging the energy store takes about three quarters of an hour.
t ≈ 45.7 min
A 10 W LED bulb uses one kilowatt-hour, that is {{u:3600000:energy:0}}. That is one hundred hours of light for the price of one kilowatt-hour.
t = 100 h
A resting human gives off about 100 W. Burning 100 kcal, that is {{u:418400:energy:0}}, takes a little over an hour.
t ≈ 70 min
The calculator finds the time to do a given work at constant power from t = W/P: it divides the work in joules by the power in watts. For 300,000 J and 2,000 W the result is 150 s, that is 2.5 minutes.
Work in joules, power in watts; the result is in seconds, as well as minutes, hours, and h:mm:ss format. Multiply kilowatt-hours by 3,600,000, kilocalories by 4,184, kilowatts by 1,000, horsepower by 745.7. After switching the header to US units the work is in ft·lb and the power is shown additionally in hp.
Because the power sits in the denominator: t = W/P. For the same work the time is inversely proportional to the power — a hyperbola, which the chart draws. The result panel shows the time at doubled power separately.
Compute it first: lifting a mass m to a height h is W = m·g·h (800 kg to 30 m is 235 kJ), heating water is Q = 4,186·m·ΔT, accelerating is ½mv². The neighboring calculators compute these values; only then enter the result in the work field.
No. The calculator assumes all the power goes into useful work. If you know the efficiency η, divide the result by it: a kettle at 90 % efficiency needs 167/0.9 = 186 s instead of 167 s. Equivalently you can enter P·η in the power field.
Because the charging power is not constant: a lithium-ion battery accepts full power only to about 80 % of capacity, after which the charger reduces it. The real time to a full charge is usually 20–40 % longer than W/P computed from the rated power.
The same work in the unit on an electricity bill: 1 kWh is the work of 1 kW over one hour, that is 3,600,000 J. For 300,000 J that is 0.0833 kWh. At 15 cents per kWh, boiling a liter of water costs about 1.4 cents.
Yes. The joule is the unit of every form of energy, so W can be heat delivered to water, energy stored in a battery, or the work of a force. A kettle, a heater, and a microwave all follow the same t = Q/P.
The formula t = W/P requires constant power. With varying power the work is the integral ∫P dt and the time must be found from the equation ∫₀ᵗ P(τ) dτ = W; for power that changes in steps you can compute the segments separately and add the times.
Hours, minutes, and seconds, as on a clock or stopwatch: 150 s is 00:02:30, and 19,636 s is 05:27:16. For times longer than a day the panel adds the number of days.
This calculator solves only for t. Power from work and time, P = W/t, is handled by the power calculator; work from power and time, W = P·t, by the engine work calculator. All three are the same definition of power solved for a different quantity.
Division by zero is undefined, and physically a device with zero power would never do the work — the time would be infinite. The calculator requires W > 0 and P > 0, and until then it displays a message instead of a result.
Use t = W/P. At W = 300,000 J and P = 2,000 W you get 150 s, or 2.5 minutes. The panel also shows 00:02:30.
About 167 s (under 3 minutes). That is the order of magnitude for boiling a litre of water from 20 °C at constant power with no losses.
W = P·t finds work (energy) from known power and time. Here the primary result is time t = W/P. Same power definition, solved for t.
Multiply by 3,600,000. 1 kWh = 3.6 MJ. Then divide by power in watts to get time in seconds.
No. It assumes all power goes into useful work. At efficiency η divide the result by η, or enter P·η in the power field.
The definition, unit, history, and derivation below come from recognized physics and metrology sources.
Page updated November 3, 2026.