How to Calculate Joules
Five Methods:Calculating Work in JoulesCalculating Joules from WattsCalculating Kinetic Energy in JoulesCalculating Heat in JoulesCalculating Electrical Energy in Joules
Named for English physicist James Prescott Joule, the joule (J) is one of the cornerstone units of the International metric system. The joule is used as a unit of work, energy, and heat, and is widely used in scientific applications. If you want your answer to be in joules, always make sure to use standard scientific units. The "foot pound" or the "British thermal unit" are still used in some fields, but they have no place in your physics homework.
Formulas
Joules are a unit of energy. Here are formulas for the most common situations where you would calculate energy. As long as you use the SI units listed beneath each formula, your answer will be in joules.

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 → joules = grams * c * ΔT
 T = temperature in terms of ºC or kelvins
 Specific heat capacity c depends on the material being heated. Its units are ^{joules}/_{gramsºC}.

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Steps
Method 1 Calculating Work in Joules
 1Understand what work means in physics. If you push a box across the room, you've done work. If you lift it upward, you've done work. There are two important qualities that have to be there for "work" to happen:^{[1]}
 You're applying constant force.
 The force is causing the object to move in the direction of the force.
 2Define work. Work is easy to calculate. Just multiply the amount of force used, and the amount of distance traveled. Usually, scientists measure force in Newtons, and distance in meters. If you use these units, your answer will be work in units of Joules.
 Whenever you read a word problem about work, stop and think where the force is being applied. If you lift a box, you're pushing upward, and the box is moving up — so the distance is however much it rises. But if you then walk forward holding the box, there's no work happening at all. You're pushing upward still, to keep the box from falling, but the box isn't moving up.^{[2]}
 3Find the mass of the object being moved. You need to know the mass to figure out how much force you need to move it. For our first example, we'll use a person lifting a weight from the floor to her chest, and calculate how much work that person exerts on the weight. Let's say the weight has a mass of 10 kilograms (kg).
 Avoid using pounds or other nonstandard units, or your final answer won't be in terms of joules.
 4Calculate the force. Force = mass x acceleration. In our example, lifting a weight straight up, the acceleration we're fighting is due to gravity, which equals 9.8 meters/second^{2}. Calculate the force required to move our weight upward by multiplying (10 kg) x (9.8 m/s^{2}) = 98 kg m/s^{2} = 98 Newtons (N).
 If the object is being moved horizontally, gravity is irrelevant. The problem may ask you to calculate the force required to overcome friction instead. If the problem tells you how fast the object is accelerating when it is pushed, you can multiply the acceleration given with the mass.
 5Measure the distance being moved. For this example, let's say the weight is being lifted 1.5 meters (m). The distance must be measured in meters, or your final answer will not be written in Joules.
 6Multiply the force by the distance. To lift a 98 Newton weight 1.5 meters upward, you'll need to exert 98 x 1.5 = 147 Joules of work.
 7Calculate work for objects moving at an angle. Our example above was simple: someone exerted a force upward on the object, and the object moved upward. Sometimes, the direction of the force and the movement of the object aren't quite the same, due to multiple forces acting on the object. In the next example, we'll calculate the amount of Joules needed for a kid to drag a sled 25 meters across flat snow by pulling on a rope angled upward at 30º. For this scenario, Work = force x cosine(θ) x distance. The θ symbol is the Greek letter "theta," and describes the angle between the direction of force and the direction of movement.^{[3]}
 8Find the total force applied. For this problem, let's say the kid is pulling on the rope with a force of 10 Newtons.
 If the problem gives you the "rightward force," "upward force," or "force in the direction of motion," it has already calculated the "force x cos(θ)" part of the problem, and you can skip down to multiplying the values together
 9Calculate the relevant force. Only some of the force is pulling the sled forward. Since the rope is at an angle upward, the rest of the force is trying to yank the sled upward, uselessly pulling against gravity. Calculate the force that applies in the direction of motion:
 In our example, the angle θ between the flat snow and the rope is 30º.
 Calculate cos(θ). cos(30º) = (√3)/2 = about 0.866. You can use a calculator to find this value, but make sure your calculator is set to the same unit as your angle measurement (degrees or radians).
 Multiply the total force x cos(θ). In our example, 10N x 0.866 = 8.66 N of force in the direction of motion.
 10Multiply force x distance. Now that we know how much force is actually going toward the direction of motion, we can calculate work as usual. Our problem tells us the sled moved 20 meters forward, so calculate 8.66 N x 20 m = 173.2 joules of work.
Method 2 Calculating Joules from Watts
 1Understand power and energy. Watts are a measure of power, or how fast energy is used (energy over time). Joules is a measure of energy. In order to convert from watts to joules, you need to specify a length of time. The longer a current flows, the more energy it uses.
 2Multiple watts by seconds to get joules. A 1 Watt device consumes 1 Joule of energy every 1 second. If you multiply the number of watts by the number of seconds, you'll end up with joules. To find out how much energy a 60W light bulb consumes in 120 seconds, simply multiply (60 watts) x (120 seconds) = 7200 Joules.^{[4]}
 This formula works for any form of power measured in watts, but electricity is the most common application.
Method 3 Calculating Kinetic Energy in Joules
 1Understand kinetic energy. Kinetic energy is the amount of energy in the form of motion. Like any unit of energy, it can be express in units of Joules.
 Kinetic energy is equivalent to the amount of work done to accelerate a stationary object to a certain speed. Once it has reached that speed, the object retains that amount of kinetic energy until that energy transforms into heat (from friction), gravitational potential energy (from moving against gravity), or other types of energy.
 2Find the mass of the object. For example, we can measure the kinetic energy of a bicycle & bicyclist. Let's say the cyclist has a mass of 50 kg, and the cycle has a mass of 20 kg, for a total mass m of 70 kg. We can now treat them as one 70 kg object, since they'll be traveling together at the same speed.
 3Calculate speed. If you already know the bicyclist's speed or velocity, just write it down and move on. If you need to calculate it yourself, use one of these methods below. Note that we care about the speed, not the velocity (which is speed in a certain direction), even though the abbreviation v is often used. Ignore any turns the bicyclist makes and pretend all distance traveled is one straight line.
 If the bicyclist moved at a constant rate (didn't accelerate), measure the distance the bicyclist traveled in meters, and divide it by the number of seconds it took to move that distance. This will give you the average speed, which in this scenario is the same as the speed at any given moment.
 If the bicyclist is accelerating at constant acceleration and doesn't change direction, calculate his speed at time t with the formula "speed at time t = (acceleration)(t) + initial speed. Use seconds to measure time, meters/second to measure speed, and m/s^{2} to measure acceleration.
 4Enter these numbers into the following formula. Kinetic energy = (1/2)mv^{2}. For instance, if the bicyclist is traveling at 15 m/s, its kinetic energy K = (1/2)(70 kg)(15 m/s)^{2} = (1/2)(70 kg)(15 m/s)(15 m/s) = 7875 kgm^{2}/s^{2} = 7875 newton meters = 7875 joules.
 The kinetic energy formula can be derived from the definition of work, W = FΔs, and the kinematic equation v^{2} = v_{0}^{2} + 2aΔs.^{[5]} Δs refers to "change in position," or the amount of distance traveled.
Method 4 Calculating Heat in Joules
 1Find the mass of the object being heated. Use a balance or spring scale for this. If the object is a liquid, first weigh the empty container the liquid will be held in and find its mass. You'll need to subtract this from the mass of the container and liquid together to find the liquid's mass. For this example, we'll assume the object is 500 grams of water.
 Use grams, not any other unit, or the result will not be in Joules.
 2Find the object's specific heat capacity. This information can be found in a chemistry reference, either in book form or online. For water, the specific heat capacity c is 4.19 joules per gram for each degree Celsius it is heated – or 4.1855, if you need to be very precise.^{[6]}
 Specific heat capacity actually varies slightly based on temperature and pressure. Different organizations and textbooks use different "standard temperatures," so you may see the specific heat capacity of water listed as 4.179 instead.
 You can use Kelvin instead of Celsius, since a difference in temperature is the same in both units (heating something by 3ºC is the same as heating by 3 Kelvin). Do not use Fahrenheit, or your result will not be in Joules.
 3Find the current temperature of the object. If the object is a liquid, you can use a bulb thermometer. For some objects, you may need a probe thermometer.
 4Heat the object and measure the temperature again. This will let use measure the amount of heat being added to the object during the heat.
 If you want to measure the total amount of energy stored as heat, you can pretend the initial temperature was absolute zero: 0 Kelvin or 273.15ºC. This is not typically useful.
 5Subtract the original temperature from the heated temperature. This will produce the degrees of temperature change in the object. Assuming the water was originally at 15 degrees Celsius and heated to 35 degrees Celsius, the temperature change would be 20 degrees Celsius.
 6Multiply the mass of the object by its specific heat capacity and by the amount of temperature change. This formula is written H = mcΔT, where ΔT means "change in temperature." For this example, this would be 500g x 4.19 x 20, or 41,900 joules.
 Heat is more commonly expressed in the metric system in terms of either calories or kilocalories. A calorie is defined as the amount of heat required to raise the temperature of 1 gram of water 1 degree Celsius, while a Kilocalorie (or Calorie) is the amount of heat required to raise the temperature of 1 kilogram of water 1 degree Celsius. In the example above, raising 500 grams of water 20 degrees Celsius would expend 10,000 calories or 10 kilocalories.
Method 5 Calculating Electrical Energy in Joules
 1Use the steps below to calculate energy flow in an electrical circuit. The steps below are written as a practical example, but you can use the method to understand written physics problems as well. First, we'll calculate the power P using the formula P = I^{2} x R, where I is the current in amperes (amps) and R is the resistance in ohms.^{[7]} These units give us the power in watts, so from there, we' can use the formula in the previous step to calculate the energy in joules.
 2Choose a resistor. Resistors are rated in ohms, with the rating either labeled directly or indicated with a series of colored bands. You can also test a resistor's resistance by connecting it to an ohmmeter or multimeter. For this example, we'll assume the resistor is rated at 10 ohms.
 3Connect the resistor to a current source. Either connect wires to the resistor with Fahnestock or alligator clips, or plug the resistor into a testing board.
 4Run a current through the circuit for a set period of time. For this example, we'll use a period of 10 seconds.
 5Measure the strength of the current. Do this with an ammeter or a multimeter. Most household current is in milliamperes, or thousandths of an ampere, so we'll assume the current is 100 milliamperes, or 0.1 ampere.
 6Use the formula P = I^{2} x R. To find the power, multiply the square of the current by the resistance. This yields the power output in watts. Squaring 0.1 gives 0.01, multiplied by 10, gives a power output of 0.1 watt, or 100 milliwatts.
 7Multiply the power by the amount of time elapsed. This gives the energy output in joules. 0.1 watt x 10 seconds equals 1 joule of electrical energy.
 As joules are small units, and because appliances commonly use watts, milliwatts, and kilowatts to indicate how much power they use, utilities commonly measure their energy output in kilowatthours. One watt equals 1 joule per second, or 1 joule equals 1 wattsecond; a kilowatt equals 1 kilojoules per second and a kilojoule equals 1 kilowattsecond. As there are 3,600 seconds in an hour, 1 kilowatthour equals 3,600 kilowattseconds, 3,600 kilojoules, or 3,600,000 joules.
Tips
 Related to the joule is another metric unit of work and energy called the erg; 1 erg equals 1 dyne of force times a distance of 1 cm. One joule equals 10,000,000 ergs.
Warnings
 Although the terms "joule" and "newtonmeter" describe the same unit, in practice "joule" is used when representing any form of energy and for work performed in a straight line, as in the example above of running up a flight of stairs. When used to measure torque, the application of force in rotating an object, the term "newtonmeter" is preferred.
Things You'll Need
Work or Kinetic Energy:
 Stopwatch or timer
 Scale or balance
 Calculator with cosine function (work only, not always needed)
Calculating Electrical Energy:
 Resistor
 Wires or test board
 Multimeter (or separate ohmmeter and ammeter)
 Fahnestock or alligator clips
Heat:
 Object to heat
 Heat source (such as a Bunsen burner)
 Thermometer (either bulb or probe thermometer)
 Chemistry reference book (for finding the specific heat capacity of the object being heated)
Sources and Citations
 ↑ http://educationportal.com/academy/lesson/workdefinitioncharacteristicsandexamples.html
 ↑ http://www.physicsclassroom.com/Class/energy/u5l1a.cfm#waiter
 ↑ http://www.physicsclassroom.com/Class/energy/u5l1a.cfm
 ↑ http://electronicsclub.info/power.htm
 ↑ http://physics.info/energykinetic/
 ↑ http://www.engineeringtoolbox.com/waterthermalpropertiesd_162.html
 ↑ http://electronicsclub.info/power.htm
Article Info
Categories: Classical Mechanics