PhysicsICSE Class 10 · Work, Energy and Power

Units of Work and Energy: Joule, erg, Calorie, kW h and eV (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Work and energy share the same units. The SI unit is the joule. Other units are the erg (1 erg = 10⁻⁷ J), the calorie (1 cal ≈ 4.2 J), the kilowatt hour (1 kW h = 3.6 × 10⁶ J) for electrical energy, and the electron volt (1 eV = 1.6 × 10⁻¹⁹ J) for atoms and nuclei.

Every unit of work or energy converts to the joule: erg is tiny, calorie about 4.2 J, kW h is huge (3.6×10⁶ J), and eV is minute.

Illustration drawn for QuestionBuddy.

Understand Units of Work and Energy (erg, calorie, kW h, eV)

Energy is work stored. A body has energy if it is able to do work, and its energy is measured by the amount of work it can do. When work is done on a body, its energy rises by exactly that work. So work and energy are measured in the same units, and the syllabus asks for the common units and their relation with the SI unit.

The joule (SI). $1 J$ is the work done when a force of 1 N moves its point of application 1 m in its own direction: $1 J = 1 N m$ .

The erg (cgs). $1 erg = 1 dyne \times 1 cm$ . Since $1 N = 1 0^{5}$ dyne and $1 m = 1 0^{2}$ cm,

$1 J = 1 0^{5} \times 1 0^{2} = 1 0^{7} erg, 1 erg = 1 0^{- 7} J .$

The calorie (heat). Heat is a form of energy. One calorie is the heat needed to raise the temperature of 1 g of water by 1 °C, and

$1 cal \approx 4.2 J, 1 kcal = 1000 cal \approx 4200 J .$

Food energy is usually given in kilocalories.

The kilowatt hour (electricity). $1$ kW h is the energy used by a device of power 1 kW working for 1 hour:

$1 kW h = 1000 W \times 3600 s = 3.6 \times 1 0^{6} J = 3.6 MJ .$

It is the 'unit' on an electricity bill. Note that kW h is energy, while kW is power.

The electron volt (atoms and nuclei). $1$ eV is the energy gained by an electron when it moves through a potential difference of 1 volt:

$1 eV = 1.6 \times 1 0^{- 19} J; 1 MeV = 1 0^{6} eV = 1.6 \times 1 0^{- 13} J .$

Nuclear energies are quoted in MeV because the joule is far too large for a single atom.

Order of size. kW h > kcal > cal > J > erg > eV:
$3.6 \times 1 0^{6} J > 4200 J > 4.2 J > 1 J > 1 0^{- 7} J > 1.6 \times 1 0^{- 19} J$ .

How to convert. To change into joules, multiply by the size of the unit in joules; to change from joules, divide. Keep powers of ten in standard form and check that a 'big' unit gives a small number and a 'small' unit gives a large one.

Units of Work and Energy (erg, calorie, kW h, eV): formulas and key facts

Joule and erg: $1 J = 1 0^{7} erg$; 1 erg = 1 dyne cm = 10⁻⁷ J.
Calorie: $1 cal \approx 4.2 J$; 1 kcal = 1000 cal ≈ 4200 J.
Kilowatt hour: $1 kW h = 3.6 \times 1 0^{6} J$; Energy = power (kW) × time (h).
Electron volt: $1 eV = 1.6 \times 1 0^{- 19} J$; 1 MeV = 1.6 × 10⁻¹³ J.
Order of size: $kW h > cal > J > erg > eV$; Largest to smallest.

How to solve Units of Work and Energy (erg, calorie, kW h, eV) questions

Every conversion is one multiplication or division by the unit's value in joules.

1Write the relation you need: e.g. $1 kW h = 3.6 \times 1 0^{6}$ J.
2Into joules: multiply by the joule value. Out of joules: divide by it.
3For kW h from power and time, first get power in kW and time in hours, then multiply.
4Keep the answer in standard form ( $a \times 1 0^{n}$ ).
5Sense-check: a larger unit gives a smaller number for the same energy.

Units of Work and Energy (erg, calorie, kW h, eV) in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Battery energy in joules

Suppose a phone battery is labelled 15 W h (watt hours).

Express this energy in joules and in kW h.

Try it first, then show the solutionSolution

$15 W h = 15 \times 3600 = 54000$ J.
In kW h: $\frac{15}{1000} = 0.015$ kW h.

Answer: 54 000 J; 0.015 kW h.

Memory hook: Watt hour × 3600 gives joules.

Units on a bill

Imagine a 2 kW geyser is used for 15 minutes every day for 30 days.

Find the energy used in kW h ('units') and in joules.

Try it first, then show the solutionSolution

Time $= 30 \times 0.25 = 7.5$ h; energy $= 2 \times 7.5 = 15$ kW h.
In joules: $15 \times 3.6 \times 1 0^{6} = 5.4 \times 1 0^{7}$ J.

Answer: 15 kW h $= 5.4 \times 1 0^{7}$ J.

Memory hook: kW × hours = units on the bill.

Tricks to remember Units of Work and Energy (erg, calorie, kW h, eV)

Short memory aids. Say each one aloud once, then use it on a question the same day.

Seven zeros to erg
Joule to erg: five zeros for N → dyne, two for m → cm: 10⁷.
Use it when: J ↔ erg.
Four point two
One calorie warms one gram of water one degree, and costs 4.2 joules.
Use it when: cal ↔ J.
Thirty-six lakh
1 kW h = 1000 × 3600 = 36 lakh joules = 3.6 × 10⁶ J.
Use it when: kW h ↔ J.
Big, big, base, small, tiny
kW h, cal, J, erg, eV: from the electricity bill down to a single electron.
Use it when: Ordering units.

Units of Work and Energy (erg, calorie, kW h, eV): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Electron volt in joules · 1 mark · MCQBoard level

One electron volt is equal to:
(a) $1.6 \times 1 0^{- 19}$ J
(b) $1.6 \times 1 0^{19}$ J
(c) $1 0^{- 7}$ J
(d) $3.6 \times 1 0^{6}$ J

Try it first, then show the solutionSolution

The eV is the energy gained by one electron through 1 V: a tiny amount.
$1 eV = 1.6 \times 1 0^{- 19}$ J.

Answer: (a)

Example 2: Two conversions · 2 marks · Short answerBoard level

Express (a) 0.5 kW h in joules, (b) 840 J in calories (1 cal = 4.2 J).

Try it first, then show the solutionSolution

(a) $0.5 \times 3.6 \times 1 0^{6} = 1.8 \times 1 0^{6}$ J. (1)
(b) $\frac{840}{4.2} = 200$ cal. (1)

Answer: (a) $1.8 \times 1 0^{6}$ J (b) 200 cal

Example 3: Ordering the units · 3 marks · Short answerHarder board level

Arrange 1 kW h, 1 cal, 1 J, 1 erg and 1 eV in decreasing order of size, giving the value of each in joules.

Try it first, then show the solutionSolution

Values: 1 kW h $= 3.6 \times 1 0^{6}$ J; 1 cal $\approx 4.2$ J; 1 J; 1 erg $= 1 0^{- 7}$ J; 1 eV $= 1.6 \times 1 0^{- 19}$ J. (2)
Order: kW h > cal > J > erg > eV. (1)

Answer: kW h > cal > J > erg > eV

Example 4: A bulb in three units · 3 marks · Short answerHarder board level

A 100 W bulb is switched on for 5 hours. Find the energy used in (a) kW h, (b) J, (c) cal (1 cal = 4.2 J).

Try it first, then show the solutionSolution

(a) $0.1 kW \times 5 h = 0.5$ kW h. (1)
(b) $0.5 \times 3.6 \times 1 0^{6} = 1.8 \times 1 0^{6}$ J. (1)
(c) $\frac{1.8 \times 1 0 ^{6}}{4.2} \approx 4.29 \times 1 0^{5}$ cal. (1)

Answer: (a) 0.5 kW h (b) $1.8 \times 1 0^{6}$ J (c) About $4.29 \times 1 0^{5}$ cal

Example 5: Food energy as work · 3 marks · Short answerHOTS / competency

A snack bar provides 250 kcal. (a) Express this in joules (1 cal = 4.2 J). (b) If all of it could be used to lift a 50 kg person, how high could she climb? ( $g = 10 m s^{- 2}$ ) (c) Express the energy in kW h.

Try it first, then show the solutionSolution

(a) $250 \times 1000 \times 4.2 = 1.05 \times 1 0^{6}$ J. (1)
(b) $m g h = 1.05 \times 1 0^{6}$ : $h = \frac{1.05 \times 1 0 ^{6}}{50 \times 10} = 2100$ m. (In reality the body uses much of it as heat.) (1)
(c) $\frac{1.05 \times 1 0 ^{6}}{3.6 \times 1 0 ^{6}} \approx 0.29$ kW h. (1)

Answer: (a) $1.05 \times 1 0^{6}$ J (b) 2100 m (c) About 0.29 kW h

Example 6: Case study: boiling water in a kettle · 4 marks · Case-basedHOTS / competency

Suppose an electric kettle heats 1 kg of water from 25 °C to 100 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ and 1 cal = 4.2 J.
(a) Find the heat needed in joules.
(b) Express it in calories and kilocalories.
(c) Express it in kW h.
(d) If the 1.5 kW kettle puts all its energy into the water, how long does it take? At ₹8 per unit, what does it cost?

Try it first, then show the solutionSolution

(a) $Q = m c Δ T = 1 \times 4200 \times 75 = 315000$ J. (1)
(b) $\frac{315 000}{4.2} = 75000$ cal $= 75$ kcal. (1)
(c) $\frac{315 000}{3.6 \times 1 0 ^{6}} = 0.0875$ kW h. (1)
(d) $t = \frac{315 000}{1500} = 210$ s (3.5 min); cost $= 0.0875 \times 8 = 0.70$ , that is ₹0.70. (1)

Answer: (a) 315 000 J (b) 75 000 cal = 75 kcal (c) 0.0875 kW h (d) 210 s; ₹0.70

Where students lose marks in Units of Work and Energy (erg, calorie, kW h, eV)

The slips examiners see most often, and what to write instead.

Mistake: Writing kW instead of kW h for energy
kW is a unit of power; energy needs power × time.
Wrong working
Energy used = 3 kW.
Right working
Energy used = 3 kW × 2 h = 6 kW h.
Do this: Check that energy answers carry 'h' (kW h) or J.
Usually costs: The unit mark.
Mistake: Multiplying when you should divide
From joules into a bigger unit you divide; into a smaller unit you multiply.
Wrong working
$840 J = 840 \times 4.2$ cal.
Right working
$840 J = \frac{840}{4.2} = 200$ cal.
Do this: Bigger unit → smaller number.
Usually costs: The answer mark.
Mistake: Using 3600 J for 1 kW h
The kilo (1000) is often forgotten.
Wrong working
1 kW h = 1 × 3600 = 3600 J.
Right working
1 kW h = 1000 × 3600 = 3.6 × 10⁶ J.
Do this: Convert kW to W first.
Usually costs: The answer mark.
Mistake: Wrong power of ten for erg or eV
Signs of exponents are easily flipped.
Wrong working
1 erg = 10⁷ J; 1 eV = 1.6 × 10¹⁹ J.
Right working
1 erg = 10⁻⁷ J; 1 eV = 1.6 × 10⁻¹⁹ J.
Do this: Both are tiny units, so both exponents are negative.
Usually costs: MCQ marks.

Exam tips

The syllabus lists erg, calorie, kW h and eV by name, with their relation to the joule: learn all four values.
In definitions, 1 kW h is 'the energy consumed by an appliance of power 1 kW in 1 hour'.
Write conversions in standard form; examiners accept $3.6 \times 1 0^{6}$ J, not '3600000'.
Energy is measured by the work a body can do: say this in 'define energy' questions.

Practice questions on Units of Work and Energy (erg, calorie, kW h, eV)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of these is NOT a unit of energy?
(a) erg
(b) calorie
(c) kilowatt
(d) electron volt
Hint
One of them is a unit of power.
Answer
(c) kilowatt
Q2 · 1 mark · Very short answerBoard level
Express $5 \times 1 0^{8}$ erg in joules.
Hint
Divide by $1 0^{7}$ .
Answer
50 J
Q3 · 2 marks · Short answerHarder board level
An electricity bill shows 150 units consumed. Express this energy in joules.
Hint
1 unit = 1 kW h.
Answer
$150 \times 3.6 \times 1 0^{6} = 5.4 \times 1 0^{8}$ J.
Q4 · 2 marks · Short answerHarder board level
A heater supplies $2.1 \times 1 0^{4}$ J of heat. Express this in calories and in kilocalories (1 cal = 4.2 J).
Hint
Divide by 4.2.
Answer
5000 cal = 5 kcal.
Q5 · 1 mark · Very short answerHarder board level
How many electron volts make one joule?
Hint
Divide 1 by $1.6 \times 1 0^{- 19}$ .
Answer
$6.25 \times 1 0^{18}$ eV.
Q6 · 2 marks · Short answerHarder board level
Starting from the definitions of the kilowatt and the hour, show that 1 kW h = 3.6 MJ.
Hint
Energy = power × time in SI units.
Answer
$1 kW h = 1000 J s^{- 1} \times 3600 s = 3.6 \times 1 0^{6} J = 3.6$ MJ.
Q7 · 3 marks · Short answerHOTS / competency
Compare the energy used by a 60 W fan running for 10 hours with the work done in lifting a 1000 kg load through 10 m. How many times larger is the bigger one? ( $g = 10 m s^{- 2}$ )
Hint
Put both in joules.
Answer
Fan: $60 \times 10 \times 3600 = 2.16 \times 1 0^{6}$ J. Lift: $1000 \times 10 \times 10 = 1 0^{5}$ J. The fan uses 21.6 times as much.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.1 kW h in joules?Show answer
$3.6 \times 1 0^{6}$ J.
2.1 J in erg?Show answer
$1 0^{7}$ erg.
3.1 cal in joules?Show answer
About 4.2 J.
4.1 eV in joules?Show answer
$1.6 \times 1 0^{- 19}$ J.
5.Energy in kW h used by a 500 W iron in 2 h?Show answer
1 kW h.

Units of Work and Energy (erg, calorie, kW h, eV): frequently asked questions

Why do work and energy have the same unit?

Energy is measured by the work a body can do, and doing work on a body increases its energy by that amount, so both are measured in joules.

What is the commercial unit of energy?

The kilowatt hour (kW h), called a 'unit' on electricity bills. 1 kW h = 3.6 × 10⁶ J.

Which is the smallest unit of energy in the ICSE syllabus?

The electron volt: 1 eV = 1.6 × 10⁻¹⁹ J, used for the energies of atoms and nuclei.