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Question

Which one of the following is the value of 1 KWh of energy converted into joules?

The correct answer is

3.6 × 10 6J

Understanding Energy Units: Converting KWh to Joules

The question asks us to find the equivalent energy value of 1 kilowatt-hour (KWh) when converted into joules (J). Both kilowatt-hour and joule are units of energy, but they belong to different systems or are commonly used in different contexts (KWh for electrical energy billing, joules in physics and SI units).

Key Concepts: Power, Energy, and Time

To perform this conversion, we need to understand the relationship between power, energy, and time. Power is the rate at which energy is transferred or used. The formula linking these is:

\(\text{Energy} = \text{Power} \times \text{Time}\)

Units Involved

  • The standard international (SI) unit of energy is the joule (J).
  • The standard international (SI) unit of power is the watt (W), where 1 Watt = 1 Joule per second (1 W = 1 J/s).
  • The kilowatt (KW) is a larger unit of power, commonly used for electrical power. 1 kilowatt = 1000 watts (1 KW = 1000 W).
  • The hour (h) is a unit of time. To convert it to the SI unit of time (second), we use: 1 hour = 60 minutes = 60 \(\times\) 60 seconds = 3600 seconds.

Step-by-Step Conversion of 1 KWh to Joules

We want to convert 1 kilowatt-hour (1 KWh) into joules (J). Using the formula Energy = Power \(\times\) Time, and substituting the values in terms of SI units (Watts and seconds):

  1. Identify the power and time components in 1 KWh: Power = 1 Kilowatt, Time = 1 Hour.
  2. Convert the power from kilowatts to watts:

    \(1 \text{ KW} = 1000 \text{ W}\)

  3. Convert the time from hours to seconds:

    \(1 \text{ h} = 3600 \text{ s}\)

  4. Substitute these values into the energy formula:

    \(\text{Energy (in joules)} = \text{Power (in watts)} \times \text{Time (in seconds)}\)

    \(1 \text{ KWh} = (1 \text{ KW}) \times (1 \text{ h})\)

    \(1 \text{ KWh} = (1000 \text{ W}) \times (3600 \text{ s})\)

  5. Perform the multiplication:

    \(1000 \times 3600 = 3,600,000\)

  6. Combine the units. Since 1 Watt = 1 Joule per second (1 W = 1 J/s), the units become W \(\times\) s = (J/s) \(\times\) s = J.

    \(1 \text{ KWh} = 3,600,000 \text{ Ws} = 3,600,000 \text{ J}\)

  7. Express the result in scientific notation:

    \(3,600,000 \text{ J} = 3.6 \times 10^6 \text{ J}\)

Therefore, 1 kilowatt-hour of energy is equal to 3.6 \(\times\) 10\(^6\) joules.

Comparison with Options

Let's compare our calculated value with the given options:

Option Value in Joules
1 \(1.8 \times 10^6 \text{ J}\)
2 \(3.6 \times 10^6 \text{ J}\)
3 \(6.0 \times 10^6 \text{ J}\)
4 \(7.2 \times 10^6 \text{ J}\)

Our calculated value, \(3.6 \times 10^6 \text{ J}\), matches Option 2.

Revision Table: Energy Unit Conversion

Unit Equivalent in SI Units Notes
1 Watt (W) 1 J/s SI unit of power
1 Kilowatt (KW) 1000 W Common unit of power
1 Joule (J) 1 W \(\times\) s SI unit of energy
1 Kilowatt-hour (KWh) 1 KW \(\times\) 1 h Common unit of electrical energy
1 KWh (conversion) \(3.6 \times 10^6\) J \(1000 \text{ W} \times 3600 \text{ s}\)

Additional Information on Energy Units

Energy can be expressed in various units depending on the context. Besides joules and kilowatt-hours, other units include:

  • Erg: A unit of energy in the CGS system. 1 erg = \(10^{-7}\) J.
  • Calorie (cal): Commonly used for heat energy. 1 calorie \(\approx\) 4.184 J. The 'food calorie' (Cal or kcal) is 1000 calories.
  • Electronvolt (eV): Used in particle physics and solid-state physics. 1 eV \(\approx\) \(1.602 \times 10^{-19}\) J.
  • British Thermal Unit (BTU): Used in heating and cooling systems. 1 BTU \(\approx\) 1055 J.
  • Foot-pound (ft-lb): An engineering unit of energy. 1 ft-lb \(\approx\) 1.356 J.

The conversion between different energy units is crucial in physics and engineering to ensure consistent calculations and understanding across different scales and applications.

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Important Questions from Power

  1. Which of the following is correct regarding electric power?

  2. An elevator weighing 400 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used? (g = 9.8 m/s 2and there is no frictional loss)

  3. An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.25 m/s. What would be the minimum horsepower of the motor to be used? G = 9.8 m/s2 and there is no frictional loss.

  4. Units of power is:

  5. One horse power is equal to

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