Question

An electric heater of power 3 kW is used for one minute. Find the energy supply by the heater.

Answer 1

This problem can be solved by the direct application of the definition of power.
The mathematical expression for power is –
$P = \dfrac{E}{t}$
where E = energy in joules and t = time in seconds.
The SI unit of power is watt (W).

Complete step by step answer:
The subject of Physics primarily deals with the concept of energy. The main law that explains the importance of energy in Physics is the law of conservation of energy.
The law of conservation of energy states that energy can neither be created nor destroyed but it can be transferred from one form to another.
This gives us a fact that energy transfer is at the core of Physics, be it any concepts in Physics such as,
i) Mechanical energy which is used to move everything in the world can be obtained from electric energy through motor, chemical energy through engines or digestion of food in humans.
ii) Electric energy can be obtained by conversion of energies such as chemical energy in cells and batteries and diesel engines, magnetic energy and mechanical energy in huge devices like generators or very small devices such as piezoelectric crystals.
The rate at which this energy transfer takes place is hence, very important concept. This is represented by the quantity known as Power.
$P = \dfrac{E}{t}$
where E = energy in joules and t = time in seconds.
Power is basically, the ratio of energy transfer to the time taken for the energy transfer.
There are several forms of power in different systems in Physics and their qualitative concepts might vary but dimensionally, they are the same.
In mechanical systems, the power is defined in terms of transfer of kinetic or potential energy per unit time.
In electric systems, the electric energy, that is defined in terms of current and resistance, per unit time is considered for calculation of the power.
In this problem, the electric power is given.
$P = 3kW$
The time for which the heater is used, $t = 1\min = 60\sec$
Substituting, we get the energy supplied by the heater.
$P = \dfrac{E}{t}$
$\implies E = P \times t$
$\implies E = 3kW \times 60\sec = 3000 \times 60 = 180000J = 180kJ$
Thus, the energy supplied by the heater, $E = 180kJ$

Note: The power calculated is the actual power used in the circuit, only when the circuit has direct current source (DC). If A.C or alternating current source is used, the real power is obtained by multiplying a constant number called as power factor.
Power factor = $\cos \phi = \dfrac{R}{Z}$
where $\phi$= phase angle, R = resistance in the circuit and Z = total impedance in the circuit (means the net obstruction offered to AC).