Joule's Law

Electric current has three effects:

A. Heating
B. Chemical 
C. Magnetic

A. Heating: W = I²Rt

Heat is produced in current-carrying conductors, resulting in an increase in temperature of the conducting material. The heating is a result of the collisions between the moving free electrons and the relatively stationary atoms of the conductor material. As a result, heating increases rapidly with increase of current flow, since a greater rate of flow results in more collisions.

Use the following apparatus to demonstrate the heating effect of an electric current:

Everyday examples of heating effect of electricity

The heating effect of an electric current has many practical applications, e.g. in radiant electric fires, cookers, hairdryers, kettles, toasters, domestic irons, immersion heaters, etc.

Joule’s Law 
The rate at which heat is produced in a resistor is proportional to the square of the current flowing through it, if the resistance is constant.
P µ I²

Verification of Joule’s Law (As  Δθµ I²)

Apparatus
Lagged beaker or calorimeter with a lid, heating coil, battery or low-voltage power supply, rheostat, ammeter or multimeter, thermometer, stopwatch, balance.

Procedure

1. Put sufficient water in a calorimeter to cover the heating coil.  Set up the circuit as shown.
2. Note the temperature. 
3. Switch on the power and simultaneously start the stopwatch. Allow a current of 0.5 A to flow for five minutes. Make sure the current stays constant throughout; adjust the rheostat if necessary.
4. Note the current, using the ammeter. 
5. Note the time for which the current flowed.
6. Stir and note the highest temperature.  Calculate the change in temperature Δθ.
7. Repeat the above procedure for increasing values of current I, taking care not to exceed the current rating marked on the rheostat or the power supply.  Take at least six readings.
8. Plot a graph of Δθ (Y-axis) against I² (X-axis).

Results

A straight-line graph through the origin verifies that Δθµ I² i.e. Joule’s Law.

Note

The heat energy produced is the mass multiplied by specific heat capacity multiplied by rise in temperature:Q  =  mcΔθ

The energy liberated per second in the device is defined as the electrical power.  This energy is P = RI².
Therefore RI² = mcΔθ /t
I² = (mc/Rt) Δθ . 

As the mass, specific heat capacity, resistance and time are constant, ΔθµI².
Hence P µ I²        i.e. Joule’s law.
 

You could do this experiment using a joulemeter to measure the power supplied, and an ammeter to measure the electric current. Then you could graph P vs I²  directly. 

Sample question

An electric kettle draws a current of 10 A when connected to the 230 V mains supply. Calculate
(a) the power of the kettle
(b) the energy produced in 5 minutes
(c) the rise in temperature
if all the energy produced in 5 minutes is used to heat 2 kg of water.
(Specific heat capacity of water = 4200 J kg^-1 K^-1.)

Solution
(a)     P = IV
= 10 × 230
= 2300 W
= 2.3 kW.

(b) Energy produced in 5 minutes = Pt
= 2300 × 5 × 60
= 690 000 J
= 690 kJ.

(c) Energy produced = energy gained by water
                690 000  = mcΔθ
= (2)(4200)(Δθ)
Δθ =690000 / ( 2 x 4200)

= 82.1 K = Rise in temperature of the water

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