Contents
Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What do you Mean by Mechanical Equivalent of Heat? What Do You Mean by J = 4.2 J/cal for electric field?
Introduction
In the preceding chapters, different types of electrical circuits, their constructions and uses have been discussed. The subject of discussion of this chapter is the measurement of electrical energy consumed in different electrical circuits and power of different effective elements of a circuit. The practical idea about electrical power and energy is very essential for systematic planning and control of the following subjects-
- how much electrical energy is to be generated in a generat-ing station,
- how the energy is to be distributed in different areas,
- how the loss of energy is to be minimised,
- what types of electrical appliances are to be used in resi-dence, office and factory and how they should be used
so that, necessary work will be obtained at minimum cost.
From the discussion in statical electricity, we know that, if Q amount of charge flows through a section under the potential difference V, then the amount of electrical work done is given by,
W = Q × V …… (1)
In this equation, if Q is measured in coulomb (C) and V in volt (V),then the unit of W is joule (J), i.e., the relation of the unit is
J = C × V
Definition: The amount of work done to carry 1 coulomb charge through a potential difference of 1 volt is called 1 joule.
Besides this, we know that, if current I flows through a conductor for time t, then the amount of charge flowing through it is given by,
Q = It …… (2)
So, combining equations (1) and (2) we have,
The electrical work done in the conductor,
W = VIt …… (3)
This work is the amount of electrical energy consumed in the conductor.
Active and Passive Electrical Devices
The electrical work done by an electrical source in a circuit, i.e., the electrical energy sent in a circuit is used up generally in two ways:
i) Electrical energy is converted only to heat energy in the con-necting wires and in some devices of the circuit. In most of the cases, this heat energy has no use, i.e., this energy is dis-sipated. These devices are called passive devices of the cir-cuit and their resistances are called passive resistances. But in our daily life, heat energy is utilised in many ways; e.g., electric heater, electric iron, etc. are very useful appliances. Yet, the resistances of these appliances will also be called passive resistances in conformity with the present discussion.
ii) Electrical energy may be converted to other forms of energy. For example, from electrical energy, mechanical energy is obtained in an electric fan and chemical energy is obtained during the charging of a storage cell. These types of devices are called active devices. However, in these appliances, some amount of heat is evolved. So, it may be said that, the active devices possess some passive resistance also.
Let V be the potential difference between the points A and C in the circuit [Fig.]. If current / flows through the circuit for time t, then electrical work,
W = VIt
In the section AB of the circuit, there is an active electrical device (e.g., an electric fan) and in the section BC there is a pas-sive device whose resistance is R. This R also includes the resis-tance that generates heat in the active device.
Now, VA – VB = V0
and, VB – VC = V’
VAC = VA – VC
= (VA – VB) + (VB – VC) = V0 + V’ = V (say)
∴ V = V0 +V’ or, VIt = V0It+ V’lt
Again, in the section BC, according to Ohm’s law,
V’ = IR
or, V’It = IR ᐧ it = I2Rt
So, VIt = V0It + I2Rt …… (4)
This equation (4) can be expressed in the following form :
Electrical work in the circuit = transformed active energy + exothermic energy
If energy and heat are expressed in the same unit (e.g., in SI, both have the same unit joule or J), the amount of exothermic energy and heat produced (H) will be equal
i. e., in that case, H = I2Rt
But in many cases, calorie (CGS) is used as the unit of heat. In that case, Joule’s law regarding the relation between heat and work, i.e., W = JH is to be used. In that case,
J = mechanical equivalent of heat or Joule’s equivalent of heat
= 4.2 ᐧ J cal-1 = 4.2 × 107 erg ᐧ cal-1
In this condition, if exothermic work or energy be W’, then heat produced, H = \(\frac{W^{\prime}}{J}\)
i.e., H = \(\frac{I^2 R t}{J}\) …… (5)
If work or energy is expressed in joule (J) and heat in calorie (cal), then J = 4.2 J ᐧ cal-1
i.e., \(\frac{1}{J}\) = \(\frac{1}{4.2}\)
So, the practical form of the equation (5) is
H = \(\frac{I^2 R t}{4.2}\) = 0.24 I2Rt ……… (6)
The phenomenon of production of heat in the passive resistance at the expense of electrical work is known as Joule effect or Joule heating. Three laws regarding Joule effect are obtained easily from equation (5) or (6) :
- m ∝ I2; when R and t are constants;
- H ∝ R ; when I and t are constants;
- H ∝ t, when I and R are constants.
Mechanical equivalent of heat: From equation (5), we have,
J = \(\frac{I^2 R t}{H}\)
Now, if I = 1, R = 1 and t = 1 we have, t = 1
From this relation, mechanical equivalent of heat is defined in current electricity.
Definition: The reciprocal of the heat produced in one sec-ond in a conductor of unit resistance, for the passage of unit current through it, is called mechanical equivalent of heat.
In electricity, J = 4.2 J ᐧ cal-1 means:
The amount of heat produced in 1 Ω resistance, for the pas-sage of 1 A current for 1 s through it = \(\frac{1}{4.2}\) cal = 0.24 cal.
Numerical Examples
Example 1.
2A current was sent through a coil of resistance 100Ω for 30 min. Determine the amount of heat produced, the quantity of charge passed and the amount of work done.
Solution:
Here, I = 2A, R = 100Ω, t = 30min = 30 × 60s
∴ The amount of heat produced,
H = \(\frac{I^2 R t}{J}\) = \(\frac{(2)^2 \times 100 \times 30 \times 60}{4.2}\)
= 1.71 × 105 cal
The quantity of charge flowing,
Q = It = 2 × 30 × 60 = 3600 C
The amount of work done,
W = QV = QIR = 3600 × 2 × 100 = 7.2 × 105J
Example 2.
Two separate circuits are made with resistances r1 and r2 connected to the same storage battery. What should be the Internal resistance (r) of the storage battery for which an equal amount of heat is produced in the external circuits?
Solution:
For the first connection, current I1 = \(\frac{E}{r+r_1}\)
[E = emf of the battery]
So, the heat produced in resistance r1 in time t,
H1 = \(\frac{I_1^2 r_1 t}{J}\) = \(\frac{E^2 r_1 t}{\left(r+r_1\right)^2 J}\)
Similarly, the heat produced in the same time in the resistance r2,
H2 = \(\frac{E^2 r_2 t}{\left(r+r_2\right)^2 J}\)
According to the question, H1 = H2
Example 3.
A heating coil of resistance 5 Ω is connected to a cell. The internal resistance of the cell Is 20 Ω. Calculate the value of the shunt to be introduced, so that, the energy consumed in the heating coil will be \(\frac{1}{9}\) of the previous value.
Solution:
Let E be the emf of the cell and S be the shunt. [Fig.]
Here, internal resistance of the cell, r = 20 Ω and external resistance, R = 5Ω.
In absence of the shunt, current flowing in the circuit,
I1 = \(\frac{E}{R+r}\)
∴ In time t, energy consumed in the resistance R,
Example 4.
The rate of energy consumed in 5 Ω resistance [shown in Fig.] is 10J ᐧ s-1. What will be the rate of energy consumed in 4 Ω resistance?
Solution:
In the adjoining figure if, VA – VB = V, then,
So, energy consumed in 1 s in 4Ω resistance
= \(\frac{1}{5}\) × energy consumed in 5Ω resistance
= \(\frac{1}{5}\) × 10 = 2J
i.e., the rate of energy consumed in 4Ω resistance 2 J ᐧ s-1
Example 5.
Water boils in an electric kettle in 10 minutes after being switched on. How will you modify the heating coil to boil water in 5 minutes using the same source of power? [WBCHSE Sample Question]
Solution:
For the source of power of the same emf, V = constant. As equal amount of heat is produced in the two cases, we have
\(\frac{V^2 t_1}{R_1}\) = \(\frac{V^2 t_2}{R_2}\), because heat produced ∝ \(\frac{V^2 t}{R}\).
∴ R2 = \(R_1 \frac{t_2}{t_1}\) = R1 × \(\frac{5 \mathrm{~min}}{10 \mathrm{~min}}\) = \(\frac{1}{2} R_1\)
This means that a heating coil with half the resistance compared to the initial one, is to be used.