# Introduction of Work, Power, and Energy with 20+ Solved Examples

Table of Contents

## Introduction:

**Introduction of Work, Power, and Energy with 20+ Solved Examples-** in this article, we will discuss Work, Power, and energy with the help of 20+ solved examples.

## Work

(i). A mechanical work is performed when a force exerts on somebody and the body starts moving as a result of that force.

(ii). If a force “F” operates on a body and that body moves some distance “S” in the direction of the influencing force, we say that a certain work has been carried out.

If a force ‘F” moves a body through a distance ‘S’ in the direction of application, then work is done. Work done = Force x distance moved in direction of force, i.e., W = F x S

### Unit of Work

(i). Under MKS system, unit of work is Joule. If a one Newton force is applied on a body and the body moves 1 meter in the direction of that force, then this quantity of work is called one joule. That’s 1 joule = 1 Newton – meter

A force required to produce an acceleration of one meter per second on a body having a mass of 1 kg, is called a Newton force.

(ii). The gravitational unit of work is meter–kilogram (i.e., m – kg). If a body can be moved up to a distance of one meter through the application of a force equivalent to one kg weight, then the work done as a result of this force is called one meter–kilogram. Normally, a meter – kg equals to 9.8-meter Newton or one joule, that’s

1 m – kg = 9.8 m – N or joule

(iii). The unit of work under the British system or F.P.S, the unit of work is foot, pound. If a body moves a distance of one foot in the direction of the force through the application of one pound weight, then work done tends to be one foot–pound.

1 ft = 0.3048 m and 1 lb. = 4.45 Nw

1 ft – lb. = 0.3048 x 4.45 = 1.356 Joules

Or 1 Joules = 0.7375 ft-lb.

## Power

The rate of doing some kind of work is called power, or power is the rate at which energy is used. That’s

P = w / t

Here, P means power, w means work done and t means time.

It is the rate of doing work or power is the rate at which energy is used or power is the rate of converting energy.

The electrical unit of power is joule per second, which is called watt. In other words, joule is also called watt per second. We can define watt as below;

If one joule energy is used in one second, then the rate of work tends to be one watt, that’s

1 watt = 1 joule / sec.

One watt is the amount of power when one joule of energy is consumed in one second.

The measurement of power can also be done in Kw (kilo watts) instead of watts.

## Energy

The capacity for doing a certain work or the ability to carry out some kind of a work, is called energy. Energy can exist in several forms and it can change from one shape to another. For example, a lead acid cell, converts chemical energy to electrical energy. A generator converts mechanical energy to electrical energy, and an electric radiator converts electrical energy to heat energy. Similarly, a bulb converts electrical energy to light energy, an electric motor converts electrical energy to mechanical energy. It should be remembered that work takes place as a result of change in energy. Under MKS system, measurement of electrical, mechanical and heat energy is done in joules.

When a coulomb charge is moved through the application of a one-volt potential difference (p.d), then one joule energy is consumed from an electrical point of view. And if 4187 joules (or 4200 joules) mechanical or electrical energy is converted to heat energy, then temperature of one kg water will rise by one degree centigrade as a result of this heat.

Moving Q coulombs quantity of electricity through potential difference of v volts, VQ joules work can be done.

W = V x Q

Work done = Volts x Q coulombs

If from a circuit, having a resistance of R ohms, ampere current flows through for a time “t” seconds, in this situation work done or omitted electrical energy equals to VQ joules, whereas V means voltage drops occurring parallel to the circuit resistance, which has a value equivalent to IR. That’s

V = IR

Q = It

Energy expended = VQ … (1)

After entering the value of Q in equation (1)

Expended energy = Vit Joules … (2)

After entering value of V in equation (2)

Energy expended = IR x It = I^{2} Rt Joules … (3)

We know that I = V / R, entering value of I in equation (3)

Energy expended = V^{2} / R^{2} x Rt = V^{2} t / R Joules … (4)

Joule or watt second is a very tiny unit of electrical energy, that’s is the reason measurement of energy for commercial purposes is done in watt hours or kilowatts (kwh). Kilowatt hour is also known as board of trade unit.

I.B.O.T unit = 1 kwh = 1000 wh = 36 x 10^{5} joules

Remember that there is a very close connection between energy and work. Energy refers to the capability to do a certain work, whereas work is a result of the conversion of energy from one form to another. That’s quantity of expended energy tends to be equal to the quantity of work done, that’s why both energy and work done are indicated a single symbol “w” and their SI unit is also the same i.e., joule, which is represented by “J”.

## Power of Electrical Circuit

The following formulae are being used for determining the power of an electrical circuit;

P = W ⨀ / t = V/ t = VI watts … (1)

P = I^{2} R watts (∵ V = IR ohm’s law) … (2)

P = V^{2} / R watts (∵ I = V / R) … (3)

The unit of power in Foot – Pound – second (F.P.S) system is horse – power (H.P). A horse – power can be defined in the following words;

It is a rate of working of 33,000 ft. lb. per minute or 550 ft. lb. per second. In other words, if a weight of 33,000 pounds is lifted up to a height of one foot in one minute or 550 pounds weight is picked up to a one-foot height in one second, then the power expended for doing so, is called a Horse – power, i.e., 1 H.P = 550 ft. lb/ sec = 550 x 1.336 Joules / Sec = 746 watts

(∵ 1 ft. lb = 1.336 joules)

1 kilo watt (kw) = 1000 watts = 1.34 h. p

### Heating Effect of Current

It is a matter of common experience that whenever a current flows through a conductor, the conductor warms up after some time. Its reason is that when current passes through any conductor, electrical energy being provided to the conductor, converts to heat energy, as a result temperature of that particular conductor tends to spike.

We know that the rate of flow of electrons through any substance is known as electric current. When these moveable electrons move through the molecules or atoms of that substance, they collide mutually as well as with other electrons. As a result, of this collision, heat is generated. The generated heat depends on the material of that particular substance. Materials which cause huge resistance in the way of flow of electrons, are known as insulators. A massive heat is produced as a result of passing of electric current through them, e.g., Nichrome, and Tungsten, etc. Whereas, materials which offer very negligible or virtually no resistance in the way of flow of electrons, are known as conductors. A very low heat is produced as a consequence of flow of current through them, e.g., Aluminum, copper, etc.

The element of a heater is generally made from a Nichrome wire. The resistance of this material tends to be extremely high. That’s the reason that when this material made element is connected to a supply, the flow of current starts through it. As a consequence of a very high material resistance, electrons have to come across substantial resistances while passing through the atoms of this material. Therefore, electrons collide with each other as well as with the material atoms. As a result of this continuous collision process, element turns out to be red, and starts providing heat. That’s electrical energy being provided to the Nichrome material, converts into a heat energy as a result of high material resistance. This entire process is construed as a heat effect of the current.

## Joule’s Law

If 1 ampere current is passed through R ohm’s resistance for “t” seconds, then the quantity of work to be done will be as follows;

WD = I^{2} Rt Joules

= Vit Joules … (∵ R = V / I)

= Wt Joules … (∵ W = VI)

= V^{2}t / R Joules … (∵ I = V / R)

This work is transformed into a heat, and heat then tends to discharge through the air. In this situation, quantity of the heat produced (H) will be as follows;

Amount of heat produced, H = Work done / Mechanical equivalent of Heat

= W. Kcal = 4200 joules / Kcal (approx.)

∴ H = I^{2}Rt / 4200 Kcal = V I t / 4200 Kcal

= Wt / 4200 Kcal = V^{2}t / 4200 Kcal

It should be remembered that one kilocalorie is that quantity of heat, which is required to raise the temperature of one kg water by one degree centigrade.

## Thermal efficiency

The ratio between actually consumed heat and total heat produced from an electrical point of view, is known as thermal efficiency, i.e.,

Thermal efficiency = Heat Actually Utilized / Total Heat Produced Electrically

Let us assume that an electric kettle is used for boiling water. Some quantity of total heat produced is lost during the process of warming up of an electric kettle, and some of its heat dissipates as a result of convection and radiation (The process, in which heat transmits by means of heat waves, is called radiation, and the process through which heat transfers due to its own movement of molecules of any particular substance, is called convection). Therefore, quantity of heat left behind of the total heat produced, is used to heat water. Thus, kettle’s thermal efficiency will be as below;

Thermal efficiency = Heat utilized for heating the water / Total Heat Produced Electrically

As such, following relation builds up between electrically produced heat and useful heat or water absorbed heat.

VI t / J x 𝜂 = ms (𝜃_{2} – 𝜃_{1})

Here, m means mass, s means time in seconds, Q_{1} initial temperature, and Q_{2 }last temperature.

## Calculations About Work, Power, Energy, And Thermal Efficiency

**Example 1;** How many kilowatts will be required to light a factory in which 250 lamps each taking 1.3 amperes at 230 volts are used.

**Solution;**

Total watts required = 230 x 1.3 x 250 = 74750 = 74.75 Kw

**Example 2:** A 100 – bolt lamp has a hot resistance of 500 ohms. Find what current it takes, its power rating in watts and the amount of energy it consumes in 40 hours.

**Solution;**

Current taken, I = V / R = 100 / 500 = 0.2 A

Power rating = V I = 100 x 0.2 = 20 W

Energy Consumed in 40 hours = 20 x 40 = 800 watt – hours

**Example 3;** An electric heater is rated at 1.5 Kw. How much heat energy does it give out each hour (3600 s)?

**Solution;**

W = Pt = 1.5 Kw x 3.6 Ks

W = 5.4 MJ

**Example 4;** How much current is drawn by a 120 – V, 100 – W light bulb when it is lit?

**Solution;**

P = V I or I = P / V

I = 100 W / 120 V = 0.8333 A

**Example 5;** An electric iron is marked 200 volts, 350 watts. What current does it take and is its hot resistance? What is the weekly cost of using it for 30 minutes daily at 50 paisa per unit?

**Solution;**

Current taken = W / V = 350 / 200 = 1.75 A

Hot resistance = V / I = 200 / 1.75 = 114.3 ohms

Energy consumed per week = 350 x 7 x 30 / 60 wh

= 350 x 7 x 0.5 / 1000 kwh = 1.225 Kwh

Cost = 1.225 x 50 = 6.1250 paisa

**Example 6:** A factory is supplied with power at 210 volts through a pair of feeders of total resistance 0.025 ohm. The load consists of 250, 60 watt lamps, and four motors each taking 40 amperes. Find (a) total current required 9b0 voltage at station end of feeders (c) power wasted in feeders.

**Solution;**

Lamp load = 250 x 60 = 15, 000 watts

Motor load = 4 x 210 x 40 = 33, 600 watts

Total load = 48, 600 watts

(a). Total current required = W / V = 48, 600 / 210 = 231.4 A

Voltage drops in feeders = I R = 231.4 x 0.025 = 5.78 or 5.8 V

(b). Voltage at station end of feeders = 210 + 5.8 = 215.8 V

©. Power wasted in feeders = I^{2} R = (231.4)^{2} x 0.025 = 1339 watts

**Example 7:** A building is supplied at 200 volts, the load being as follows;

50 in number 100 – watts lamps; 10 in number 2 Kw radiators, 2 motors each taking 50 amperes. Find;

(a) total current (b) cost of energy consumed in a week of 48 hours at 50 paisa per unit (or Kwh)

**Solution;**

(a). Voltage of the supply is 200 V

Current taken by 100-watt lamp = W / V = 100 / 200 = 0.5 A

Current taken by the radiator = 2 x 1000 / 200 = 2000 / 200 = 10 A

Motor current = 50 A

Now total load current is the sum of currents drawn by the lamps + radiators + motors, i.e.,

Total current = 50 x 0.5 + 10 x 10 + 2 x 50 = 25 + 100 + 100 = 225 A

(b). Cost of energy consumed in a week of 48 hrs.

Energy = W x Time in hours = V I T

= 200 x 225 x 48 Wh

Per unit rate of energy = 50 Paisa = Rs 50 / 100

Then, cost of energy = 200 x 225 x 48 x 50 / 1000 x 100 = Rs. 1080 Ans.

**Example 8;** A house consist of 4 lamps of 40 watt, 2 lamps of 60 watts, 2 fans of 150 watts, which works for 3 hours daily. Calculate the total energy bill of the house for one month if per unit rate is 50 paisa. Take 1 month = 30 days.

**Solution;**

40 watts lamps load = 4 x 40 = 160 W

Fans load = 2 x 150 = 300 W

Total load = 160 + 120 + 300 = 580 W

Daily consumption 9if all the load work for 3 hours daily)

3 x 580 = 1740 Wh

Monthly consumption = 30 x 1740 = 52200 Wh

Or 52200 / 1000 = 52.2 units or Kwh

Monthly energy bill = 52.2 x 50 / 100 = Rs. 26 approx.

**Example 9;** Two lamps (A) 100-volt 60 W (B) 100 volts 200 W. are connected in series across 200 V mains. What current will flow through the lamps? What resistance must be placed in parallel with (A), so that each lamp will get its correct current at the proper voltage? Determine the watts absorbed by this resistor?

**Solution;**

Figure 3.1

The resistance of 60W lamp is;

W = V^{2} / R or R = V^{2} / W = 100 x 100 / 60

= 166.7 Ω

The resistance of 200 W lamp = 100 x 100 / 200 = 50 Ω

Both resistances are connected in series so the total resistance will be r_{1 }+ r_{2} = 166.7 + 50 = 216.7 Ω

The current will be, I = V / R = 200 / 216.7 = 0.092 Amp. Ans

In case of the same voltage, the resistance contributed by both lamps should also be same;

V_{AB} = 100 V

Fig. 3.2

Here, the wattage of both should also be same i.e., 200 W. One lamp of 60W is there the other resistance should be of 200 – 60 = 140 watts at a voltage of 100V.

R = V2 / W = 100 x 100 / 140 = 71.4 Ω Ans.

**Example 10:** Three resistors are connected in parallel across 100-volt supply. One resistor absorbs 1600 watts and the second 1200 watts. The total energy used by the whole circuit in 10 minutes is 276 x 10^{4} joules. Find;

(a). Current in each resistor

(b). Resistance of third wire

**Solution;**

The energy consumed is 276 x 10^{4} joules in 10 minutes

Power consumed one second = 276 x 10^{4} / 10 x 60 joules

And joules/ sec is nothing but watts

So, power consumed is = 276 x 10000 / 10 x 60 = 4600 watts

The combination is a parallel combination, so difference power will be added up.

W = W_{1} + W_{2} + W_{3 }(let W_{3} be unknown)

4600 = 1600 + 1200 + W_{3}

Or W_{3} = 4600 – 1600 -1200 = 1800 watts

(a) Now, current taken by each will be as;

i_{1} = 1600 / 100 = 16 Amp.

i_{2} = 1200 / 100 = 12 Amp.

i_{3} = 1800 / 100 = 18 Amp.

(b). The resistance of third, R = E^{2} / W = 100 x 100 / 1600 = 5.55 Ω Ans.

**Example 11;** Find the resistance of a 220-volt, 100-watt lamp and the cost of running it for 20 hours at 20 paisa per unit.

**Solution;**

Voltage of lamp = 220 V

Wattage of lamp = 100 W

Current = W / V = 100 / 220 Amp.

Now, resistance = V / I = 220 x 220 / 100 = 484 ohms. Ans.

It can also be calculated directly R = V^{2} / W = 48400 / 100 = 484 Ω Ans.

Now, running costs at 20 P/unit;

Energy consumed in 20 hours = wt = 100 x 20

= 2000 Wh or 2 units

Cost = cost of per unit x No. of units = 20 x 2 = 40 paisa

**Example 12;** A certain class of lamp takes 0.45 ampere at 220 volts. How many such lamps can be run for an expenditure of power equal to 1 ^{1/3} h. p?

**Solution;**

The power as given = 1 ^{1/3} = 4/3 H.P = 4/3 x 746 watts

The lamp wattage = (0.45 x 220) watts

Now, number of lamps to be connected = total wattage / wattage of each lamp

= 4 x 746 / 3 x 0.45 x 220 = 10 lamp, Ans.

**Example 13;** How long will it take to raise the temperature of 880 grams of water from 16°C to boiling point? The heater takes 2 amperes and 220 volts and its efficiency is 90 per cent.

**Solution;**

Heat required = 0.88 x (100 – 16) = 0.88 x 84 kcal

Heat generated = V/ T / 4187 = 220 x 2 x t / 4187 = 440 x t / 4187 kcal

Useful heat = 90 / 100 x 440 x t / 4187 = 396 x t / 4187 kcal

Useful heat = Heat required

396 x t / 4187 = 0.88 x 84

t = 0.88 x 84 x 4187 / 396 seconds

= 782 seconds = 13.03 minutes

**Example 14;** An electric kettle is required to raise the temperature of ½ gallon of water from 20°C to 100°C in 15 minutes. Calculate the resistance of the heating elements if the kettle is to be used on a 240 volts supply. Assume an efficiency of 80 per cent.

**Solution;**

Weight of ½ gallon of water = 5 lb

= 5 x 0.4536 kg = 2.268 kg

Heat required = 2.268 = 80 kcal

Useful heat generated in kettle = 80 / 100 x V^{2} t / 4187 R kcal

So, useful heat = heat required

∴ 80 x 240^{2} x 15 x 60 / 100 x 4187 x R = 2.268 x 80

R = 80 x 240^{2} x 15 x 60 / 100 x 4187 x 2.268 x 80 = 54.6 ohms

**Example 15;** It is required to heat 22.68 kg. of water from 15°C to boiling point in 15 minutes. If the thermal efficiency of the heater is 80%, find the current taken, the supply voltage being 220V.

**Solution;**

The water is to be heated so the heat given to water to boil is;

H = m x s x t

Where, H = the heat in cal, m = mass in gm, s the specific heat, and t is the temperature.

H = 22.68 x 1000 x 1 x (100 – 15)

= 22680 x 85 cal

Heat is taken from the heater operating at 220 volts, so = V x i x t, where V is the voltage, i is the current and t is the time in seconds.

= 220 x I x 15 x 60 joules

Now heat given = heat taken, and heat given by the heater having 80% efficiency

= [220 x I x 15 x 60/ 4.2 x 80 / 100] cal.

So, equating we have;

22680 x 85 = 220 x I x 15 x 60 x 80/ 4.2 x 100

I = 22680 x 85 x 4.2 x 100 / 220 x 15 x 60 x 89 = 50. 9 Amp. Ans.

**Example 16;** An electric heater rated at 2kw, contains 68.04 kg of water at 20°C. Assuming the heater losses to be 15% of input, find the temperature of the water after the heater has been switched on for 2 ½ hours.

**Solution;**

Heat produced = V i t = W t

= (2000 x 2.5 x 60) / 4.2 cal

Heat required to raise the temperature of water to t°C,

= m x s x t = 68.04 x 1000 x 1 x (t – 20) cal

Let, T = t – 20, so the heat will be 68040 x T cal

As 15% being the heat η loss i.e., η = 85%

Heat produced = heat taken by water

2000 x 2.5 x 3600 / 4.2 x 85 / 100 = 68040 T

Or, T = 2000 x 2.5 x 3600 x 8.5 / 4.2 x 100 x 68040 = 53.7°

Now, the actual temperature of the water is

T = t – 20 53.7 = t – 20, t = 73.7°C Ans.

**Example 17;** Find the time taken to boil 576 gm. of water by means of an electric heater whose efficiency is 80%. The heater is rated ½ kw, and the initial temperature of water is 11°C.

**Solution;**

Heat required = 576 x 1 x (100 – 11) = (576 x 89) cal

Heat produced = [500 x t / 4.2 x 80 / 100] cal

Equating, we have 500 x t x 80 / 4.2 x 100 = 576 x 89

t = 576 x 89 x 4.2 x 100 / 500 x 80 x 60 = 8.8 minutes Ans.

**Example 18;** An electric kettle contains 1.5 kg of water at 15°C. It takes 15 minutes to raise the temperature to 95°C. Assuming the heat losses due to radiation and heating the kettle to be 14 kg, calories, find the current taken if the supply voltage is 100 V.

**Solution;**

Heat required by the water to raise the temp. to 95°C,

= 1500 x 1 x (95 – 15) cal.

Heat lost = 14 x 100- cal.

So, total heat required = 1500 x 80 + 14000 = 134000 cal

Heat produced = 100 x I x 15 x 60 / 4.2 cal.

Heat produced = Heat taken

100 x I x 15 x 60/ 4.2 = 134000

I = 134000 x 4.2 / 100 x 15 x 60 = 6.2 Amp. Ans.

**Example 19;** The heater element of an electric kettle has a constant resistance of 100 Ω. The applied voltage is 240 V and it is found that it takes 27.3 minutes to raise the temperature of 2.268 kg of water from 20°C to boiling point. What is the efficiency of the kettle?

**Solution;**

Heat produced = 240 x 2.4 x 27.3 x 60 / 4.2 cal.

Heat required = 2.268 x 1000 x (100 – 20) cal

And internal efficiency = output / input

= 2268 x 80 x 4.2 / 240 x 2.4 x 27.3 x 60 = 80.6% Ans.

**Example 20:** A heater contains 1.6 kg of water at 20°C. It takes 12 minutes to raise the temperature to 100°C. Assume the losses due to radiation and heating the kettle to be 10 kg calories. Find the power rating of the heater.

**Solution;**

Heat required to raise the temperature of 1.6 kg of water to boiling point = 1.6 x 1000 x 1 x (100 – 20) cal

Heat lost = 10 x 1000 = 10000 cal.

Total heat = (1600 x 80) + 10000 = 138000 cal.

Now, heat produced = Wt = W x 12 x 60 / 4.2 cal.

Heat produced = heat taken by heater

W x 12 x 60 / 4.2 = 138,000

W = 138000 x 4.2 / 12 x 60 = 805 W = 0.8 kw Ans.

**Example 21;** The heater element of an electric kettle has a constant resistance of 100 Ω and the applied voltage is 250 V. Calculate the time taken to raise the temperature of one liter of water from 15°C to 90°C assuming that 85% of the power input to the kettle is usefully employed. If the water equivalent of the kettle is 100g, find how long will it take to raise a second liter of water through the same temperature range immediately after the first?

**Solution;**

Mass of water = 1000 g = 1 kg (∴ 1 cm^{2} weighs 1 gram)

Heat taken by water = 1 x (90 – 15) = 75 Kcal.

Heat taken by the kettle = 0.1 x (90 – 15) = 7.5 Kcal

Total heat taken = 75 + 7.5 = 82.5 kcal

Heat produced electrically H = I^{2} Rt / J kcal

Now, I = 250 / 100 = 2.5 A, J = 4,200 J / kcal, H = 2.5^{2} x 100 x t / 4200 kcal

Heat electrically utilized for heating one liter of water and kettle

= 0.85 x 2.5^{2} x 100 x t/ 4200 kcal

0.85 x 6.25 x 100 x t/ 4200 ∴ t = 10 min 52 second

In the second case, heat would be required only for heating the water because kettle would be already hot;

75 x 0.85 x 6.25 x 100 x t / 4200 ∴ t = 9 min 53 seconds

**Example 22;** Two heaters A and B are in parallel across supply voltage V. Heater A produces 500 kcal in 20 min and B produces 1000 kcal in 10 min. The resistance of A is 10 ohms. What is the resistance of B? If the same heaters are connected in series across the voltage V, how much heat will be produced in kcal in 5 min?

**Solution;**

Heat produced = V^{2}t / JR kcal

For heater A, 500 = V^{2} x (20 x 60)/ 10 x J … (i)

For heater B, 1000 = V^{2 }x (10 x 60)/ R x J … (ii)

From equation (i) and (ii), we get, R = 2.5 Ω

When the two heaters are connected in series, let H be the amount of heat produced in kcal

Since combined resistance is (10 + 2.5) = 12.5 Ω, hence

H = V^{2 }x (5 x 60)/ 12.5 x J … (iii)

Dividing equation (iii) by equation (i), we have H = 100 kcal

**Example 23;** An electric kettle needs six minutes to boil 2 kg of water from the initial temperature of 20°C. The cost of electrical energy required for this operation is 12 paise, the rate being 40 paise per kWh. Find the kW – rating and the overall efficiency of the kettle.

**Solution;**

Input energy to the kettle = 12 paise / 40 paise / kWh – 0.3 kWh

Input power = Energy in kWh / Time in hours = 0.3 / (6 / 60) = 3 kW

Hence, the power rating of electric kettle is 3 kW.

Energy utilized in heating the water = mst = 2 x 1 x (100 – 20) = 160 kcal = 160 / 860 kWh = 0.186 kWh [ 1 kcal = 1 / 860 kwh]

Efficiency = Output / Input = 0.186 / 0.3 = 0.62 = 62%

**Example 24;**

(a). Find the amount of electrical energy expended in raising the temperature of 45 liters of water by 75°C. Assume a heater efficiency of 90%.

(b). For how long an electric heater with a resistance of 50 Ω will have to work on a 200 – V supply for the same input energy as in (a) above?

**Solution;**

(a). Mass of 45 liters of water = 45 kg

Heat absorbed = 45 x 75 = 3.375 kcal

Heat produced = 3.375 / 0.9 = 3750 kcal = 3.750 / 860 = 4.38 kwh

(b). energy input = 3750 kcal = 3750 x 4200 J

Now, energy produced electrically is

= V^{2 }/ R x t joules

= 200^{2 }/ 50 x t = 3750 x 4200

t = 19,687 = 5.47 hours

**Example 25;** An electric immersion heater is placed in a calorimeter containing 192 g of water and is connected to 210 V mains. The temperature rise is 10°C in one minute. Find the resistance of the heater, assuming the water equivalent of the calorimeter and the heater to be 18 g. If the mains voltage drops to 200 V, what is the new rise of temperature per minute?

**Solution;**

Effective amount of water = 192 + 18 = 210 g

Heat produced = 210 x 10 = 2100 cal / min = 2100 / 60 = 35 cal/ s

Now, heat produced electrically = V^{2} / 4.2 R = 210^{2} / 4.2 R cal/ s

210^{2} / 4.2R = 35 ∴ R = 210^{2} / 4.2 x 35 = 300 Ω

When the voltage drops, the rate of production of heat decreases because it is directly proportional to (voltage)^{2}

New rise of temp. = 200^{2} / 210^{2} x 10 = 9.1°C / min

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