CBSE Class 10 Science-Sources of Energy

This CBSE Class 10 Science chapter 14 page covers Sources of Energy. It explains conventional and non-conventional sources, the characteristics of a good source of energy, fossil fuels, biogas and its composition, wind energy, solar cells, nuclear energy and the forms of energy, followed by a crossword and the chapter question bank with answers and Teacher's Notes.

CHAPTER 14: SOURCES OF ENERGY

1. Our energy requirements increase with our standard of living.
2. Characteristics of a good source of energy: high efficiency, easy to use, easy to store and transport, economical and easily accessible.
3. Sources of energy are divided into conventional sources (traditional sources) and non-conventional sources (not used widely).

[Figure: Flow chart titled Sources of energy. Conventional source of energy (Traditional source) leads to Fossil fuels (like coal, petroleum, natural gas), thermal power plant, hydro energy, bio energy. Non conventional source of energy (Not used widely) leads to Solar energy, energy from sea, geothermal energy, nuclear energy]

FOSSIL FUELS

Fossil fuels are non-renewable, exhaustible and cause air pollution, acid rain and global warming.

BIOGAS

Biogas is formed by the decomposition of cow dung, agricultural and domestic wastes.
Composition of biogas: Methane, Hydrogen sulphide, Hydrogen, Carbon dioxide.
Advantages of biogas: pollution free, wealth from wastes, no residue and high efficiency.

[Figure: Figure 14.1 Schematic diagram of a bio-gas plant. Labels: Slurry, Gas outlet, Manure, Soil, Gas tank, Digester, Outlet, Fertiliser]

WIND ENERGY

Wind energy: the kinetic energy of the wind is used to do useful work.
Advantages: renewable, inexhaustible, pollution free and no residue.
Disadvantages: high cost of installation, minimum wind speed of 15 km/h, high maintenance cost, and requirement of a large area of land.

SOLAR CELLS

Solar cells are devices which convert light energy into electric energy.
Advantages: inexhaustible, pollution free, used in artificial satellites, can be used in remote areas and low cost of maintenance.
Disadvantages: requirement of a special grade of silicon, high cost of installation and moderate efficiency.

[Figure: Figure 14.2 A solar cooker. Labels: Sun rays being reflected, Mirror, Glass sheet]

NUCLEAR ENERGY

In nuclear power plants, nuclear energy is used for the production of electricity.
Advantage: very high efficiency.
Disadvantages: radioactive emissions, fear of nuclear power plant accidents and disposal of radioactive wastes.

DIAGRAMS

[Figure: Concept chart titled Sources of Energy. Non-Renewable Resources: Fossil Fuels (Coal, Petroleum, Natural Gas) and Nuclear Energy. Renewable Resources: Solar Energy, Wind Energy, Hydro Energy and Geothermal energy. Forms of Energy: Potential Energy, Kinetic Energy, Chemical Energy, Heat energy and Sound energy]

CROSSWORD

Find the hidden words related with the chapter sources of energy (you can go across and down).

[Figure: 12 by 16 letter grid crossword as printed in the PDF]

The hidden words that can be found in the grid are: SOLAR, ENERGY, BIOGAS, BIOMASS, GEOTHERMAL, METHANE, GAS, OTEC and CELL.

QUESTION BANK

Very short answer type questions (1 Mark)

1. Expand OTEC. [1 Mark]

Answer: OTEC stands for Ocean Thermal Energy Conversion.

Teacher's Note:
a) OTEC uses the temperature difference between warm surface sea water and cold deep sea water to produce electricity.

2. Thermal power plants are set up near coal or oil fields. Give reason [1 Mark]

Answer: Thermal power plants use huge quantities of coal or oil, so placing them near the fields reduces the cost and effort of transporting the fuel.

Teacher's Note:
a) Fuel is bulky and needed in large amounts, so transport is the main cost.

Short answer type questions (2 Marks)

1. Hydrogen has been used as a rocket fuel. Would you consider it a cleaner fuel than CNG? Why or why not? [2 Marks]

Answer:
1. Yes, hydrogen is a cleaner fuel than CNG.
2. Hydrogen burns in oxygen to form only water, so it produces no carbon dioxide or other pollutants.
3. CNG is a hydrocarbon that burns to form carbon dioxide, which adds to global warming.

Teacher's Note:
a) Hydrogen gives only water on burning, so its exhaust is clean.
b) Hydrogen is costly to store and transport, which is why CNG is still widely used on roads.

2. Fire wood is our conventional fuel. List any four reasons of replacing it with the alternate sources of energy. [2 Marks]

Answer:
1. Burning firewood causes air pollution and smoke, which harms health.
2. Collecting firewood leads to deforestation.
3. Firewood has a low calorific value, so it gives less heat per kilogram.
4. Its supply is limited and it is difficult to store and transport.

Teacher's Note:
a) Deforestation is the most serious long-term reason.
b) Biogas and LPG burn more cleanly than firewood, so they are better household fuels.

3. List two advantages and disadvantages each for using wind as a source of energy. [2 Marks]

Answer:
Advantages:
1. Wind energy is renewable and inexhaustible.
2. It is pollution free and leaves no residue.
Disadvantages:
1. The cost of installation is high.
2. It needs a minimum wind speed of 15 km/h and a large area of land.

Teacher's Note:
a) Wind is free, so the running cost is low once the turbine is installed.
b) Wind farms can only be built where the wind blows steadily.

Short answer type questions (3 Marks)

1. What is solar cell panel? List two advantages and disadvantages each of using solar cells for producing electricity. [3 Marks]

Answer:
1. A solar cell panel is an arrangement of many solar cells that converts light energy directly into electric energy.
2. Advantages: solar energy is inexhaustible and pollution free; it can be used in remote areas and has a low cost of maintenance.
3. Disadvantages: it needs a special grade of silicon, the installation cost is high and the efficiency is moderate.

Teacher's Note:
a) A single solar cell gives a small voltage, so many cells are joined in a panel.
b) Solar panels produce power only in daylight, so storage is needed for night use.
c) Silicon of a special purity is the main reason the cost is high.

VALUE BASED QUESTION

Rohan, a class 10 student was helping his mother in her cooking. His mother stressed the importance of using copper bottom vessels for cooking and switching off of gas cylinder regulator after the use.

a) What value is shown by his mother?

Answer: The value shown by his mother is responsibility towards the safety of the family and care for saving energy and the environment.

Teacher's Note:
a) Switching off the regulator prevents gas leakage and accidents at home.
b) Saving fuel reduces pollution and the load on natural resources.
c) Such habits teach children responsibility.

b) Why is Rohan's mother so particular of using copper bottom vessel for cooking?

Answer:
1. Copper is an excellent conductor of heat.
2. Copper bottom vessels spread heat evenly and cook food faster.
3. Faster cooking uses less gas, which saves fuel.

Teacher's Note:
a) Good conductivity is the key property here.
b) Less fuel means lower cost and lower emissions.
c) Copper vessels must be cleaned well, since they can tarnish.

Long answer type questions (5 Marks)

8. Describe the design and function of each part of a solar cooker with the help of a neat labelled diagram. [5 Marks]

[Figure: Figure 14.2 A solar cooker. Labels: Sun rays being reflected, Mirror, Glass sheet]

Answer:
1. Box: a wooden or metal box with insulated walls, with the inside painted black to absorb heat. The insulation reduces heat loss.
2. Glass sheet: a glass sheet covers the top of the box. It allows sunlight to enter and traps the heat inside, like a greenhouse.
3. Black cooking vessel: food is kept in a black vessel, which absorbs more heat from the sunlight and cooks the food.
4. Mirror (reflector): a mirror is fixed to the lid. It reflects additional sunlight into the box, which increases the temperature inside.

Teacher's Note:
a) Black surfaces absorb heat, while shiny surfaces reflect it.
b) The glass cover works on the greenhouse principle and keeps heat from escaping.
c) Mirror and glass together can raise the temperature to about 100 degrees Celsius or more.
d) A solar cooker needs no fuel, so it is a good example of using a renewable source.

PRACTICAL BASED QUESTIONS

1. Four solutions of i, ii, iii, iv were given to test their acidic or basic nature by using paper. It was observed that the colour of pH paper turned to red, blue, green, and orange respectively when dipped in four solutions. The correct conclusion would be:
(a) i, ii, and iii are acidic
(b) i, and iv are acidic
(c) ii, iii, and iv are acidic
(d) iii is acidic

Answer: (b) i, and iv are acidic

1. Red colour on pH paper shows an acidic solution, so i is acidic.
2. Blue colour shows a basic solution, so ii is basic.
3. Green colour shows a neutral solution, so iii is neither acidic nor basic.
4. Orange colour shows an acidic solution, so iv is acidic.
5. Hence, the correct option is (b).

Teacher's Note:
a) pH paper turns red or orange in acidic solutions and blue in basic solutions.
b) Green is the neutral colour, and a neutral solution is neither acidic nor basic.

2. When a student added zinc granules to dil. HCl, a colourless and odourless gas evolved. On testing with a burning matchstick it was observed that the matchstick:
(a) Continued to burn brilliantly
(b) Burnt slowly with a blue flame
(c) Extinguished
(d) Burnt with a pop sound

Answer: (d) Burnt with a pop sound

1. Zinc reacts with dilute HCl: Zn + 2HCl → ZnCl₂ + H₂.
2. The colourless and odourless gas evolved is hydrogen.
3. Hydrogen burns with a pop sound when a burning matchstick is brought near it.
4. Hence, the correct option is (d).

Teacher's Note:
a) Hydrogen gives a squeaky pop, which is its test.
b) Oxygen makes the matchstick burn brilliantly, so do not confuse the two gases.
c) Zinc is above hydrogen in the reactivity series, so it displaces hydrogen from acids.

3. A small amount of quick lime is taken in a beaker. Water is added slowly to the beaker.
(a) Hissing sound and the solution becomes hot
(b) A hissing sound and the solution becomes cold
(c) Hissing sound and the solution becomes cold
(d) No sound and solution becomes hot

Answer: (a) Hissing sound and the solution becomes hot

1. Quick lime (CaO) reacts with water to form slaked lime: CaO + H₂O → Ca(OH)₂ + heat.
2. The reaction releases a large amount of heat, so the solution becomes hot.
3. The heat turns some water into steam, which escapes with a hissing sound.
4. Hence, the correct option is (a).

Teacher's Note:
a) This reaction is exothermic, so options describing a cold solution are wrong.
b) Quick lime becomes slaked lime after reacting with water.
c) Add water slowly and carefully, because the reaction is vigorous.

4. Reddish brown deposit observed on iron nails, when these are kept in aqueous solution of Copper sulphate is that of
(a) Copper oxide
(b) Copper
(c) Copper sulphide
(d) Iron sulphate

Answer: (b) Copper

1. Iron is more reactive than copper, so it displaces copper from copper sulphate.
2. Fe + CuSO₄ → FeSO₄ + Cu.
3. The copper deposited on the nail is reddish brown.
4. Hence, the correct option is (b).

Teacher's Note:
a) A more reactive metal displaces a less reactive metal from its salt solution.
b) The solution turns light green because iron sulphate is formed.

5. When a solution of barium chloride in water is added to an aqueous solution of sodium sulphate, the following happens:
(a) A white precipitate is formed.
(b) A red precipitate is formed
(c) The colour of the solution turns blue
(d) A pungent smelling gas is evolved.

Answer: (a) A white precipitate is formed.

1. BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.
2. Barium sulphate is insoluble in water.
3. It separates out as a white precipitate.
4. Hence, the correct option is (a).

Teacher's Note:
a) This is a double displacement reaction.
b) No gas is evolved and no colour change occurs in this reaction.

6. A strip of copper was placed in a beaker containing zinc sulphate solution. On observing the strip the next day, it was noticed that
(a) The copper strip remained as it was
(b) The copper strip became thinner
(c) The copper strip became thicker
(d) The colour of copper strip changed

Answer: (a) The copper strip remained as it was

1. Zinc is more reactive than copper.
2. Copper cannot displace zinc from zinc sulphate, so no reaction takes place.
3. The copper strip therefore remains unchanged.
4. Hence, the correct option is (a).

Teacher's Note:
a) A less reactive metal cannot displace a more reactive metal from its salt.
b) Check the reactivity series before predicting a displacement reaction.

7. A student puts one big iron nail in four test tubes containing solutions of zinc sulphate, aluminium sulphate, copper sulphate and iron sulphate. A reddish brown coating was observed only on the surface of iron nail which was put in the solution of
(a) Zinc sulphate
(b) Iron sulphate
(c) Copper sulphate
(d) Aluminium sulphate

Answer: (c) Copper sulphate

1. Iron displaces copper from copper sulphate: Fe + CuSO₄ → FeSO₄ + Cu.
2. The copper deposited on the nail forms a reddish brown coating.
3. Iron cannot displace zinc or aluminium, which are more reactive than iron.
4. Iron sulphate contains iron itself, so no displacement takes place.
5. Hence, the correct option is (c).

Teacher's Note:
a) The coating is copper, not rust.
b) Aluminium and zinc are more reactive than iron, so no coating forms in those solutions.
c) Use the reactivity series to predict the result.

8. The current flowing through a resistor connected in an electric circuit and the potential difference developed across its ends are shown in the given diagrams. The value of resistance is
a. 25
b. 20
c. 15
d. 10

[Figure: Two meter dials. Left dial labelled mA with scale 0, 100, 200, 300, pointer just below 200 (about 180 mA). Right dial labelled V with scale 0 to 3, pointer between 1 and 2, close to 2 (about 1.8 V)]

Answer: d. 10

1. Current from the mA meter is about 180 mA, which is 0.18 A.
2. Potential difference from the voltmeter is about 1.8 V.
3. R = V / I = 1.8 / 0.18 = 10 Ω.
4. Hence, the correct option is d.

Teacher's Note:
a) Convert mA to A by dividing by 1000 before using Ohm's law.
b) Read the pointer against the scale carefully, since the two scales are different.

9. Function of the rheostat in the circuit arrangement to find V-I relationship for a conductor is
(a) To prevent the battery from short circuiting
(b) To adjust the value of the current passing in the circuit
(c) To adjust potential difference in the circuit
(d) To adjust the value of the resistance of the given conductor

Answer: (b) To adjust the value of the current passing in the circuit

1. A rheostat is a variable resistance connected in series with the conductor.
2. Changing the rheostat changes the current through the conductor.
3. The readings of current and potential difference at different settings give the V-I values.
4. Hence, the correct option is (b).

Teacher's Note:
a) The rheostat does not change the resistance of the conductor itself.
b) Its job is to vary the current, so that several V and I readings can be taken.

10. The component x, y, and z shown in the circuit respectively represents
(a) Rheostat, resistor, ammeter
(b) Ammeter, voltmeter, rheostat
(c) Voltmeter, ammeter, rheostat
(d) Rheostat, ammeter, voltmeter

[Figure: Circuit with a battery, a key K, and two resistors R1 and R2. X is a circle placed in series in the main line; Y is a circle with V placed across R1 and R2; Z is a variable resistor with a sliding arrow]

Answer: (b) Ammeter, voltmeter, rheostat

1. X is in series in the main circuit, so it is an ammeter.
2. Y is connected across R1 and R2, so it is a voltmeter.
3. Z is a variable resistor with a sliding arrow, so it is a rheostat.
4. Hence, the correct option is (b).

Teacher's Note:
a) An ammeter is always connected in series with the component.
b) A voltmeter is connected in parallel across the component.
c) The arrow through a resistor marks a rheostat.

10. To determine equivalent resistance of two resistors when connected in series, a student arranged the circuit as shown below. But he did not get the required result because
(a) Position of ammeter is not correct
(b) Position of voltmeter is not correct
(c) Terminals of voltmeter are wrongly connected
(d) Terminals of ammeter are wrongly connected

[Figure: Circuit with a battery, key K, an ammeter A at the top right, a rheostat and resistors R1 and R2 in series in the lower branch, and a voltmeter V connected across R1 and R2 with its + terminal on the left junction and its - terminal on the right junction]

Answer: (c) Terminals of voltmeter are wrongly connected

1. The ammeter is in series in the main line, so its position is correct.
2. Current flows from the battery positive, through the ammeter, into the right end of R2, so the right junction is at a higher potential.
3. The voltmeter's positive terminal is joined to the left junction, which is at a lower potential.
4. The voltmeter is therefore connected with reversed polarity, and its needle does not read correctly.
5. Hence, the correct option is (c).

Teacher's Note:
a) Connect the positive terminal of a voltmeter to the point nearer the positive of the battery.
b) A reversed voltmeter gives a negative or wrong reading.
c) Check the ammeter and voltmeter polarity separately before blaming the position.

11. In the experiment for finding equivalent resistance of two resistors connected in parallel, three students connected the voltmeter in three ways X, Y, Z as shown below. The voltmeter has been correctly connected in cases
(a) X and Y only
(b) Y and Z only
(c) X and Z only
(d) All the three

[Figure: Three circuits labelled X, Y and Z, each with a battery, an ammeter A in series, resistors R1 and R2, and a voltmeter V. In X, V is across the parallel pair R1 and R2 below them. In Y, V is across the parallel pair R1 and R2 at the top of the arrangement. In Z, V is connected between the two ends of R1 and R2 in the middle branch]

Answer: (d) All the three

1. In each case the voltmeter is connected across the two resistors, which are in parallel.
2. In each case the positive terminal of the voltmeter is joined to the side connected to the positive of the battery through the ammeter.
3. Hence, all three cases are correct, and the correct option is (d).

Teacher's Note:
a) A voltmeter must be in parallel with the resistors, not in series.
b) Check the + and - marks on every voltmeter before taking a reading.
c) Parallel resistors share the same potential difference, so one voltmeter reading serves for both.

13. To determine the equivalent resistance of three resistors when connected in a parallel arrangement. Four students connected the resistors as follows. The correct setup is that of student
(a) A
(b) B
(c) C
(d) D

[Figure: Four circuits. (A) R1, R2 and R3 in series in a single line. (B) R1, R2 and R3 as three branches joined between the same two vertical wires. (C) R1 in series with R2 and R3 connected in parallel. (D) R1 and R2 in series, with this pair in parallel with R3]

Answer: (b) B

1. In a parallel arrangement all three resistors have both ends joined to the same two points.
2. In (A) all three resistors are in series.
3. In (C) and (D) one resistor is in series with the parallel part.
4. In (B) R1, R2 and R3 are all connected across the same two points.
5. Hence, the correct setup is (b).

Teacher's Note:
a) Parallel means every resistor has its two ends on the same two junctions.
b) Series arrangements add resistances, while parallel arrangements reduce the total.

14. In the following diagram of the stomatal apparatus, which parts are incorrectly labelled?
(a) I and IV
(b) II and III
(c) IV and V
(d) I and V

[Figure: Stomatal apparatus with labels (I) Guard cell, (II) Chloroplast, (III) Cell wall of stoma, (IV) Stomata, (V) Epidermal cell]

Answer: (a) I and IV

1. The stomata is the pore between two guard cells, and its label must point to the opening.
2. Labels I and IV do not point to the structures they name in the diagram.
3. Hence, the incorrectly labelled parts are I and IV, and the correct option is (a).

Teacher's Note:
a) Two guard cells surround each stoma and contain chloroplasts.
b) The epidermal cell is the large cell around the guard cells, which is correctly labelled.
c) Stomata are used for gas exchange and transpiration.

15. The steps involved in making a slide of epidermal peel of leaf are given as follows:
i. Pull out a thin peel from the lower surface of the leaf
ii. Place a drop of glycerine on the slide
iii. Place the peel in saffron
iv. Place the stained peel on the glycerine
v. Remove the extra stain by washing with water
vi. Place the cover slip over the peel
(a) i, ii, iii, iv, v, vi
(b) i, iii, v, ii, iv, vi
(c) i, iii, iv, ii, v, vi
(d) i, ii, iv, iii, v, vi

Answer: (b) i, iii, v, ii, iv, vi

1. Pull out a thin peel from the lower surface of the leaf (i).
2. Place the peel in saffron to stain it (iii).
3. Wash off the extra stain with water (v).
4. Place a drop of glycerine on the slide (ii).
5. Place the stained peel on the glycerine (iv).
6. Place the cover slip over the peel (vi).
7. Hence, the correct option is (b).

Teacher's Note:
a) Staining must come before washing and mounting.
b) The glycerine goes on the slide before the peel is placed.
c) The cover slip is always the last step.

16. The teacher instructed a student to place a healthy potted plant in a dark room for 24 hours prior to an experiment on photosynthesis. The purpose of placing it in a dark room is
(a) To increase the intake of carbon dioxide
(b) To activate the chloroplast in the leaves
(c) To de-starch the leaves
(d) To denature the enzymes in the leaves

Answer: (c) To de-starch the leaves

1. Leaves store the food made in photosynthesis as starch.
2. In the dark, photosynthesis stops and the stored starch is used up by the plant.
3. After 24 hours the leaves are de-starched, so any starch found later is made during the experiment.
4. Hence, the correct option is (c).

Teacher's Note:
a) A plant with starch already stored would give a misleading result.
b) Darkness does not activate or destroy chloroplasts; it only stops photosynthesis.

17. In an experiment to show that sunlight is necessary for photosynthesis, the leaf is boiled in alcohol for a few minutes using water bath. It is essential because
(a) Alcohol is highly volatile
(b) Steam from water bath heats the leaf
(c) Steam from water bath dissolves the chlorophyll
(d) Alcohol is inflammable

Answer: (d) Alcohol is inflammable

1. Alcohol dissolves the chlorophyll, so the leaf becomes colourless.
2. Alcohol catches fire easily, so it must not be heated over a direct flame.
3. A water bath heats the alcohol safely without a naked flame.
4. Hence, the correct option is (d).

Teacher's Note:
a) It is the alcohol, not the steam, that removes chlorophyll.
b) The water bath is used for safety, because the alcohol is inflammable.

18. Potassium hydroxide is used during the respiration experiment because
(a) It releases oxygen
(b) It absorbs oxygen
(c) It absorbs carbon dioxide
(d) It releases carbon dioxide

Answer: (c) It absorbs carbon dioxide

1. Respiration releases carbon dioxide.
2. Potassium hydroxide absorbs carbon dioxide: 2KOH + CO₂ → K₂CO₃ + H₂O.
3. Hence, the correct option is (c).

Teacher's Note:
a) Potassium hydroxide removes carbon dioxide, so only the oxygen used in respiration affects the result.
b) It does not absorb oxygen, which is the gas used up in respiration.

20. A student sets up the apparatus for the experiment to show that carbon dioxide is released during respiration. After 2 hours, he observes that
(a) KOH turns milky
(b) Water level rises in the bent tube in the beaker
(c) Water level decreases in the bent tube in the beaker
(d) Water turns turbid in the beaker

Answer: (d) Water turns turbid in the beaker

1. Carbon dioxide released during respiration passes into the lime water in the beaker.
2. Carbon dioxide reacts with lime water to form insoluble calcium carbonate.
3. The lime water therefore turns turbid (milky).
4. Hence, the correct option is (d).

Teacher's Note:
a) Potassium hydroxide is not used to test for carbon dioxide in this experiment; lime water is.
b) A turbid lime water is the test for carbon dioxide.

6. Solution of CuSO₄ and FeSO₄ are placed in I and II test tubes respectively. Few pieces of Al are added to each solution. What will be the observations?

Answer:
1. Aluminium is more reactive than both iron and copper, so it displaces both metals from their salts.
2. In test tube I, 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu. The blue colour of the solution fades and a reddish brown deposit of copper forms on the aluminium.
3. In test tube II, 2Al + 3FeSO₄ → Al₂(SO₄)₃ + 3Fe. The green colour of the solution fades and a dark grey deposit of iron forms on the aluminium.

Teacher's Note:
a) The aluminium sulphate formed is colourless.
b) Aluminium is placed above iron and copper in the reactivity series.
c) The colour change in each tube shows that a displacement reaction has taken place.

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