Forms of energy and conservation of energy

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 1-1Core

   A stone is at rest high up on a cliff, and its potential energy is $50\,\text{J}$. The stone falls to the ground, where the potential energy is zero. Air resistance is ignored.

   What is the kinetic energy of the stone just before it hits the ground?

   JCheck
   Answer

   $50\,\text{J}$

   Show the solution

   Given

   potential energy of the stone on the cliff $E_\text{p} = 50\,\text{J}$, kinetic energy on the cliff $0\,\text{J}$, potential energy on the ground $0\,\text{J}$

   Find

   kinetic energy of the stone $E_\text{k}$ just before it hits the ground in $\text{J}$

   Principle

   Air resistance is ignored, so no energy enters or leaves the system formed by the stone and the Earth. The system is isolated, so energy is conserved: $E_\text{initial} = E_\text{final}$. Without friction no thermal energy is produced.

   Formula

   At the start there is only potential energy and at the end only kinetic energy: $E_\text{p} = E_\text{k}$, so $E_\text{k} = E_\text{p}$

   Conversions

   The energy is given in joules and the answer is asked in joules, so no conversions are needed.

   Substitution

   $E_\text{k}$$= E_\text{p}$

   $= 50\,\text{J}$

   Answer

   $50\,\text{J}$

   Two significant figures as in the potential energy $50\,\text{J}$.

   Reasonableness

   In the fall the potential energy changes into kinetic energy, and because nothing changes into thermal energy, the amount of energy is the same $50\,\text{J}$ at the start and at the end. The kinetic energy cannot be greater than the initial energy, because energy is not created.

2. Exercise 1-2Core

   A plant leaf absorbs radiant energy of $245\,\text{J}$ from the Sun. Through photosynthesis the leaf changes $6.0\,\text{J}$ of it into chemical energy, and the rest changes into thermal energy.

   How much energy is changed into thermal energy?

   JCheck
   Answer

   $239\,\text{J}$

   Show the solution

   Given

   radiant energy absorbed by the leaf $E_\text{rad} = 245\,\text{J}$, chemical energy produced $E_\text{chem} = 6.0\,\text{J}$

   Find

   energy changed into thermal energy $E_\text{thermal}$ in $\text{J}$

   Principle

   The leaf changes the absorbed radiant energy into chemical energy and thermal energy, and energy is neither created nor destroyed. By the law of conservation of energy $E_\text{initial} = E_\text{final}$.

   Formula

   $E_\text{rad} = E_\text{chem} + E_\text{thermal}$, so $E_\text{thermal} = E_\text{rad} - E_\text{chem}$

   Conversions

   All energies are in joules, so no conversions are needed.

   Substitution

   $E_\text{thermal}$$= 245\,\text{J} - 6.0\,\text{J}$

   $= 239\,\text{J}$

   Answer

   $239\,\text{J}$

   In whole joules, because the radiant energy $245\,\text{J}$ is given to the nearest joule and this limits the precision of the difference.

   Reasonableness

   The chemical energy and thermal energy together are $6.0\,\text{J} + 239\,\text{J} = 245\,\text{J}$, as at the start. Only $6.0\,\text{J} : 245\,\text{J} \approx 0.02$, about two per cent, changes into chemical energy, and plants are known to capture only a small part of the energy of sunlight; the leaf warms up in the sun.

3. Exercise 1-3

   A skater starts from rest at point A on the rim of a half-pipe and slides down a slope with friction, past point B, to point C at the bottom. The bars in the figure show his potential energy and kinetic energy at each point, with the bottom as the zero level of potential energy.

   Read the energies from the figure. How much energy has friction changed into thermal energy when the skater has moved from point A to point C?

   JCheck
   Answer

   $200\,\text{J}$

   Show the solution

   Given

   energies read from the figure at points A and C: $E_\text{p,A} = 700\,\text{J}$, $E_\text{k,A} = 0\,\text{J}$, $E_\text{p,C} = 0\,\text{J}$ and $E_\text{k,C} = 500\,\text{J}$

   Find

   energy changed into thermal energy $E_\text{thermal}$ between points A and C in $\text{J}$

   Principle

   The skater, the half-pipe and the air form an isolated system, so energy is conserved: $E_\text{initial} = E_\text{final}$. Friction changes part of the mechanical energy into thermal energy, so at the end the energy is mechanical energy and thermal energy.

   Reading the graph

   At point A there is only the blue potential-energy bar, and its top is at the gridline $700\,\text{J}$; there is no kinetic energy because the skater is at rest. At point C there is only the orange kinetic-energy bar, and its top is at the line $500\,\text{J}$; there is no potential energy because the bottom is the zero level. The bars of point B are not needed.

   Formula

   $E_\text{p,A} + E_\text{k,A}$$= E_\text{p,C} + E_\text{k,C} + E_\text{thermal}$

   so

   $E_\text{thermal}$$= (E_\text{p,A} + E_\text{k,A}) - (E_\text{p,C} + E_\text{k,C})$

   Conversions

   All energies are in joules, so no conversions are needed.

   Substitution

   $E_\text{thermal}$$= (700\,\text{J} + 0\,\text{J}) - (0\,\text{J} + 500\,\text{J})$

   $= 200\,\text{J}$

   Answer

   $200\,\text{J}$

   To the nearest hundred joules, because the lengths of the bars are read from the $100\,\text{J}$ gridlines.

   Reasonableness

   The mechanical energy decreases from $700\,\text{J}$ to $500\,\text{J}$, as it must on a track with friction. At point B the mechanical energy is $300\,\text{J} + 300\,\text{J} = 600\,\text{J}$, so by then only $100\,\text{J}$ has changed into thermal energy, and the thermal energy grows along the whole way. Check: $500\,\text{J} + 200\,\text{J} = 700\,\text{J}$.

4. Exercise 1-4

   In each situation, name the form of energy that the energy is in at the start and the useful form of energy that it changes into. a) Firewood burns in a fireplace and warms a room. b) A solar panel produces electricity. c) Water flows down from a reservoir into a turbine that turns a generator.

   I have done my solution.
   Answer

   a) chemical energy → thermal energy; b) radiant energy → electrical energy; c) potential energy of water → (kinetic energy →) electrical energy

   Show the solution

   Find

   in each situation, the form of energy at the start and the useful form of energy into which the energy changes

   Principle

   The form of energy tells you where the energy is stored or how it shows itself. In fuels the energy is in chemical bonds, sunlight is radiation, and water high up has potential energy because of its position. Energy does not come from nothing, so the useful energy always comes from the form of energy at the start, and in every change part of it also changes into thermal energy.

   Model answer

   a) The chemical energy of the wood changes in combustion into thermal energy (and a little light), which warms the room. b) The radiant energy of the Sun changes in the panel into electrical energy; however, most of it changes into thermal energy in the panel. c) The potential energy of the water in the reservoir changes into the kinetic energy of the flowing water, the turbine and generator change it into electrical energy, and a little changes into thermal energy. So the forms of energy at the start are a) chemical energy, b) radiant energy and c) potential energy, and the useful forms of energy at the end are a) thermal energy, b) electrical energy and c) electrical energy.

5. Exercise 1-5

   A sledge starts from rest at the top of a hill. At the top its potential energy is $600\,\text{J}$, and on the way down friction changes $90\,\text{J}$ of energy into thermal energy. At the bottom of the hill the potential energy is zero.

   What is the kinetic energy of the sledge at the bottom of the hill?

   JCheck
   Answer

   $510\,\text{J}$

   Show the solution

   Given

   potential energy of the sledge at the top $E_\text{p} = 600\,\text{J}$, energy changed into thermal energy by friction $E_\text{thermal} = 90\,\text{J}$, potential energy at the bottom of the hill $0\,\text{J}$

   Find

   kinetic energy $E_\text{k}$ at the bottom of the hill in $\text{J}$

   Principle

   The sledge, the hill and the surroundings form an isolated system, so energy is conserved: $E_\text{initial} = E_\text{final}$. At the start all the energy is potential energy, and at the end it is kinetic energy and thermal energy, into which friction has changed part of it.

   Formula

   $E_\text{p} = E_\text{k} + E_\text{thermal}$, so $E_\text{k} = E_\text{p} - E_\text{thermal}$

   Conversions

   All energies are in joules, so no conversions are needed.

   Substitution

   $E_\text{k}$$= 600\,\text{J} - 90\,\text{J}$

   $= 510\,\text{J}$

   Answer

   $510\,\text{J}$

   Three significant figures as in the potential energy $600\,\text{J}$.

   Reasonableness

   The kinetic energy is smaller than the initial energy $600\,\text{J}$, because part of it changed into thermal energy, as it should with friction. Checking backwards: $510\,\text{J} + 90\,\text{J} = 600\,\text{J}$.

6. Exercise 1-6

   A charged phone is on a table and is not being used. What form of energy is the energy stored in the battery of the phone?

   Choose an option.

   • Potential energy
   • Chemical energy
   • Kinetic energy
   • Radiant energy

   Answer
   I have done my solution.
   Answer

   Chemical energy

   Show the solution

   Find

   the form of energy of the energy stored in the battery

   Principle

   The form of energy tells you in what form the energy is. A battery is a device in which energy is stored in the bonds of chemical substances, and it is released as electrical energy when the device is put to use. Kinetic energy belongs to a moving object, potential energy depends on the height of an object, and radiant energy is light and other radiation.

   Model answer

   Chemical energy. A phone at rest does not move (no kinetic energy), its energy does not depend on height (no potential energy) and it does not emit light (no radiant energy). The energy is in the substances of the battery as chemical energy, and when the phone is used it changes into electrical energy and from that into light, sound and thermal energy.

7. Exercise 1-7

   A car drives along a level road and brakes to a stop. Which of the following describes correctly what happens to the kinetic energy of the car?

   Choose an option.

   • The kinetic energy changes into thermal energy in the brakes, the tyres and the road, and the total amount of energy stays the same.
   • The kinetic energy disappears, because the car stops.
   • The kinetic energy changes into thermal energy in the brakes, but the total energy decreases, because the heat is transferred to the surroundings.
   • The kinetic energy is still in the car as kinetic energy, even though the car has stopped.

   Answer
   I have done my solution.
   Answer

   The kinetic energy changes into thermal energy in the brakes, the tyres and the road, and the total amount of energy stays the same.

   Show the solution

   Find

   what happens to the kinetic energy of the car in braking

   Principle

   Energy is neither created nor destroyed but changes form: in an isolated system $E_\text{initial} = E_\text{final}$. In braking, friction changes kinetic energy into thermal energy. In everyday language "energy is used up" means in physics that the energy has changed into a form that is hard to use.

   Model answer

   The correct answer is the first one: the kinetic energy changes into thermal energy in the brakes, the tyres and the road, and the total amount of energy stays the same. When the car has stopped it no longer has kinetic energy, but the same energy is thermal energy in the brake discs, the tyres and the road, which have warmed up a little. The claim that energy disappears or decreases breaks the law of conservation of energy.

8. Exercise 1-8

   A phone is being charged with a charger from a wall socket, and the charger feels warm in the hand. Describe the change of energy as a chain, from the energy coming along the cable from the wall socket to the moment it is in the battery. Use the law of conservation of energy to explain why the charger warms up.

   I have done my solution.
   Answer

   Electrical energy → (in the charger) electrical energy and thermal energy → chemical energy in the battery; part changes into thermal energy, so the charger warms up, and energy is not lost

   Show the solution

   Find

   the change of energy as a chain from the charger to the battery, and why the charger warms up

   Principle

   Energy is neither created nor destroyed but changes form: $E_\text{initial} = E_\text{final}$. In every change part of the energy changes into thermal energy, which is transferred as heat to the surroundings.

   Model answer

   Electrical energy comes from the wall socket. The charger changes it into electrical energy suitable for the battery of the phone, and in the battery the electrical energy is stored as chemical energy. So the chain is electrical energy → electrical energy → chemical energy. The change is not perfect: part of the electrical energy from the wall socket changes into thermal energy in the charger, and that is why the charger warms up and heat is transferred to the surroundings. By the law of conservation of energy, the energy taken from the wall socket equals the chemical energy stored in the battery and the thermal energy together, so no energy has been lost. However, the thermal energy can no longer be used to charge the phone.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 1-9Core

   An electric scooter sets off. Chemical energy of $3.00\,\text{kJ}$ is released from its battery, and the scooter and rider gain kinetic energy of $2150\,\text{J}$ from it. The rest changes into thermal energy in the battery, the motor and the tyres.

   How many joules of energy are changed into thermal energy?

   JCheck
   Answer

   $850\,\text{J}$

   Show the solution

   Given

   chemical energy released from the battery $E_\text{chem} = 3.00\,\text{kJ}$, kinetic energy of the scooter and rider $E_\text{k} = 2150\,\text{J}$

   Find

   energy changed into thermal energy $E_\text{thermal}$ in $\text{J}$

   Principle

   The chemical energy of the battery changes through electrical energy into kinetic energy and thermal energy, and energy is neither created nor destroyed. By the law of conservation of energy $E_\text{initial} = E_\text{final}$.

   Formula

   $E_\text{chem} = E_\text{k} + E_\text{thermal}$, so $E_\text{thermal} = E_\text{chem} - E_\text{k}$

   Conversions

   The chemical energy is in kilojoules and the kinetic energy in joules. The prefix kilo means a thousand, so we substitute

   $3.00\,\text{kJ}$$= 3.00 \cdot 1000\,\text{J}$

   $= 3000\,\text{J}$

   Substitution

   $E_\text{thermal}$$= 3000\,\text{J} - 2150\,\text{J}$

   $= 850\,\text{J}$

   Answer

   $850\,\text{J}$

   To the nearest ten joules, because the chemical energy $3.00\,\text{kJ}$ is given to the nearest ten joules.

   Reasonableness

   $2150\,\text{J} : 3000\,\text{J} \approx 0.72$, a little under three quarters, ends up as kinetic energy and the rest is thermal energy. This suits an electric motor, which is much more efficient than a combustion engine. Checking backwards: $2150\,\text{J} + 850\,\text{J} = 3000\,\text{J}$.

2. Exercise 1-10Core

   A child starts from rest at the top of a water slide. At the bottom her kinetic energy is $2400\,\text{J}$, and on the way friction has changed $0.60\,\text{kJ}$ of energy into thermal energy. The bottom is the zero level of potential energy.

   What was the potential energy of the child at the top? Answer in kilojoules.

   kJCheck
   Answer

   $3.0\,\text{kJ}$

   Show the solution

   Given

   kinetic energy at the bottom $E_\text{k} = 2400\,\text{J}$, energy changed into thermal energy by friction $E_\text{thermal} = 0.60\,\text{kJ}$, potential energy at the bottom $0\,\text{J}$, the child starts from rest

   Find

   potential energy $E_\text{p}$ at the top in $\text{kJ}$

   Principle

   The child, the slide and the water form an isolated system, so energy is conserved: $E_\text{initial} = E_\text{final}$. At the top the child is at rest, so all the energy is potential energy.

   Formula

   $E_\text{p} = E_\text{k} + E_\text{thermal}$. The formula is already solved for the quantity asked.

   Conversions

   The answer is asked in kilojoules, and the kinetic energy is in joules. The joule is a thousandth of a kilojoule, so we substitute

   $2400\,\text{J}$$= 2400 \cdot 0.001\,\text{kJ}$

   $= 2.4\,\text{kJ}$

   Substitution

   $E_\text{p}$$= 2.4\,\text{kJ} + 0.60\,\text{kJ}$

   $= 3.0\,\text{kJ}$

   Answer

   $3.0\,\text{kJ}$

   Two significant figures as in the kinetic energy $2400\,\text{J}$, that is $2.4\,\text{kJ}$.

   Reasonableness

   The initial energy must be greater than the kinetic energy at the bottom, $2.4\,\text{kJ}$, because part changed into thermal energy, and $3.0\,\text{kJ} > 2.4\,\text{kJ}$. Energy changed into thermal energy: $0.60\,\text{kJ} : 3.0\,\text{kJ} = 0.20$, that is a fifth.

3. Exercise 1-11

   A child sitting on a swing is pulled to the side and released from rest. At the highest point the potential energy of the child and the swing is $0.36\,\text{kJ}$, when the zero level is the lowest point of the swing. The swing keeps swinging in smaller and smaller swings until it stops at the lowest point. Sound is ignored.

   How many joules of energy are changed into thermal energy in total?

   JCheck
   Answer

   $360\,\text{J}$

   Show the solution

   Given

   potential energy at the highest point $E_\text{p} = 0.36\,\text{kJ}$, the swing starts from rest and finally stops at the lowest point, sound is ignored

   Find

   total energy changed into thermal energy $E_\text{thermal}$ in $\text{J}$

   Principle

   The child, the swing, its fixings and the surrounding air form an isolated system, so energy is conserved. At the end the swing is at rest at the lowest point, so it has neither kinetic energy nor potential energy. All the initial energy has changed into thermal energy through air resistance and friction in the fixings.

   Formula

   $E_\text{p} = E_\text{thermal}$, so $E_\text{thermal} = E_\text{p}$

   Conversions

   The prefix kilo means a thousand, so we substitute

   $0.36\,\text{kJ}$$= 0.36 \cdot 1000\,\text{J}$

   $= 360\,\text{J}$

   Substitution

   $E_\text{thermal}$$= E_\text{p}$

   $= 360\,\text{J}$

   Answer

   $360\,\text{J}$

   Or $3.6 \cdot 10^{2}\,\text{J}$, two significant figures as in the given $0.36\,\text{kJ}$.

   Reasonableness

   Each swing is lower than the one before, because part of the mechanical energy changes into thermal energy, and finally all $360\,\text{J}$ is thermal energy. It spreads into the air and the fixings so widely that the warming cannot be noticed, but no energy has been lost.

4. Exercise 1-12

   Electrical energy of $360\,\text{kJ}$ goes into a kettle. Of this, $331\,\text{kJ}$ is transferred to the water as heat, and the rest warms the kettle and the surroundings.

   How many joules of energy end up in the kettle and the surroundings?

   JCheck
   Answer

   $29\,000\,\text{J}$

   Show the solution

   Given

   electrical energy going into the kettle $E_\text{elec} = 360\,\text{kJ}$, energy transferred to the water as heat $E_\text{water} = 331\,\text{kJ}$

   Find

   energy ending up in the kettle and the surroundings $E_\text{other}$ in $\text{J}$

   Principle

   The electrical energy changes into thermal energy, and part of it ends up in the water and the rest in the kettle and the surroundings. Energy is neither created nor destroyed, so $E_\text{initial} = E_\text{final}$.

   Formula

   $E_\text{elec} = E_\text{water} + E_\text{other}$, so $E_\text{other} = E_\text{elec} - E_\text{water}$

   Conversions

   The prefix kilo means a thousand, so we substitute

   $360\,\text{kJ}$$= 360 \cdot 1000\,\text{J}$

   $= 360\,000\,\text{J}$

   $331\,\text{kJ}$$= 331 \cdot 1000\,\text{J}$

   $= 331\,000\,\text{J}$

   Substitution

   $E_\text{other}$$= 360\,000\,\text{J} - 331\,000\,\text{J}$

   $= 29\,000\,\text{J}$

   Answer

   $29\,000\,\text{J}$

   Or $2.9 \cdot 10^{4}\,\text{J}$, two significant figures, because the data are given to the nearest kilojoule and the difference is $29\,\text{kJ}$.

   Reasonableness

   $331\,\text{kJ} : 360\,\text{kJ} \approx 0.92$, over $90\,\%$, goes into the water, and the rest, $29\,\text{kJ}$, is a small part, as it should be with a well-insulated kettle. Checking backwards: $331\,000\,\text{J} + 29\,000\,\text{J} = 360\,000\,\text{J}$.

5. Exercise 1-13

   The table gives, as example figures, the yearly energy consumption of a detached house by use. Calculate what percentage of the total consumption is space heating.

   Which use would be the biggest target for saving, and why?

   Yearly energy consumption of a detached house by use

   Use	$E\,(\text{GJ})$
   Space heating	36.0
   Hot water	10.5
   Electrical appliances and lighting	9.0
   Sauna	4.5

   %Check
   Answer

   $60\,\%$

   Show the solution

   Given

   space heating $36.0\,\text{GJ}$, hot water $10.5\,\text{GJ}$, electrical appliances and lighting $9.0\,\text{GJ}$, sauna $4.5\,\text{GJ}$

   Find

   the share of space heating in the total consumption as a percentage, and the biggest target for saving

   Principle

   A share is the ratio of a part to the whole. The total consumption is the sum of the energies of all the uses, because the uses do not overlap.

   Formula

   $E_\text{tot}$$= E_\text{heating} + E_\text{water} + E_\text{elec} + E_\text{sauna}$

   and share $= \dfrac{E_\text{heating}}{E_\text{tot}}$

   Conversions

   All energies are in gigajoules, so no conversions are needed. The share is changed into a percentage by multiplying by one hundred.

   Substitution

   The sum is

   $E_\text{tot}$$= 36.0\,\text{GJ} + 10.5\,\text{GJ} + 9.0\,\text{GJ} + 4.5\,\text{GJ}$

   $= 60.0\,\text{GJ}$

   so the share is

   $\dfrac{36.0\,\text{GJ}}{60.0\,\text{GJ}}$$= 0.600$

   $= 60.0\,\%$

   Answer

   The share of heating is $60\,\%$ of the consumption. The biggest target for saving is space heating, because it uses the most energy: a ten percent saving in heating would reduce the consumption by $3.6\,\text{GJ}$, but the same saving in the sauna by only $0.45\,\text{GJ}$.

   Reasonableness

   The share is between $0$ and $100\,\%$, and heating is the biggest item, so more than half is believable in a residential building in the north. Checking the sum of the shares: $60\,\% + 17.5\,\% + 15\,\% + 7.5\,\% = 100\,\%$.

6. Exercise 1-14

   An inventor demonstrates a water device that he claims works forever without outside energy.

   Water falling from the upper pool turns a generator and loses potential energy of $100\,\text{J}$ per cycle. The generator changes all of this energy into electrical energy. A pump uses $80\,\text{J}$ of it and lifts the water back to the upper pool, and the rest of the electrical energy goes to a lamp.

   How many joules of energy does the device create from nothing in one cycle, if the claim is true? Explain which principle of energy is broken.

   JCheck
   Answer

   $20\,\text{J}$

   Show the solution

   Given

   decrease of the potential energy of the falling water $100\,\text{J}$ per cycle, electrical energy of the generator $100\,\text{J}$, electrical energy used by the pump $80\,\text{J}$, the water returns to the upper pool

   Find

   the energy created from nothing in a cycle $E_\text{created}$ in $\text{J}$ and which principle of energy is broken

   Principle

   Energy is conserved: in an isolated system $E_\text{initial} = E_\text{final}$. At the end of the cycle the water is in the upper pool again, so the energy of the water is the same as at the start. The total energy of the whole device must not change when nothing comes from outside.

   Formula

   The energy left for the lamp is $E_\text{lamp} = E_\text{elec} - E_\text{pump}$, and because the water is unchanged after the cycle, the same energy is energy created in the device: $E_\text{created} = E_\text{lamp}$

   Conversions

   All energies are in joules, so no conversions are needed.

   Substitution

   $E_\text{created}$$= E_\text{lamp}$

   $= 100\,\text{J} - 80\,\text{J}$

   $= 20\,\text{J}$

   Answer

   $20\,\text{J}$

   Per cycle, two significant figures as in the data. The claim breaks the law of conservation of energy, because energy cannot be created from nothing: the device would have to create $20\,\text{J}$ more energy in every cycle.

   Reasonableness

   The pump should lift the water back up, and that needs at least as much energy as the water loses in falling, that is $100\,\text{J}$. However, the pump gets only $80\,\text{J}$, so the claim requires $20\,\text{J}$ of energy from nothing. In a real device friction and electricity also change part of the energy into thermal energy, so nothing would be left for the lamp and the device would stop. A perpetual motion machine is not possible.

7. Exercise 1-15

   The table gives, as example figures, the electrical energy taken by three lamps in one hour and how much of it changes into light; for comparison, the lamps take the same electrical energy. The rest of the energy changes into thermal energy.

   How many kilojoules more thermal energy does the incandescent lamp produce than the LED lamp? Justify which lamp is the wiser one to use in terms of energy.

   Electrical energy of the lamps and the energy that changes into light in an hour

   Lamp	$E_\text{elec}\,(\text{kJ})$	$E_\text{light}\,(\text{kJ})$
   Incandescent lamp	216	11
   Halogen lamp	216	16
   LED lamp	216	54

   kJCheck
   Answer

   $43\,\text{kJ}$

   Show the solution

   Given

   electrical energy of each lamp $E_\text{elec} = 216\,\text{kJ}$; energy that changes into light $11\,\text{kJ}$ for the incandescent lamp, $16\,\text{kJ}$ for the halogen lamp and $54\,\text{kJ}$ for the LED lamp (example figures)

   Find

   the difference between the thermal energies of the incandescent lamp and the LED lamp $\Delta E_\text{thermal}$ in $\text{kJ}$

   Principle

   A lamp changes electrical energy into light and thermal energy, and energy is neither created nor destroyed, so $E_\text{elec} = E_\text{light} + E_\text{thermal}$.

   Formula

   $E_\text{thermal} = E_\text{elec} - E_\text{light}$ for both lamps and $\Delta E_\text{thermal} = E_\text{thermal,incand} - E_\text{thermal,LED}$

   Conversions

   All energies are in kilojoules, so no conversions are needed.

   Substitution

   $E_\text{thermal,incand}$

   $= 216\,\text{kJ} - 11\,\text{kJ}$

   $= 205\,\text{kJ}$

   $E_\text{thermal,LED}$

   $= 216\,\text{kJ} - 54\,\text{kJ}$

   $= 162\,\text{kJ}$

   so

   $\Delta E_\text{thermal}$$= 205\,\text{kJ} - 162\,\text{kJ}$

   $= 43\,\text{kJ}$

   Answer

   The incandescent lamp produces $43\,\text{kJ}$ more thermal energy. Two significant figures match the accuracy of the energy of light $11\,\text{kJ}$.

   Reasonableness

   The thermal energy of the incandescent lamp is greater because its share of light is smaller ($11\,\text{kJ}$ is only $5\,\%$ of the electrical energy, while the $54\,\text{kJ}$ of the LED is $25\,\%$). Checking backwards: LED $54\,\text{kJ} + 162\,\text{kJ} = 216\,\text{kJ}$. Justification: when the same amount of light is wanted, the LED lamp needs less electrical energy, and less energy changes into useless thermal energy, so the LED is the wiser choice in terms of energy.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 1-16

   The energies of a skateboarder are measured at three points in a half-pipe.

   At point A the skateboarder is at rest and her potential energy is $900\,\text{J}$. At point B the potential energy is $400\,\text{J}$ and the kinetic energy $480\,\text{J}$. At point C the potential energy is zero and the kinetic energy $750\,\text{J}$. Friction keeps changing mechanical energy into thermal energy.

   How much energy is changed into thermal energy between points B and C?

   JCheck
   Answer

   $130\,\text{J}$

   Show the solution

   Given

   at point A $E_\text{p} = 900\,\text{J}$ and $E_\text{k} = 0\,\text{J}$; at point B $E_\text{p} = 400\,\text{J}$ and $E_\text{k} = 480\,\text{J}$; at point C $E_\text{p} = 0\,\text{J}$ and $E_\text{k} = 750\,\text{J}$

   Find

   energy changed into thermal energy between points B and C, $\Delta E_\text{thermal}$, in $\text{J}$

   Principle

   The total energy of the system is conserved, so at every point $E_\text{p} + E_\text{k} + E_\text{thermal}$ is equal to its value at point A, which is $900\,\text{J}$. The thermal energy is what the mechanical energy is missing.

   Formula

   $E_\text{thermal} = E_\text{A} - E_\text{p} - E_\text{k}$ at each point, and between points B and C $\Delta E_\text{thermal} = E_\text{thermal,C} - E_\text{thermal,B}$

   Conversions

   All energies are in joules, so no conversions are needed.

   Substitution

   $E_\text{thermal,B}$$= 900\,\text{J} - 400\,\text{J} - 480\,\text{J}$

   $= 20\,\text{J}$

   $E_\text{thermal,C}$$= 900\,\text{J} - 0\,\text{J} - 750\,\text{J}$

   $= 150\,\text{J}$

   so

   $\Delta E_\text{thermal}$$= 150\,\text{J} - 20\,\text{J}$

   $= 130\,\text{J}$

   Answer

   $130\,\text{J}$

   Two significant figures, because the initial data are given to the nearest ten joules.

   Reasonableness

   The thermal energy grows from point to point ($0$, $20\,\text{J}$, $150\,\text{J}$), as it should when friction changes mechanical energy into thermal energy, and it never decreases. The mechanical energy is $880\,\text{J}$ at B and $750\,\text{J}$ at C, and the difference $880\,\text{J} - 750\,\text{J} = 130\,\text{J}$ confirms the result.

2. Exercise 1-17

   A rubber ball is dropped on the floor. The mechanical energy of the ball is $5.00\,\text{J}$ when dropped, $4.00\,\text{J}$ after the first bounce and $3.20\,\text{J}$ after the second bounce.

   How much more energy is changed into thermal energy in the first bounce than in the second?

   JCheck
   Answer

   $0.20\,\text{J}$

   Show the solution

   Given

   mechanical energy of the ball when dropped $E_0 = 5.00\,\text{J}$, after the first bounce $E_1 = 4.00\,\text{J}$ and after the second bounce $E_2 = 3.20\,\text{J}$

   Find

   how much more energy is changed into thermal energy in the first bounce than in the second, $\Delta E$, in $\text{J}$

   Principle

   The system formed by the ball, the floor and the air is isolated, so the total energy is conserved. In each bounce the mechanical energy the ball is missing has changed into thermal energy. We calculate the energy balance for each bounce separately and compare.

   Formula

   In the first bounce $E_0 = E_1 + E_\text{thermal,1}$ and in the second $E_1 = E_2 + E_\text{thermal,2}$, so $E_\text{thermal,1} = E_0 - E_1$, $E_\text{thermal,2} = E_1 - E_2$ and $\Delta E = E_\text{thermal,1} - E_\text{thermal,2}$

   Conversions

   All energies are in joules, so no conversions are needed.

   Substitution

   $E_\text{thermal,1}$$= 5.00\,\text{J} - 4.00\,\text{J}$

   $= 1.00\,\text{J}$

   $E_\text{thermal,2}$$= 4.00\,\text{J} - 3.20\,\text{J}$

   $= 0.80\,\text{J}$

   so

   $\Delta E$$= 1.00\,\text{J} - 0.80\,\text{J}$

   $= 0.20\,\text{J}$

   Answer

   $0.20\,\text{J}$

   To two decimal places like the energies in the table.

   Reasonableness

   In each bounce a fifth of the ball's energy at that moment changes into thermal energy ($1.00\,\text{J}$ is a fifth of $5.00\,\text{J}$ and $0.80\,\text{J}$ is a fifth of $4.00\,\text{J}$). In the second bounce the ball has less energy, so less energy changes into thermal energy, and so the bounces get lower and lower but the decrease slows down. Checking backwards: $3.20\,\text{J} + 0.80\,\text{J} + 1.00\,\text{J} = 5.00\,\text{J}$.

3. Exercise 1-18

   In a nuclear power plant, the nuclear reactions in the reactor heat water into steam, the steam turns a turbine and the turbine turns a generator, which produces electricity. The steam leaving the turbine is cooled with seawater, which warms up.

   a) Name the forms of energy at each stage of the chain.

   b) The plant turns only about a third of the nuclear energy released in the reactor into electrical energy. Use the law of conservation of energy to explain where the rest of the energy goes and why it has not been lost.

   I have done my solution.
   Answer

   a) nuclear energy → thermal energy → kinetic energy of steam and turbine → electrical energy; b) about two thirds ends up as thermal energy in the sea and the surroundings, and the total energy is conserved

   Show the solution

   Find

   a) the forms of energy at each stage of the nuclear power plant chain; b) where about two thirds of the nuclear energy goes and why the energy has not been lost

   Principle

   Energy changes form and moves, but is neither created nor destroyed: in the system formed by the power plant, the cooling water and the surroundings $E_\text{initial} = E_\text{final}$. In every change part of the energy changes into thermal energy, which spreads into the surroundings and is hard to use.

   Model answer

   a) The nuclear energy stored in the atomic nuclei of uranium is released in the reactor and changes into the thermal energy of water and steam. The steam flows into the turbine and turns it, so the energy is kinetic energy. The generator changes the kinetic energy of the turning turbine into electrical energy. b) Only about a third becomes electricity, because the steam leaving the turbine is still warm: energy is transferred from it as heat to the cooling water, which carries it to the sea. In addition, in the turbine, the generator and the pipes part of the energy changes into thermal energy, which is transferred as heat into the surroundings. The energy has not been lost: the electrical energy and the thermal energy transferred to the sea and the surroundings together equal the nuclear energy released. So about two thirds ends up as thermal energy that has spread widely and can no longer easily be used.

4. Exercise 1-19

   The table gives the total yearly energy consumption of an imaginary country by source (rounded example figures, not the statistics of a real country). The country has $5.5 \cdot 10^{6}$ inhabitants. Oil, coal and natural gas are fossil fuels.

   a) Calculate the share of fossil fuels in the total consumption as a percentage.

   b) Calculate the total consumption per inhabitant in gigajoules; $1\,\text{PJ} = 10^{15}\,\text{J}$ and $1\,\text{GJ} = 10^{9}\,\text{J}$.

   c) The country wants to bring the share of fossil fuels below a third of the total consumption by replacing part of the fossil energy with hydro and wind power. The total consumption stays the same. At least how many petajoules of fossil energy must be replaced? Justify with a calculation and use the law of conservation of energy to explain why the total consumption can stay the same.

   Total yearly energy consumption of an example country by source

   Energy source	$E\,(\text{PJ})$
   Bioenergy	380
   Oil	290
   Nuclear energy	230
   Hydro and wind power	120
   Coal	100
   Natural gas	80

   %Check
   Answer

   $39\,\%$

   Show the solution

   Given

   consumption by source: bioenergy $380\,\text{PJ}$, oil $290\,\text{PJ}$, nuclear energy $230\,\text{PJ}$, hydro and wind power $120\,\text{PJ}$, coal $100\,\text{PJ}$, natural gas $80\,\text{PJ}$; inhabitants $N = 5.5 \cdot 10^{6}$; $1\,\text{PJ} = 10^{15}\,\text{J}$ and $1\,\text{GJ} = 10^{9}\,\text{J}$

   Find

   a) the share of fossil fuels in the total consumption as a percentage; b) the total consumption per inhabitant in $\text{GJ}$; c) the minimum replacement of fossil energy in petajoules so that the share of fossil fuels stays below a third

   Principle

   The total consumption is the sum of the energies of all the sources, and a share is the ratio of a part to the whole. The amount per inhabitant is obtained by dividing the total energy by the number of inhabitants. Energy is conserved, so when fossil energy is replaced with hydro and wind power, the energy needed for the same use does not change, only its source does.

   Formula

   $E_\text{tot} = \sum E$, share $= \dfrac{E_\text{fossil}}{E_\text{tot}}$ and

   $E_\text{person}$$= \dfrac{E_\text{tot}}{N}$

   condition c):

   $E_\text{fossil} - E_\text{replaced}$

   $< \dfrac{E_\text{tot}}{3}$

   Conversions

   Prefixes: $1\,\text{PJ} = 10^{15}\,\text{J}$ and $1\,\text{GJ} = 10^{9}\,\text{J}$, so

   $1\,\text{PJ}$$= 10^{15} : 10^{9}\,\text{GJ}$

   $= 10^{6}\,\text{GJ}$

   Substitution

   The total consumption is

   $E_\text{tot}$$= 380\,\text{PJ} + 290\,\text{PJ} + 230\,\text{PJ} + 120\,\text{PJ} + 100\,\text{PJ} + 80\,\text{PJ}$

   $= 1200\,\text{PJ}$

   The sum of the fossil fuels is

   $E_\text{fossil}$$= 290\,\text{PJ} + 100\,\text{PJ} + 80\,\text{PJ}$

   $= 470\,\text{PJ}$

   so the share is

   $\dfrac{470\,\text{PJ}}{1200\,\text{PJ}}$$= 0.392$

   $= 39.2\,\%$

   In gigajoules the total consumption is

   $1200\,\text{PJ}$$= 1200 \cdot 10^{6}\,\text{GJ}$

   $= 1.2 \cdot 10^{9}\,\text{GJ}$

   and per inhabitant

   $E_\text{person}$$= \dfrac{1.2 \cdot 10^{9}\,\text{GJ}}{5.5 \cdot 10^{6}}$

   $= 218\,\text{GJ}$

   A third of the total consumption is $1200\,\text{PJ} : 3 = 400\,\text{PJ}$, so the amount to be replaced is

   $E_\text{replaced}$$> 470\,\text{PJ} - 400\,\text{PJ}$

   $= 70\,\text{PJ}$

   Answer

   a) The share of fossil fuels is about $39\,\%$. b) The consumption is about $2.2 \cdot 10^{2}\,\text{GJ}$, that is $220\,\text{GJ}$ per inhabitant per year (two significant figures as in the number of inhabitants). c) More than $70\,\text{PJ}$ of fossil energy must be replaced (two significant figures as in the data), for example $71\,\text{PJ}$, which leaves $399\,\text{PJ}$ of fossil energy, that is $33.25\,\%$, and the share is below a third. The total consumption can stay the same because energy is neither created nor destroyed: the same energy needed for use is simply produced from another source, so the replacing hydro and wind power adds as much energy as the fossil energy decreases.

   Reasonableness

   The share is below $100\,\%$. Checking backwards: $218\,\text{GJ} \cdot 5.5 \cdot 10^{6} \approx 1.2 \cdot 10^{9}\,\text{GJ}$. Limit value: if exactly $70\,\text{PJ}$ is replaced, there is $400\,\text{PJ}$ of fossil energy, exactly a third, so a little more is needed.

5. Exercise 1-20

   The LED lamp on a desk gets its electricity from a solar panel and a battery. The table gives the energies of one day at the stages of the chain: the input of each stage and the useful energy that goes on from it. The input of the next stage is the useful energy of the previous stage. The model assumes that the rest of the energy of a stage changes into thermal energy.

   a) Calculate the energy that changes into thermal energy at each stage and in the whole chain in total.

   b) At which stage is the most thermal energy produced, and what percentage of all the thermal energy is that?

   c) Check the result with the law of conservation of energy and explain why the radiant energy arriving at the panel has not been used up, even though the light of the lamp is only a small part of it.

   From the solar panel to the lamp: energies during one day

   Stage	$E_\text{in}\,(\text{MJ})$	$E_\text{out}\,(\text{MJ})$
   Solar panel	5.00	0.80
   Charge controller and battery	0.80	0.64
   Battery discharge and wires	0.64	0.56
   LED lamp	0.56	0.14

   MJCheck
   Answer

   $4.86\,\text{MJ}$

   Show the solution

   Given

   input and useful energy of the stages: panel $5.00\,\text{MJ} \to 0.80\,\text{MJ}$, charge controller and battery $0.80\,\text{MJ} \to 0.64\,\text{MJ}$, battery discharge and wires $0.64\,\text{MJ} \to 0.56\,\text{MJ}$, LED lamp $0.56\,\text{MJ} \to 0.14\,\text{MJ}$

   Find

   a) the thermal energy $E_\text{thermal}$ at each stage and in total in $\text{MJ}$; b) the stage where the most thermal energy is produced, and its share as a percentage; c) the check with the law of conservation and the justification

   Principle

   At each stage energy is conserved: $E_\text{in} = E_\text{out} + E_\text{thermal}$. The model assumes that the rest of the energy changes into thermal energy, so the thermal energy is calculated as the part that does not continue onwards as useful energy. The thermal energy of the whole chain is the sum of the thermal energies of the stages.

   Formula

   $E_\text{thermal} = E_\text{in} - E_\text{out}$ at each stage and $E_\text{thermal,tot}$ is the sum of the stages

   Conversions

   All energies are in megajoules, so no conversions are needed.

   Substitution

   In the panel $5.00\,\text{MJ} - 0.80\,\text{MJ} = 4.20\,\text{MJ}$, in the charge controller and battery $0.80\,\text{MJ} - 0.64\,\text{MJ} = 0.16\,\text{MJ}$, in the battery discharge and wires $0.64\,\text{MJ} - 0.56\,\text{MJ} = 0.08\,\text{MJ}$ and in the lamp $0.56\,\text{MJ} - 0.14\,\text{MJ} = 0.42\,\text{MJ}$. The sum is

   $E_\text{thermal,tot}$

   $= 4.20\,\text{MJ} + 0.16\,\text{MJ} + 0.08\,\text{MJ} + 0.42\,\text{MJ}$

   $= 4.86\,\text{MJ}$

   The most is produced in the panel, and its share is

   $\dfrac{4.20\,\text{MJ}}{4.86\,\text{MJ}}$$= 0.864$

   $= 86\,\%$

   Answer

   a) In total $4.86\,\text{MJ}$ changes into thermal energy (three significant figures as in the table). b) The most thermal energy is produced in the solar panel, about $86\,\%$ of all of it. c) Check: $0.14\,\text{MJ} + 4.86\,\text{MJ} = 5.00\,\text{MJ}$, which equals the radiant energy that arrived at the panel. The energy has not been used up: it has changed into light and thermal energy, which is transferred to the surroundings. The light is only $2.8\,\%$ of the initial energy, because at every stage part of it changes into thermal energy, which cannot be turned back into a useful form.

   Reasonableness

   The output energy of each stage is smaller than its input energy, for example $0.80 < 5.00$ and $0.14 < 0.56$, as it should be. The light of the lamp is $0.14\,\text{MJ} : 5.00\,\text{MJ} = 0.028$, that is $2.8\,\%$ of what arrived at the panel. The large thermal share of the solar panel is believable, because most of the sunlight does not change into electricity.