High School Physics · Grades 10–12

9Energy: Forms and Conservation

A moving car, a charged phone, a hot oven and a raised hammer share one property: each can make something happen. Physics measures that capacity with a single number — energy — and keeps its accounts under one iron rule: the total never changes, it only moves and changes costume. This chapter opens the ledger: the forms energy takes, the chains it travels, and what a kilowatt-hour buys.

9.1 Energy: the common currency

Definition 9.1 (Energy and the joule)

Energy measures a system’s capacity to produce changes: set matter in motion, lift it, heat it, light it. Its unit is the joule ( $\mathrm{J}$ ). Stored, transferred, converted — every branch of physics counts in this one currency.

Definition 9.2 (Kinetic energy)

A body of mass $m$ (in $\mathrm{kg}$ ) moving at speed $v$ (in $\mathrm{m}/\mathrm{s}$ ) carries the kinetic energy $E_k = \tfrac12\, m v^2$ : doubling the mass doubles it, doubling the speed quadruples it.

Example 9.3 (Car versus pedestrian)

A $1300\,\mathrm{kg}$ car at $100\,\mathrm{km}/\mathrm{h}$ $= 27.8\,\mathrm{m}/\mathrm{s}$ carries $E_k = \tfrac12 \times 1300 \times 27.8^2 \approx 5.0 \times 10^{5}\,\mathrm{J}$ ; a $70\,\mathrm{kg}$ pedestrian at $1.5\,\mathrm{m}/\mathrm{s}$ , $\tfrac12 \times 70 \times 1.5^2 \approx 79\,\mathrm{J}$ — six thousand times less. Speed is expensive: the square sees to it.

Definition 9.4 (Gravitational potential energy)

A body of mass $m$ at height $h$ (in $\mathrm{m}$ ) above a chosen reference level stores the gravitational potential energy $E_p = m g h$ , with $g = 9.81\,\mathrm{N}/\mathrm{kg}$ — valid near the ground, where $g$ is effectively constant. Only differences of height matter: the level where $E_p = 0$ is chosen freely.

Example 9.5 (Water tower)

Each cubic metre ( $1000\,\mathrm{kg}$ ) of water stored $30\,\mathrm{m}$ up a water tower holds $1000 \times 9.81 \times 30 \approx 2.9 \times 10^{5}\,\mathrm{J}$ — about six tenths of the car’s kinetic energy above. Cities stockpile energy overhead.

Definition 9.6 (The other forms)

Energy also hides in less visible costumes: thermal energy, the disordered agitation of a body’s molecules (the hotter, the more); chemical energy, stored in molecular bonds — food, wood, gasoline, batteries; electrical energy, carried by currents from plant to socket; radiative energy, carried by light even across empty space — how the Sun’s energy reaches us; nuclear energy, stored inside atomic nuclei, released in stars (Chapter 33).

9.2 Transfers and conversion chains

Definition 9.7 (Transfer, conversion, chain)

An energy transfer moves energy between systems — reservoirs; an energy conversion changes its form. A device doing either (turbine, motor, lamp, muscle) is a converter. A conversion chain pictures the journey: boxes for reservoirs and forms, labeled arrows for transfers.

Conversion chain of a hydroelectric dam: boxes are reservoirs, arrows transfers — every real arrow leaks heat.

Method 9.8 (Drawing a conversion chain)

Identify the initial reservoir and form (what is consumed?).
Follow the energy converter by converter: one arrow per transfer, the form named in each box.
Add a heat arrow at every real converter — some energy always leaks as heat (friction, wires, exhaust).

A pendulum runs the simplest chain of all — all potential at the top of the swing, all kinetic through the lowest point (taking $E_p = 0$ there):

The pendulum’s budget at two positions: E_p and E_k swap while their sum (dashed) stays put. — The pendulum’s budget at two positions: $E_p$ and $E_k$ swap while their sum (dashed) stays put.

9.3 Conservation of energy

Theorem 9.9 (Conservation of energy)

The total energy of an isolated system — one exchanging nothing with the outside — is constant. Energy is never created and never destroyed: it only changes form and place, and what one reservoir loses, the others gain, joule for joule.

Proof. Admitted at this level. ∎

Remark 9.10

No experiment has ever caught this principle failing; here we take it as given and use it numerically. Chapter 29 derives its mechanical part ( $E_k + E_p$ ) from Newton’s laws; the full accounting waits for the university volumes.

Example 9.11 (Free fall, priced in joules)

A $2.0\,\mathrm{kg}$ ball drops from $10\,\mathrm{m}$ (air resistance negligible): it starts with $E_p = 2.0 \times 9.81 \times 10 \approx 196\,\mathrm{J}$ , lands with $E_k = 196\,\mathrm{J}$ , so $v = \sqrt{2 \times 196/2.0} = 14\,\mathrm{m}/\mathrm{s} \approx 50\,\mathrm{km}/\mathrm{h}$ . No forces, no stopwatch: conservation alone prices the impact.

9.4 Efficiency

Definition 9.12 (Efficiency)

A converter consumes $E_{\text{consumed}}$ and delivers the wanted form $E_{\text{useful}}$ ; its efficiency is

\eta = \frac{E_{\text{useful}}}{E_{\text{consumed}}} < 1 .

Conservation forbids $\eta > 1$ , and real converters never reach $1$ : the missing energy is not destroyed — it leaves as heat.

Example 9.13 (Where the gasoline goes)

A kilogram of gasoline stores $45\,\mathrm{MJ}$ of chemical energy; a car engine converts it with $\eta \approx 0.30$ : $13.5\,\mathrm{MJ}$ of motion, $31.5\,\mathrm{MJ}$ heating the exhaust, the radiator and the street. Two thirds of every tank warms the atmosphere.

$The car engine’s budget, arrow widths to scale: of 45\, MJ consumed, 13.5\, MJ moves the car; the rest is heat.$

The car engine’s budget, arrow widths to scale: of

45\,\mathrm{MJ}

consumed,

13.5\,\mathrm{MJ}

moves the car; the rest is heat.

Remark 9.14 (Efficiencies multiply)

Chained converters multiply their efficiencies: a power plant ( $0.38$ ) feeding the grid ( $0.94$ ) feeding a charger ( $0.85$ ) delivers $\eta = 0.38 \times 0.94 \times 0.85 \approx 0.30$ . Long chains are leaky pipes: each arrow taxes the flow.

9.5 Power, the watt and the kilowatt-hour

Definition 9.15 (Power)

Power is the rate at which energy flows, $P = E/t$ , measured in watts ( $\mathrm{W}$ ), with $1\,\mathrm{W} = 1\,\mathrm{J}/\mathrm{s}$ . Energy is an amount, power a flow — a bathtub versus its tap.

Example 9.16 (Kettle versus human)

A $2200\,\mathrm{W}$ kettle running $150\,\mathrm{s}$ uses $E = 2200 \times 150 = 3.3 \times 10^{5}\,\mathrm{J}$ . A human runs on about $10\,\mathrm{MJ}$ of food per day, an average $1.0 \times 10^{7}/86\,400 \approx 116\,\mathrm{W}$ : you idle like a bright old light bulb.

Definition 9.17 (The kilowatt-hour)

The kilowatt-hour ( $\mathrm{kW}\,\mathrm{h}$ ) is the energy of a $1\,\mathrm{kW}$ flow running for one hour: $1\,\mathrm{kW}\,\mathrm{h} = 1000 \times 3600\,\mathrm{J} = 3.6\,\mathrm{MJ}$ — the unit electricity meters count: power in $\mathrm{kW}$ times hours.

Example 9.18 (Reading the bill)

A bill is a subtraction and a multiplication: new meter reading minus old gives the kilowatt-hours; times the unit price (about $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ ), plus a fixed subscription, gives the total. A $2\,\mathrm{kW}$ heater running $2.5\,\mathrm{h}$ nightly uses $5\,\mathrm{kW}\,\mathrm{h}$ a day: $38$ euros a month on its own.

Example 9.19 (Orders of magnitude)

Worth knowing by heart: lifting an apple one metre, about $1\,\mathrm{J}$ ; a phone battery, $10\,\mathrm{W}\,\mathrm{h} \approx 4 \times 10^{4}\,\mathrm{J}$ ; a cereal bowl, $1.5\,\mathrm{MJ}$ ; your food for a day, $10\,\mathrm{MJ}$ ; a kilogram of gasoline, $45\,\mathrm{MJ}$ — almost exactly a household’s daily electricity ( $\approx 12\,\mathrm{kW}\,\mathrm{h}$ ); a lightning bolt, $5\,\mathrm{GJ}$ : a hundred household-days, delivered in a flash.

An energy ladder: each rung is a factor of 100. Everyday life spans ten orders of magnitude. — An energy ladder: each rung is a factor of $100$ . Everyday life spans ten orders of magnitude.

9.6 Exercises

Exercise 9.1 ★

A $1200\,\mathrm{kg}$ car drives at $90\,\mathrm{km}/\mathrm{h}$ , i.e. $25\,\mathrm{m}/\mathrm{s}$ : compute its kinetic energy. Redo it at $130\,\mathrm{km}/\mathrm{h}$ ( $36.1\,\mathrm{m}/\mathrm{s}$ ): what did $44\%$ more speed do to $E_k$ ?

Solution

Solution of Exercise 9.1.

$E_k = \tfrac12 \times 1200 \times 25^2 = 3.75 \times 10^{5}\,\mathrm{J}$ . At $36.1\,\mathrm{m}/\mathrm{s}$ : $\tfrac12 \times 1200 \times 36.1^2 \approx 7.8 \times 10^{5}\,\mathrm{J}$ — $44\%$ more speed, $(130/90)^2 \approx 2.1$ times the kinetic energy: the square at work.

Exercise 9.2 ★

A hiker of mass $72\,\mathrm{kg}$ climbs from altitude $1200\,\mathrm{m}$ to $1650\,\mathrm{m}$ . Compute the potential energy gained, convert it to kilowatt-hours, and price it at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ .

Solution

Solution of Exercise 9.2.

$\Delta h = 450\,\mathrm{m}$ : $E_p = 72 \times 9.81 \times 450 \approx 3.2 \times 10^{5}\,\mathrm{J} = 3.2 \times 10^{5}/3.6 \times 10^{6} \approx 0.088\,\mathrm{kW}\,\mathrm{h}$ , worth about $0.022$ euros. A day’s climbing is two cents of electricity.

Exercise 9.3 ★

Convert: $1\,\mathrm{kW}\,\mathrm{h}$ to joules; a $12\,\mathrm{W}\,\mathrm{h}$ phone battery to kilojoules; $45\,\mathrm{MJ}$ (one kilogram of gasoline) to kilowatt-hours; your $10\,\mathrm{MJ}$ food day to kilowatt-hours.

Solution

Solution of Exercise 9.3.

$1\,\mathrm{kW}\,\mathrm{h} = 3.6 \times 10^{6}\,\mathrm{J}$ ; $12\,\mathrm{W}\,\mathrm{h} = 12 \times 3600\,\mathrm{J} \approx 43\,\mathrm{kJ}$ ; $45\,\mathrm{MJ} = 45/3.6 = 12.5\,\mathrm{kW}\,\mathrm{h}$ ; $10\,\mathrm{MJ} = 10/3.6 \approx 2.8\,\mathrm{kW}\,\mathrm{h}$ .

Exercise 9.4 ★

A $2200\,\mathrm{W}$ kettle runs for $150\,\mathrm{s}$ . Compute the energy used, in joules and in kilowatt-hours. Then compute the average power of a human running on $10\,\mathrm{MJ}$ per day.

Solution

Solution of Exercise 9.4.

$E = 2200 \times 150 = 3.3 \times 10^{5}\,\mathrm{J} = 3.3 \times 10^{5}/3.6 \times 10^{6} \approx 0.092\,\mathrm{kW}\,\mathrm{h}$ . Human: $P = 1.0 \times 10^{7}/86\,400 \approx 116\,\mathrm{W}$ .

Exercise 9.5 ★

Write the conversion chain (Method 9.8) of: a bicycle dynamo powering a lamp; a gas water heater; a solar panel charging a phone. Mark the heat leaks.

Solution

Solution of Exercise 9.5.

Dynamo: chemical (cyclist’s food) $\to$ kinetic (wheel) $\to$ electrical (dynamo) $\to$ radiative $+$ thermal (lamp); heat leaks at muscles, tire contact, dynamo, lamp. Gas heater: chemical (gas) $\to$ thermal (water), leak in the flue gases. Panel: radiative (Sun) $\to$ electrical (panel) $\to$ chemical (battery), heat leaks at panel and charger.

Exercise 9.6 ★★

An engine burns $2.0\,\mathrm{kg}$ of gasoline ( $45\,\mathrm{MJ}/\mathrm{kg}$ ) and delivers $27\,\mathrm{MJ}$ of motion. Compute its efficiency and the energy released as heat. Where does that heat go?

Solution

Solution of Exercise 9.6.

Consumed $2.0 \times 45 = 90\,\mathrm{MJ}$ ; $\eta = 27/90 = 0.30$ . Heat: $90 - 27 = 63\,\mathrm{MJ}$ , carried off by the exhaust gases and the radiator into the surrounding air.

Exercise 9.7 ★★

A $70\,\mathrm{kg}$ cliff diver steps off a $20\,\mathrm{m}$ cliff (air resistance negligible). Using conservation, compute the kinetic energy at the water and the entry speed, in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ .

Solution

Solution of Exercise 9.7.

$E_k = E_p = 70 \times 9.81 \times 20 \approx 1.37 \times 10^{4}\,\mathrm{J}$ ; $v = \sqrt{2 \times 13734/70} = \sqrt{392} \approx 19.8\,\mathrm{m}/\mathrm{s} \approx 71\,\mathrm{km}/\mathrm{h}$ .

Exercise 9.8 ★★

A meter read $41\,236\,\mathrm{kW}\,\mathrm{h}$ on 1 March, $41\,562\,\mathrm{kW}\,\mathrm{h}$ 30 days later: energy used, daily average, bill at $0.26$ euros per $\mathrm{kW}\,\mathrm{h}$ plus $12$ euros fixed, and average power in watts?

Solution

Solution of Exercise 9.8.

$41562 - 41236 = 326\,\mathrm{kW}\,\mathrm{h}$ ; daily $326/30 \approx 10.9\,\mathrm{kW}\,\mathrm{h}$ . Bill: $326 \times 0.26 + 12 = 96.76$ euros. Average power: $326\,\mathrm{kW}\,\mathrm{h}/720\,\mathrm{h} \approx 453\,\mathrm{W}$ .

Exercise 9.9 ★★

An old $60\,\mathrm{W}$ bulb turns only $5\%$ of its consumption into light; an LED gives the same light from $9\,\mathrm{W}$ . Compute the bulb’s light power and the LED’s efficiency; then, at $4\,\mathrm{h}$ per day, each one’s yearly consumption and the saving at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ .

Solution

Solution of Exercise 9.9.

1. Light: $0.05 \times 60 = 3.0\,\mathrm{W}$ ; LED: $\eta = 3/9 \approx 0.33$ . 2. Yearly ( $1460\,\mathrm{h}$ ): bulb $60 \times 1460 = 87.6\,\mathrm{kW}\,\mathrm{h}$ , LED $13.1\,\mathrm{kW}\,\mathrm{h}$ ; saving $74.5\,\mathrm{kW}\,\mathrm{h} \approx 19$ euros per bulb per year.

Exercise 9.10 ★★

A dam passes $150\,\mathrm{m}^{3}$ of water ( $1.5 \times 10^{5}\,\mathrm{kg}$ ) per second down a $40\,\mathrm{m}$ drop. Compute the potential energy released each second — the available power — then the electrical power at $\eta = 0.90$ , and how many $525\,\mathrm{W}$ -average households it supplies.

Solution

Solution of Exercise 9.10.

1. $P = 1.5 \times 10^{5} \times 9.81 \times 40 \approx 5.9 \times 10^{7}\,\mathrm{W} = 59\,\mathrm{MW}$ . 2. $0.90 \times 59 \approx 53\,\mathrm{MW}$ ; $5.3 \times 10^{7}/525 \approx 1.0 \times 10^{5}$ households.

Exercise 9.11 ★★

A $0.20\,\mathrm{kg}$ pendulum is released $5.0\,\mathrm{cm}$ above its lowest point: compute its speed there. After a few minutes it hangs still: where did its energy go, and why does this not contradict Theorem 9.9?

Solution

Solution of Exercise 9.11.

$v = \sqrt{2 \times 9.81 \times 0.050} \approx 0.99\,\mathrm{m}/\mathrm{s}$ (the mass cancels: $\tfrac12 m v^2 = mgh$ ). The initial $E_p = 0.20 \times 9.81 \times 0.050 \approx 0.098\,\mathrm{J}$ ends as thermal energy in the air and the pivot: the pendulum is not isolated, and the total — swing plus heat — is exactly conserved.

Exercise 9.12 ★★★

A coal plant ( $\eta_1 = 0.38$ ), the grid ( $\eta_2 = 0.94$ ) and a charger ( $\eta_3 = 0.85$ ) recharge a $12\,\mathrm{W}\,\mathrm{h}$ phone battery. Compute the overall efficiency; the chemical energy consumed at the plant, in $\mathrm{W}\,\mathrm{h}$ and kilojoules; and the coal ( $25\,\mathrm{MJ}/\mathrm{kg}$ ) per charge, then per year of daily charges.

Solution

Solution of Exercise 9.12.

1. $\eta = 0.38 \times 0.94 \times 0.85 \approx 0.304$ ; consumed $12/0.304 \approx 39.5\,\mathrm{W}\,\mathrm{h} = 39.5 \times 3600 \approx 142\,\mathrm{kJ}$ . 2. $1.42 \times 10^{5}/2.5 \times 10^{7} \approx 5.7\,\mathrm{g}$ of coal per charge; $\times 365 \approx 2.1\,\mathrm{kg}$ per year.

Exercise 9.13 ★★★

A lightning bolt carries about $5\,\mathrm{GJ}$ . Convert to kilowatt-hours and to days of a $12.6\,\mathrm{kW}\,\mathrm{h}$ -per-day household. It is delivered in about $100\,\text{µ}\mathrm{s}$ : compute the power, compare it with humanity’s average electric power ( $3\,\mathrm{TW}$ ), and explain why nobody harvests lightning.

Solution

Solution of Exercise 9.13.

1. $5 \times 10^{9}/3.6 \times 10^{6} \approx 1400\,\mathrm{kW}\,\mathrm{h}$ ; $1400/12.6 \approx 110$ days. 2. $P = 5 \times 10^{9}/1 \times 10^{-4} = 5 \times 10^{13}\,\mathrm{W} = 50\,\mathrm{TW}$ — seventeen times humanity’s average electric power, for $100\,\text{µ}\mathrm{s}$ , at an unpredictable place. Lightning is much power but little energy, delivered uselessly fast: no converter accepts such a spike.

Exercise 9.14 ★★★

A $1300\,\mathrm{kg}$ car brakes from $110\,\mathrm{km}/\mathrm{h}$ ( $30.6\,\mathrm{m}/\mathrm{s}$ ) to rest. Compute the kinetic energy dissipated; what form does it take, and where? An electric car recovers $60\%$ into its battery: energy per stop, and stops needed to refill one kilowatt-hour?

Solution

Solution of Exercise 9.14.

1. $E_k = \tfrac12 \times 1300 \times 30.6^2 \approx 6.1 \times 10^{5}\,\mathrm{J}$ , converted to thermal energy in the brake discs and pads (then the air). 2. $0.60 \times 6.1 \times 10^{5}\,\mathrm{J} \approx 3.7 \times 10^{5}\,\mathrm{J}$ per stop; $3.6 \times 10^{6}/3.7 \times 10^{5} \approx 10$ stops per kilowatt-hour.

Exercise 9.15 ★★★

A cyclist and bicycle ( $85\,\mathrm{kg}$ in all) climb a mountain pass, gaining $1200\,\mathrm{m}$ of altitude; muscles convert food with $\eta \approx 0.25$ . Compute the mechanical energy of the climb, the food energy it demands, the equivalent in $1.5\,\mathrm{MJ}$ cereal bowls, and the gasoline mass ( $45\,\mathrm{MJ}/\mathrm{kg}$ ) storing the same energy. Comment.

Solution

Solution of Exercise 9.15.

$E_p = 85 \times 9.81 \times 1200 \approx 1.0\,\mathrm{MJ}$ ; food: $1.0/0.25 = 4.0\,\mathrm{MJ}$ , i.e. $4.0/1.5 \approx 2.7$ cereal bowls — or $4.0/45 \approx 0.09\,\mathrm{kg}$ of gasoline. A whole morning of honest effort fits in a coffee cup of fuel: chemical energy is absurdly concentrated.

9.7 Problem: Powering a house for a day

Problem 9.1

Weekend problem — from cereal bowl to rooftop panels: a full energy audit of one ordinary day at home, and what a joule costs depending on who sells it

One ordinary Tuesday, a family sets out to find where its electricity goes, armed with this chapter, the hallway meter and the back of the bill. Tonight’s verdict: their day equals one kilogram of gasoline, sixteen square metres of sunshine — or five cyclists pedaling around the clock.

Part I — The audit. The day’s inventory: fridge $100\,\mathrm{W}$ for $24\,\mathrm{h}$ ; water heater $2400\,\mathrm{W}$ for $2.5\,\mathrm{h}$ ; oven $2000\,\mathrm{W}$ for $1.0\,\mathrm{h}$ ; washing machine $500\,\mathrm{W}$ for $2.0\,\mathrm{h}$ ; lighting $60\,\mathrm{W}$ for $5.0\,\mathrm{h}$ ; screens $150\,\mathrm{W}$ for $6.0\,\mathrm{h}$ .

Compute each appliance’s energy for the day, in $\mathrm{kW}\,\mathrm{h}$ .
Total the day in kilowatt-hours, then in megajoules; compare with one kilogram of gasoline ( $45\,\mathrm{MJ}$ ).
Compute the house’s average power over the $24\,\mathrm{h}$ ; what fraction of a running kettle is that?
Which single appliance dominates? Recompute the daily total if the heater ran only $1.25\,\mathrm{h}$ .
Standby lights draw a permanent $15\,\mathrm{W}$ : daily energy, share of the total, yearly cost at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ ?

Part II — The bill.

In 30 days the meter went from $56\,214\,\mathrm{kW}\,\mathrm{h}$ to $56\,604\,\mathrm{kW}\,\mathrm{h}$ : consumption, and daily average checked against the audit?
Compute the bill: $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ plus a fixed $14$ euros subscription.
Price of a megajoule from the grid? From gasoline, at $1.80$ euros per litre storing $34\,\mathrm{MJ}$ ?
A $1.5\,\mathrm{MJ}$ cereal bowl costs about $0.50$ euros: its megajoule price? Rank the three sellers of joules.

Part III — The rooftop.

Full sunlight delivers about $1000\,\mathrm{W}$ per square metre; panels convert it with $\eta = 0.20$ : electrical power of $1\,\mathrm{m}^{2}$ in full sun?
The region averages the equivalent of $4.0\,\mathrm{h}$ of full sun per day: daily yield of $1\,\mathrm{m}^{2}$ ?
What panel area covers the family’s $12.6\,\mathrm{kW}\,\mathrm{h}$ day? The south roof offers $20\,\mathrm{m}^{2}$ : does it fit?
In December the equivalent drops to $1.5\,\mathrm{h}$ : the panels’ production, its share of the need, and what fills the gap?
Manufacturing the panels cost about $700\,\mathrm{kW}\,\mathrm{h}$ per square metre: energy payback time, using question 11?

Part IV — The cereal bowl.

A generator bicycle yields $100\,\mathrm{W}$ of electricity: how many pedaling hours make $12.6\,\mathrm{kW}\,\mathrm{h}$ — how many people pedaling $24\,\mathrm{h}$ non-stop?
Muscles run at $\eta \approx 0.25$ : the food energy behind those $12.6\,\mathrm{kW}\,\mathrm{h}$ , in megajoules and in cereal bowls?
Cost of the pedaled day (bowls at $0.50$ euros), against what the grid charges for the same day?
A $5\,\mathrm{GJ}$ lightning bolt equals how many house-days? Given its $100\,\text{µ}\mathrm{s}$ duration, why can it still not power the house?
Joules are never lost, yet grid, gasoline and food joules sell at very different prices: what do you actually pay for?
Finale: state the day’s verdict in one sentence — one kilogram of gasoline, sixteen square metres of sunny roof or five round-the-clock cyclists — and name the two quantities (one stock, one flow) never to confuse again.

Solution

Solution of Problem 9.1.

1. Fridge $0.1 \times 24 = 2.4\,\mathrm{kW}\,\mathrm{h}$ ; heater $2.4 \times 2.5 = 6.0\,\mathrm{kW}\,\mathrm{h}$ ; oven $2.0\,\mathrm{kW}\,\mathrm{h}$ ; washing machine $1.0\,\mathrm{kW}\,\mathrm{h}$ ; lighting $0.3\,\mathrm{kW}\,\mathrm{h}$ ; screens $0.9\,\mathrm{kW}\,\mathrm{h}$ . 2. Total $12.6\,\mathrm{kW}\,\mathrm{h}$ $= 12.6 \times 3.6 \approx 45\,\mathrm{MJ}$ : one kilogram of gasoline, to the joule. 3. $12\,600\,\mathrm{W}\,\mathrm{h}/24\,\mathrm{h} = 525\,\mathrm{W}$ — the house idles at a quarter of a kettle. 4. The water heater ( $6.0\,\mathrm{kW}\,\mathrm{h}$ , nearly half). At $1.25\,\mathrm{h}$ : heater $3.0\,\mathrm{kW}\,\mathrm{h}$ , total $9.6\,\mathrm{kW}\,\mathrm{h}$ . 5. $15 \times 24 = 0.36\,\mathrm{kW}\,\mathrm{h}$ per day, about $2.8\%$ of the metered total; yearly $0.36 \times 365 \approx 131\,\mathrm{kW}\,\mathrm{h} \approx 33$ euros — for lights nobody watches. 6. $56604 - 56214 = 390\,\mathrm{kW}\,\mathrm{h}$ ; $390/30 = 13.0\,\mathrm{kW}\,\mathrm{h}$ per day: the audit’s $12.6\,\mathrm{kW}\,\mathrm{h}$ plus the $0.36\,\mathrm{kW}\,\mathrm{h}$ of standby — confirmed. 7. $390 \times 0.25 + 14 = 111.50$ euros. 8. Grid: $0.25/3.6 \approx 0.07$ euros per megajoule. Gasoline: $1.80/34 \approx 0.05$ euros per megajoule. 9. Bowl: $0.50/1.5 \approx 0.33$ euros per megajoule — five times the grid. Cheapest first: gasoline, grid, then food, far behind. 10. $0.20 \times 1000 = 200\,\mathrm{W}$ per square metre. 11. $200 \times 4.0 = 800\,\mathrm{W}\,\mathrm{h} = 0.80\,\mathrm{kW}\,\mathrm{h}$ per square metre per day. 12. $12.6/0.80 = 15.75 \approx 16\,\mathrm{m}^{2}$ : it fits the $20\,\mathrm{m}^{2}$ roof. 13. $16 \times 0.2 \times 1.5 = 4.8\,\mathrm{kW}\,\mathrm{h}$ : about $38\%$ of the need; the grid (or a battery charged in better months) must supply the rest. 14. One square metre yields $0.80 \times 365 \approx 292\,\mathrm{kW}\,\mathrm{h}$ per year; payback $700/292 \approx 2.4$ years — short against a 25-year panel lifetime. 15. $12.6\,\mathrm{kW}\,\mathrm{h}/0.1\,\mathrm{kW} = 126\,\mathrm{h}$ : more than five people ( $126/24 = 5.25$ ) pedaling day and night. 16. Food $= 12.6/0.25 = 50.4\,\mathrm{kW}\,\mathrm{h} \approx 181\,\mathrm{MJ}$ , i.e. $181/1.5 \approx 121$ cereal bowls. 17. $121 \times 0.50 \approx 60$ euros, against $12.6 \times 0.25 \approx 3.15$ euros from the grid: pedal power costs about $19$ times more — before paying the cyclists. 18. $5000/45 \approx 110$ house-days. But $5 \times 10^{9}/1 \times 10^{-4} = 5 \times 10^{13}\,\mathrm{W}$ for $100\,\text{µ}\mathrm{s}$ , at a random place: no converter or battery can drink from that hose. A house needs a steady $525\,\mathrm{W}$ flow, not a spike. 19. Not joules — they are conserved and identical. You pay for their form and availability: energy that is concentrated, storable and on demand costs the whole conversion chain that made it so. 20. One family day $= 12.6\,\mathrm{kW}\,\mathrm{h} \approx 45\,\mathrm{MJ}$ : one kilogram of gasoline, sixteen square metres of sunny roof, or five round-the-clock cyclists. The stock is energy (joules, kilowatt-hours); the flow is power (watts) — the bathtub and the tap, never again confused.