More problems that combine several ideas. As before, try each part on your own first, and only then open the two boxes below it: the setup (starting point, goal and conversion information) and the computation.
13.1 digging for digital gold
Bitcoin (BTC) is a digital currency. One bitcoin can be split into a hundred million smaller units, called satoshis (sat), after Satoshi Nakamoto, the person (or group) who invented Bitcoin.
New bitcoins are created in a process called mining, in analogy to gold mining. To mine, computers all over the world repeat a huge number of computations, called hashes (we abbreviate a single hash calculation as H). Mining is a competition, and every ten minutes the winner is rewarded with newly created bitcoins.
Here is the data we’ll need (all of it is correct for October 10, 2020):
1 \text{ BTC} is worth 38{,}300 \text{ NIS} (new Israeli shekels), and 1 \text{ BTC} = 10^8 \text{ sat}.
All the computers in the world that are mining bitcoin, together, calculate 140 \cdot 10^{12} hashes per second.
As a result, 6.25 new bitcoins are created every 10 minutes.
Experts estimate that the electricity used by the mining of bitcoin is about 65 \text{ TWh} per year. For comparison, the electricity consumption of Israel in 2018 was 69 \text{ TWh}.
The price of one kilowatt-hour of electricity is 0.05 \text{ USD}.
1 \text{ USD} = 3.4 \text{ NIS}.
a. how many hashes for one satoshi?
In the process of mining bitcoin, how many hash calculations (H) are needed, on average, to create one satoshi?
Tipsetup: starting point, goal, conversion
We know how many hashes are calculated every second, and how many bitcoins are created in ten minutes. Time is the bridge between the two.
Starting point:140 \cdot 10^{12} hashes every second.
That’s about 34 million shekels of new bitcoin, every single day. It also means that 900 new bitcoins are created every day (the chain says 24 \cdot 60 \cdot 6.25/10 = 900).
c. the electricity bill
How much money, in dollars, did the electricity for bitcoin mining cost the whole world in one year?
Tipsetup: starting point, goal, conversion
The energy is given in terawatt-hours, but the price is per kilowatt-hour, so we need to convert. A watt-hour is a unit of energy: it’s a power (watts) multiplied by a time (hours).
About 3.7 billion dollars of new bitcoin are created every year.
e. how much goes to electricity?
What percentage of the value of the newly created bitcoins is spent on electricity alone?
Tipsetup: starting point, goal, conversion
This is a ratio of two amounts of money that we already know, both per year: the cost of the electricity (part c) and the value that is created (part d). We must make sure that they are in the same unit, dollars, which they are.
Almost nine out of every ten dollars of new bitcoin go to paying for electricity. And this is only the electricity. It doesn’t include the cost of the computers themselves, or of the buildings that hold them.
13.2 galileo’s breath
Galileo, the father of modern physics, died in 1642 at the age of 78. Every time we breathe in, we take in some air molecules that were once in Galileo’s lungs. But how many? Let’s find out. We’ll assume that:
A person breathes 16 times per minute, on average.
The average volume of a breath is 0.5 \text{ L}.
One liter of air contains 2.7 \cdot 10^{22} molecules.
The atmosphere contains 1.0 \cdot 10^{44} molecules (mostly nitrogen and oxygen).
Galileo always breathed in fresh air that had never been in his lungs, and over the centuries, all the air that he breathed out spread evenly over the entire atmosphere. Also, the molecules that Galileo breathed out did not break down or take part in any chemical reactions.
All these assumptions are unrealistic, but they make the calculation much simpler, and we’re only after the order of magnitude.
a. galileo’s breaths
How many molecules did Galileo breathe out during his life?
Tipsetup: starting point, goal, conversion
We start from his age and go step by step: from years to minutes, then from minutes to breaths, then from breaths to liters of air, and finally from liters to molecules.
Along the way, this is about 6.6 \cdot 10^8 breaths (more than half a billion!), which is about 3.3 \cdot 10^8 \text{ L} of air.
b. galileo’s share of the atmosphere
What fraction of the molecules in the atmosphere are molecules that Galileo breathed out?
Tipsetup: starting point, goal, conversion
A fraction is the part divided by the whole. The part is the number of molecules from part a, and the whole is the number of molecules in the atmosphere. Both are just numbers of molecules, so the units cancel out and the result has no units.
That’s a tiny number: less than one in ten trillion. In the language of concentration, it’s about 0.09 parts per trillion.
c. a breath of galileo
You take one breath (an average one). How many of the molecules that you now have in your lungs were once in Galileo’s lungs?
Tipsetup: starting point, goal, conversion
Because the air is evenly mixed, every sample of air has the same fraction of Galileo’s molecules: the fraction that we found in part b. So we count how many molecules there are in one breath, and then we take the fraction of Galileo’s.
Starting point:1 breath.
Goal: the number of Galileo’s molecules in it.
Conversion information:
1 \text{ breath} \longleftrightarrow 0.5 \text{ L of air} \\
1 \text{ L of air} \longleftrightarrow 2.7 \cdot 10^{22} \text{ molecules} \\
\text{a fraction } 8.9 \cdot 10^{-14} \text{ of the molecules are Galileo's}
So in every breath that you take, there are about a billion molecules that Galileo once breathed out. Notice the two sides of this result: a billion is a huge number, but a single breath has 1.35 \cdot 10^{22} molecules, so Galileo’s molecules are only about one in ten trillion of them.
The fraction 8.9 \cdot 10^{-14} has no units, so multiplying by it changes the number of molecules, but not the unit.
13.3 the sverdrup
The sverdrup (Sv) is a unit of flow rate, named after the Norwegian oceanographer Harald Ulrik Sverdrup. It is used mainly to talk about ocean currents, and it is defined as one million cubic meters of water per second:
1 \text{ Sv} = 10^6 \text{ m}^3 \text{ s}^{-1}
A flow rate tells us what volume of water passes through a section of the current or pipe in every second. For example, the flow of the Gulf Stream as it leaves the Gulf of Mexico is 30 \text{ Sv}, while all the rivers of the world together bring to the seas only 1.2 \text{ Sv}.
a. water in a pipe
Water flows at a speed of 0.5 \text{ m/s} in a cylindrical pipe with a radius of 3 \text{ cm}. What is the flow rate? Give the answer in sverdrups.
Tipsetup: starting point, goal, conversion
A speed of 0.5 \text{ m/s} means that in every second the water advances 0.5 \text{ m} along the pipe. How much water is in 1 \text{ m} of pipe? That’s a cylinder of length 1 \text{ m} and radius 3 \text{ cm}. This is our conversion factor from length of pipe to volume of water.
Starting point:0.5 \text{ m} of pipe every second.
A pipe of this size carries only about a billionth of a sverdrup.
In general, this is how the flow rate is calculated: it is the cross-section area of the flow multiplied by its speed,
Q = A v
which is the same idea as \Delta x = v \Delta t, with the area turning a length into a volume.
b. filling the oceans with rivers
How many years would we have to wait in order to collect, from all the rivers of the world, a volume of water equal to the volume of all the oceans? The volume of the oceans is 1.3 billion cubic kilometers.
Tipsetup: starting point, goal, conversion
Starting point:1.3 \cdot 10^{9} \text{ km}^3 of water.
The first line needs care: a cubic kilometer is a cube with sides of 1 \text{ km} = 10^3 \text{ m}, so its volume is (10^3 \text{ m})^3 = 10^9 \text{ m}^3.
We’d have to wait about 34{,}000 years. For comparison, the Gulf Stream, at 30 \text{ Sv}, is 25 times stronger than all the rivers together, and would fill the oceans in about 1400 years.
c. how fast is the gulf stream?
What is the average speed of the Gulf Stream as it flows between Florida and Cuba? Assume that its width is 100 \text{ km} and its depth is 1000 \text{ m}. Give the answer in meters per second.
Tipsetup: starting point, goal, conversion
We use Q = A v from part a, but this time we know the flow rate and want the speed: v = Q/A. The area A is the cross-section of the current, a rectangle with a width and a depth. We need to bring everything to meters.
That’s about 1 \text{ km/h}, slower than a walk. It seems slow for such a famous current, but this is the average over the whole cross-section. In reality, the current is much faster in its core and much slower near the edges and at the bottom.
13.4 the world’s energy bill
In 2019, the energy consumption of the whole world was 580 \cdot 10^{18} \text{ J} = 5.8 \cdot 10^{20} \text{ J}. That’s a big number, but how big? In this exercise we’ll compare it with some familiar things. Each part comes with the extra information that you need.
a. hiroshima bombs
The atomic bomb dropped on Hiroshima in 1945 had a power of 15 kilotons of TNT. A ton of TNT is a unit of energy, equal to 4.184 \cdot 10^9 \text{ J}. In terms of energy, how many Hiroshima bombs is the energy that the world consumed in 2019?
That’s more than nine million Hiroshima bombs in a year. In other words, the world consumes the energy of one Hiroshima bomb about every 3.4 seconds.
b. around the world in a jumbo jet
One liter of jet fuel contains 30 \text{ MJ} of energy. A Boeing 747 flies at a speed of 933 \text{ km/h}, and burns 4 \text{ L} of fuel every second. How many times could a Boeing 747 fly around the Earth, using an amount of energy equal to the world’s energy consumption in 2019? The radius of the Earth is 6371 \text{ km}.
Tipsetup: starting point, goal, conversion
The plane burns fuel, and fuel carries energy, so the energy is our bridge to a distance: energy, then liters of fuel, then time, then distance, and finally the number of laps.
About thirty one million laps around the Earth. To do this, the plane would need to fly non-stop for about 1.5 \cdot 10^5 years, that’s about a hundred and fifty thousand years!
c. a solar solution
On average, every square meter of the surface of the Earth receives 342 \text{ J} of solar radiation every second. This is the average over the whole planet, both the lit side and the dark side. What area would we need to cover with solar panels, in order to collect the energy that the world consumed in 2019 within one year? Assume that the panels absorb 100\% of the energy that reaches them. Express the area also as a multiple of the area of Israel, which is about 22{,}000 \text{ km}^2.
Tipsetup: starting point, goal, conversion
We do it in two steps. First we find how much energy one square meter of panel collects in a year. Then we find how many square meters are needed to collect 5.8 \cdot 10^{20} \text{ J}.
Energy collected by 1 \text{ m}^2 in a year
Starting point:1 year.
Goal: the energy that one \text{m}^2 collects, in J.
In square kilometers, that’s about 5.4 \cdot 10^4 \text{ km}^2.
That’s smaller than one might expect. But remember our assumptions: the panels absorb all the energy that reaches them, and they are placed where the average sunlight is like the average of the whole planet. Real solar panels convert only about a fifth of the light into electricity, and the sunlight that reaches the ground is affected by clouds and the atmosphere, so the real area would be several times larger. Still, it’s a small fraction of the land area of the Earth.
13.5 how heavy is a cloud?
A cloud looks light and fluffy, but it is made of water, and water has mass. How much? Let’s find out. Here is the data we’ll need:
One cubic centimeter of a typical cumulus cloud contains between 50 and 500 water droplets, and the typical radius of a droplet is 10 \, \mu\text{m}.
The volume of a sphere with radius r is V = \frac{4}{3}\pi r^3, and the volume of a cylinder with radius R and height h is V = \pi R^2 h.
The density of water is 1000 \text{ kg/m}^3.
Our cloud is a cylinder, 3.0 \text{ km} tall and with a radius of 1.0 \text{ km}. Since the number of droplets is given as a range, our answers will also be ranges: a smaller value, for 50 droplets per \text{cm}^3, and a larger value, for 500.
a. one droplet
What is the volume of a single water droplet, in \text{cm}^3?
Tipsetup: starting point, goal, conversion
Starting point: the radius of the droplet, 10 \, \mu\text{m}.
First we convert the radius to centimeters, and then we use the formula for the volume of a sphere. Remember that the radius is cubed in the formula, so a small change in the radius has a big effect.
How many \text{cm}^3 of liquid water are there in every cubic meter of cloud?
Tipsetup: starting point, goal, conversion
We know the number of droplets in a \text{cm}^3 of cloud, and the volume of each droplet. We multiply them, and then convert from a \text{cm}^3 of cloud to a \text{m}^3 of cloud.
Starting point:50 droplets (or 500) in 1 \text{ cm}^3 of cloud.
Goal:\text{cm}^3 of water in 1 \text{ m}^3 of cloud.
With 500 droplets per \text{cm}^3, the result is ten times larger, 2.1 \frac{\text{cm}^3 \text{ water}}{\text{m}^3 \text{ cloud}}.
So each cubic meter of cloud contains between 0.2 and 2 \text{ cm}^3 of water, that is less than half a teaspoon! The cloud is almost entirely air. This also means that water is only about 0.2 to 2 parts per million (ppm) of the volume of a cloud.
c. the water in the cloud
How many cubic meters of water are there inside the cylindrical cloud?
Tipsetup: starting point, goal, conversion
First we find the volume of the cloud, and then we use the result from part b.
Starting point: the volume of the cloud, V = \pi R^2 h = \pi (1.0 \text{ km})^2 (3.0 \text{ km}).
That’s between 2000 and 20{,}000 tons of water floating over our heads!
For a different perspective, the air in this same cloud, at about 1.2 \text{ kg/m}^3, weighs 1.2 \cdot 9.4 \cdot 10^{9} \approx 1 \cdot 10^{10} \text{ kg}. So the water is just 0.02\% to 0.2\% of the mass of the cloud’s air. Even a cloud that weighs thousands of tons is mostly air.