Temperature – From The Smallest To The Largest

For temperature there is a definite and incontrovertible lower limit: 0 K. Among the closest things to absolute zero in the universe is the temperature of supermassive black holes (10-18 K). At this temperature it will take them 10100 years and more to evaporate their mass. Yes, that’s a one with one-hundred zeros. If the universe really does keep on expanding as believed by most scientist today, supermassive black holes will be the last remaining objects in the fading universe. Compared to their temperature, the lowest temperature ever achieved in a laboratory (10-12 K) is a true hellfire, despite it being many orders of magnitudes lower than the background temperature of the universe (2.73 K and slowly decreasing).

In terms of temperature, helium is an exceptional element. The fact that we almost always find it in the gaseous state is a result of its low boiling point (4.22 K). Even on Uranus (53 K), since the downgrading of Pluto the coldest planet in the solar system and by far the planet with the most inappropriate name, it would appear as a gas. Another temperature you definitely should remember is 92 K. Why? Because at this temperature the material Y-Ba-Cu-oxide becomes superconductive and there is no material known to man that is superconductive at higher temperatures. Note that you want a superconductor to do what it does best at temperatures as close to room temperature as possible because otherwise making use of this effect will require enormous amounts of energy for cooling.

The lowest officially recorded air temperature on Earth is 184 K ≈ -89 °C, so measured in 1983 in Stántsiya Vostók, Antarctica. Just recently scientists reported seeing an even lower temperature, but at the time of writing this is still unconfirmed. The next two values are very familiar to you: the melting point (273 K ≈ 0 °C) and the boiling point (373 K ≈ 100 °C) of water. But I would not advise you to become too familiar with burning wood (1170 K ≈ 900 °C) or the surface of our Sun (5780 K ≈ 5500 °C).

Temperatures in a lightning channel can go far beyond that, up to about 28,000 K. This was topped on August 6, 1945, when the atomic bomb “Little Boy” was dropped on Hiroshima. It is estimated that at a distance of 17 meters from the center of the blast the temperature rose to 300,000 K. Later and more powerful models of the atomic bomb even went past the temperature of the solar wind (800,000 K).

If you are disappointed about the relatively low surface temperature of the sun, keep in mind that this is the coldest part of the sun. In the corona surrounding it, temperatures can reach 10 million K, the center of the Sun is estimated to be at 16 million K and solar flares can be as hot as 100 million K. Surprisingly, mankind managed to top that. The plasma in the experimental Tokamak Fusion Test Reactor was recorded at mind-blowing 530 million K. Except for supernova explosions (10 billion K) and infant neutron stars (1 trillion K), there’s not much beyond that.

The Jeans Mass, or: How are stars born?

No, this has nothing to do with pants. The Jeans mass is a concept used in astrophysics and its unlikely name comes from the British physicist Sir James Jeans, who researched the conditions of star formation. The question at the core is: under what circumstances will a dark and lonely gas cloud floating somewhere in the depth of space turn into a shining star? To answer this, we have to understand what forces are at work.

One obvious factor is gravitation. It will always work towards contracting the gas cloud. If no other forces were present, it would lead the cloud to collapse into a single point. The temperature of the cloud however provides an opposite push. It “equips” the molecules of the cloud with kinetic energy (energy of motion) and given a high enough temperature, the kinetic energy would be sufficient for the molecules to simply fly off into space, never to be seen again.

It is clear that no star will form if the cloud expands and falls apart. Only when gravity wins this battle of inward and outward push can a stable star result. Sir James Jeans did the math and found that it all boils down to one parameter, the Jeans mass. If the actual mass of the interstellar cloud is larger than this critical mass, it will contract and stellar formation occurs. If on the other hand the actual mass is smaller, the cloud will simply dissipate.

The Jeans mass depends mainly on the temperature T (in K) and density D (in kg/m³) of the cloud. The higher the temperature, the larger the Jeans mass will be. This is in line with our previous discussion. When the temperature is high, a larger amount of mass is necessary to overcome the thermal outward push. The value of the Jeans mass M (in kg) can be estimated from this equation:

M ≈ 1020 · sqrt(T³ / D)

Typical values for the temperature and density of interstellar clouds are T = 10 K and D = 10-22 kg/m³. This leads to a Jeans mass of M = 1.4 · 1032 kg. Note that the critical mass turns out to be much greater than the mass of a typical star, indicating that stars generally form in clusters. Rather than the cloud contracting into a single star, which is the picture you probably had in your mind during this discussion, it will fragment at some point during the contraction and form multiple stars. So stars always have brothers and sisters.

(This was an excerpt from the Kindle book Physics! In Quantities and Examples)

Estimating Temperature Using Cricket Chirps

I stumbled upon a truly great formula on the GLOBE scientists’ blog. It allows you to compute the ambient air temperature from the number of cricket chirps in a fixed time interval and this with a surprising accuracy. The idea actually quite old, it dates back 1898 when the physicist Dalbear first analyzed this relationship, and has been revived from time to time ever since.

Here’s how it works: count the number of chirps N over 13 seconds. Add 40 to that and you got the outside temperature T in Fahrenheit.

T = N + 40

From the picture below you can see that the fit is really good. The error seems to be plus / minus 6 % at most in the range from 50 to 80 °F.

Computing the Surface Area of a Person – Mosteller Formula

While doing research for my new book “More Great Formulas Explained”, I came across a neat formula that can be used to calculate the surface area of a person. It goes by the name Mosteller formula and requires two inputs: the mass m (in kg) and the height h (in cm). The surface area S (in m²) is proportional to the square root of m times h:

S = sqrt (m * h / 3600)

For example, a person with the weight m = 75 kg and height h = 175 cm can be expected to have the body surface area S = 1.91 m². A note for American readers: you can use this table to easily convert the height in feet / inches to centimeters.

What’s the use of this? In my book I needed to know this quantity to compute heat loss. According to Newton’s law of cooling, the heat loss rate P (in Watt = Joules per second) is proportional to the surface area S and the temperature difference ΔT (in °C or K):

P = a * S *ΔT

with a being the so called heat transfer coefficient. For calm air it has the value a = 10 W/(m² * K). A person’s body temperature is around 37 °C. So the m = 75 kg and h = 175 cm person from above would lose this amount of heat every second at an air temperature of 20 °C:

P = 10 W/(m² * K) * 1.91 m² * 17 °C = 325 Watt

That is of course assuming the person is naked, clothing will reduce this value significantly. So the surface area formula indeed is useful.

Physics (And The Formula That Got Me Hooked)

A long time ago, in my teen years, this was the formula that got me hooked on physics. Why? I can’t say for sure. I guess I was very surprised that you could calculate something like this so easily. So with some nostalgia, I present another great formula from the field of physics. It will be a continuation of and a last section on energy.

To heat something, you need a certain amount of energy E (in J). How much exactly? To compute this we require three inputs: the mass m (in kg) of the object we want to heat, the temperature difference T (in °C) between initial and final state and the so called specific heat c (in J per kg °C) of the material that is heated. The relationship is quite simple:

E = c · m · T

If you double any of the input quantities, the energy required for heating will double as well. A very helpful addition to problems involving heating is this formula:

E = P · t

with P (in watt = W = J/s) being the power of the device that delivers heat and t (in s) the duration of the heat delivery.


The specific heat of water is c = 4200 J per kg °C. How much energy do you need to heat m = 1 kg of water from room temperature (20 °C) to its boiling point (100 °C)? Note that the temperature difference between initial and final state is T = 80 °C. So we have all the quantities we need.

E = 4200 · 1 · 80 = 336,000 J

Additional question: How long will it take a water heater with an output of 2000 W to accomplish this? Let’s set up an equation for this using the second formula:

336,000 = 2000 · t

t ≈ 168 s ≈ 3 minutes


We put m = 1 kg of water (c = 4200 J per kg °C) in one container and m = 1 kg of sand (c = 290 J per kg °C) in another next to it. This will serve as an artificial beach. Using a heater we add 10,000 J of heat to each container. By what temperature will the water and the sand be raised?

Let’s turn to the water. From the given data and the great formula we can set up this equation:

10,000 = 4200 · 1 · T

T ≈ 2.4 °C

So the water temperature will be raised by 2.4 °C. What about the sand? It also receives 10,000 J.

10,000 = 290 · 1 · T

T ≈ 34.5 °C

So sand (or any ground in general) will heat up much stronger than water. In other words: the temperature of ground reacts quite strongly to changes in energy input while water is rather sluggish. This explains why the climate near oceans is milder than inland, that is, why the summers are less hot and the winters less cold. The water efficiently dampens the changes in temperature.

It also explains the land-sea-breeze phenomenon (seen in the image below). During the day, the sun’s energy will cause the ground to be hotter than the water. The air above the ground rises, leading to cooler air flowing from the ocean to the land. At night, due to the lack of the sun’s power, the situation reverses. The ground cools off quickly and now it’s the air above the water that rises.


I hope this formula got you hooked as well. It’s simple, useful and can explain quite a lot of physics at the same time. It doesn’t get any better than this. Now it’s time to leave the concept of energy and turn to other topics.

This was an excerpt from my Kindle ebook: Great Formulas Explained – Physics, Mathematics, Economics. For another interesting physics quicky, check out: Intensity (or: How Much Power Will Burst Your Eardrums?).