Consider the 2 °C average global atmospheric temperature rise as discussed by the IPCC. Assume an oven with an the interior that is a perfect cube having a volume of one cubic meter, empty of everything except air. What would be the temperature inside the oven if the energy represented by said 2 °C global warming were instead applied to the atmosphere inside the oven?

The problem I'm trying to solve is to make climate change understandable to friends and family.

We talk about trying to limit global warming to 1.5 °C (if that's even possible any more) or 2 °C. The problem is that 1.5 °C or even 2 °C doesn't sound so bad on a local level. The temperature goes up and down by that much every day from one sunrise to the next. Some days it's so cold that a 2 °C rise would be welcome. What's wrong with a little warming?

What's wrong is that it just seems like a little warming; it's actually a lot of warming spread out over a large area or volume; it's really quite a bit of energy.

To illustrate the amount, suppose we confined it to the volume of an average domestic kitchen oven, say, a cubic meter to make it easy.

That's my question.

For the brief period of time that the oven would continue to exist.

  • $\begingroup$ The trouble with your idea is that the energy from even a very slight increase in global temperature (say 0.01 C) is enough to blow up your oven. So what are you proving? $\endgroup$ – Stéphane Rollandin Apr 24 '18 at 17:43
  • $\begingroup$ @StéphaneRollandin I think what you said goes to my objective of communicating the large amount of energy. $\endgroup$ – compton Apr 25 '18 at 0:18
  • $\begingroup$ Yes, there is indeed a lot of energy, but since even a small increase in temperature (which would not change anything to the climate) represents already a large amount of energy, it follows that showing that the energy is huge does not prove that the climate is going to change. $\endgroup$ – Stéphane Rollandin Apr 25 '18 at 12:01
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    $\begingroup$ Maybe a better approach is to say that an average 4 degree change is the difference between an ice age and our "usual" climate: xkcd.com/1379 $\endgroup$ – Rococo Apr 28 '18 at 0:14

Let's rearrange the definition of heat capacity $C$ as follows:

$$\Delta T =\frac{Q}{mC}$$

$\Delta T$ = the change in the temperature of the mass (in degrees Celsius)
$Q$ = the heat that causes the rise in temperature (in Joules)
$m$ = the mass that is having its temperature raised (in kg)

We'll next create a ratio of the change in temperature of the atmosphere (abbreviated $atm$) $\Delta T_{atm}$ to the change in temperature in the oven $\Delta T_{oven}$:

$$\frac{\Delta T_{oven}}{\Delta T_{atm}}=\frac{Q_{oven}}{Q_{atm}}\cdot \frac{m_{atm}C_{atm}}{m_{oven}C_{oven}}$$

Since both the atmosphere and the oven contain air, let's assume that their heat capacities are the same so $C_{atm} = C_{oven}$ and those terms cancel out. The assumption is that the heat is also equal so $Q_{oven} = Q_{atm}$ and those cancel out as well. Rearranging, we are left with:

$$\Delta T_{oven} = \Delta T_{atm} \frac{m_{atm}}{m_{oven}}$$

Increase in temperature of atmosphere $\Delta T_{atm}$ = $2 \text{ °C}$
Mass of the atmosphere $m_{atm}$ = $5.15×10^{18}\ kg $
Mass of air in oven $m_{oven}$ = $1.275\ kg$

Plugging in results in $\Delta T_{oven} = 8.08×10^{18}\text{ °C}$. For reference the center of the Sun is only $1.5×10^{6}\text{ °C}$. More massive stars can get up to $1×10^{10}\text{ °C}$. Both are still way colder than your hypothetical oven.


Start with the heat transfer over the entire earth. This simplification uses convection, but you can add conductive and radiative heat transfer if you wanted.

Q=UAdT; where Q is the amount of heat, U is the convective heat transfer coefficient (0.5 - 1000 (W/(m^2K) air), A is area, and dT is the temperature change in kelvin.

After finding Q from the surface area of the earth, use this Q value along with the surface area of the oven to find the temperature change. Add that to the ambient value.


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