We cover some basics in this post. The subject was inspired by one commenter on the blog.
- When we look at a “radiative forcing” what does it mean?
- What immediate and long-term impact does it have on temperature?
- What is the new equilibrium temperature?
The IPCC, drawing on the work of many physicists over the years, states that the radiative forcing from the increase in CO2 to about 380ppm is 1.7 W/m2. You can see how this is all worked out in the series CO2 – An Insignificant Trace Gas.
What is “radiative forcing”? At the top of atmosphere (TOA) there is an effective downward increase in radiation. So more energy reaches the surface than before..
If you put very cold water in a pot and heat it on a stove, what happens? Let’s think about the situation if the water doesn’t boil because we don’t apply so much heat..
I used simple concepts here.
T= water temperature and the starting temperature of the water, T (t=0) = 5°C
Air temperature, T1 = 5°C
Energy in per second = constant (=1000W in this example)
Energy out per second = h x (T – T1), where h is just a constant (h=20 in this example)
And the equation for temperature increase is:
Energy per second, Q = mc.ΔT
m = mass, and c= specific heat capacity (how much heat is required to raise 1kg of that material by 1’C) – for water this is 4,200 J kg-1 K-1. I used 1kg.
ΔT is change in temperature (and because we have energy per second the result is change in temperature per second)
The simple and obvious points that we all know are:
- the liquid doesn’t immediately jump to its final temperature
- as the liquid gets closer to its final temperature the rate of temperature rise slows down
- as the temperature of the liquid increases it radiates or conducts or convects more energy out, so there will be a new equilibrium temperature reached
In this case, the heat calculation is by some kind of simple conduction process. And is linearly proportional to the temperature difference between the water and the air.
It’s not a real world case but is fairly close – as always, simplifying helps us focus on the key points.
What might be less obvious until attention is drawn to it (then it is obvious) – the final temperature doesn’t depend on the heat capacity of the liquid. That only affects how long it takes to reach its equilibrium – whatever that equilibrium happens to be.
Heating the World
Suppose we take the radiative forcing of 1.7W/m2 and heat the oceans. The oceans are the major store of the climate system’s heat, around 1000x more energy stored than in the atmosphere. We’ll ignore the melting of ice which is a significant absorber of energy.
Ocean mean depth = 4km (4000m) – the average around the world
Only 70% of the earth’s surface is covered by ocean and we are going to assume that all of the energy goes into the oceans so we need to “scale up” – energy into the oceans = 1.7/0.7 = 2.4 W/m2 going into the oceans.
The density of ocean water is approximately 1000 kg/m3 (it’s actually a little more because of salinity and pressure..)
Each square meter of ocean has a volume of 4000 m3 (thinking about a big vertical column of water), and therefore a mass of 4×106 kg.
Q = mc x dT
Q is energy, m is mass, c is specific heat capacity = 4.2 kJ kg-1 K-1,
dT = change in temperature
We have energy per second (W/m2), so change in temperature per second, dT = Q/mc
dT per second = 2.4 / (4×106 x 4.2×103)
= 1.4 x10-10 °C/second
dT per year = 0.004 °C/yr
That’s really small! It would take 250 years to heat the oceans by 1°C..
Let’s suppose – more realistically – that only the top “well-mixed” 100m of ocean receives this heat, so we would get (just scaling by 4000m/100m):
dT per year = 0.18 ‘C per year.
An interesting result, which of course, ignores the increase in heat lost due to increased radiation, and ignores the heat lost to the lower part of the ocean through conduction.
If we took this result and plotted it on a graph the temperatures would just keep going up!
Calculating the new Equilibrium Temperature
The climate is slightly complicated. How do we work out the new equilibrium temperature?
Do we think about the heat lost from the surface of the oceans into the atmosphere through conduction, convection and radiation? Then what happens to it in the atmosphere? Sounds tricky..
Fortunately, we can take a very simple view of planet earth and say energy in = energy out. This is the “billiard ball” model of the climate, and you can see it explained in CO2 – An Insignificant Trace Gas – Part One and subsequent posts.
What this great and simple model lets us do is compare energy in and out at the top of atmosphere (TOA). Which is why “radiative forcing” from CO2 is “published” at TOA. It helps us get the big picture.
Energy radiated from a body per unit area per second is proportional to T4, where T is temperature in Kelvin (absolute temperature). Energy radiated from the earth has to be balanced by energy we absorb from the sun.
This lets us do a quick comparison, using some approximate numbers.
Energy absorbed from the sun, averaged over the surface of the earth, we’ll call it Pold = 239 W/m2.
Surface temperature, we’ll call it Told = 15°C = 288K
If we add 1.7W/m2 at TOA what does this do to temperature? Well, we can simply divide the old and new values, making the equation slightly easier..
So Tnew=288 x (239+1.7/239)1/4
Therefore, Tnew = 288.5K or 15.5°C – a rise of 0.5°C
I don’t want to claim this represents some kind of complete answer, but just for some element of completeness, if we redo the calculation with the radiative forcing for all of the “greenhouse” gases, excluding water vapor, we have a radiative forcing of 2.4W/m2.
Tnew = 288.7 or 15.7°C – a rise of 0.7°C.
(Note for the purists, I believe the only way to actually calculate the old and new surface temperature is using the complete radiative transfer equations, but the results aren’t so different)
The aim of this post is to clarify a few basics, and in the process we looked at how quickly the oceans might warm as a result of increased radiative forcing from CO2.
It does demonstrate that depending on how well-mixed the oceans are, the warming can be extremely slow (250 years for 1°C rise) or very quick (5 years for 1°C rise).
So from the information presented so far, temperatures we currently experience at the surface might be the new equilibrium from increased CO2, or a long way from it – this post doesn’t address that huge question! Or any feedbacks.
What we ignored in the calculation of temperature rise was the increased energy lost as the temperature rose – which would slow the rise down (like the heated water in the graph). But at least it’s possible to get a starting point.
We can also see a rudimentary calculation of the final increase in temperature – the new equilibrium – as a result of this forcing (we are ignoring any negative or positive feedbacks).
And the new equilibrium doesn’t depend on the thermal lag of the oceans.
Of course, calculations of feedback effects in the real climate might find thermal lag parameters to be extremely important.