<p>I'm not sure how accurate you need, but beginning NASA's global 398.2 watts per square meter estimate from their energy budget a rounding down based on temperature variation should give a pretty good estimate.    A lot of people have a problem with that number because it's more energy than we get from the sun, but that's because our atmosphere is a kind of blanket that traps heat, which warms the Earth above it's planetary eqilibrium temperature.</p>

<p>It's also (but less so), higher because Earth is warmed from below as well as above.</p>

<p><a href="https://i.sstatic.net/VPbI4.jpg" rel="noreferrer"><img src="https://i.sstatic.net/VPbI4.jpg" alt="enter image description here"></a></p>

<p><a href="https://en.wikipedia.org/wiki/Diurnal_temperature_variation" rel="noreferrer">Diurnal temperature variation</a> is between highest day temperature and coldest night temperature and you're looking for average.  Using that number won't give a correct answer, but using, perhaps 1/2 that number might be in the range and that's over land.   The diurnal temperature variation over water is much lower.</p>

<p>So, 288 Kelvin Earth, say 12 degree variation day to night, so 294 / 282.  Ratio of 1.0425, to the 4th power means 18% more heat is radiated from land during the day than at night.   Adjust the temperature if you think mine is off.    That's 18% x 29% of the surface area.</p>

<p>For oceans, 71% of the surface area the temperature variation is smaller, say 2.5 degrees.   289.25 / 286.75, ratio of 1.0087.   4th power, means the oceans radiate about 3.5% more energy during the day than at night.</p>

<p>(18% x 29%) + (3.5% x 79%) = about 7.7%</p>

<p>So if Earth's surface radiation global average is about 398.2 w/m^2, then the day/night numbers work out to about 7.7% variation between them, or +/- 15.5 degrees, about 413.7 watts/m^2 during the day and about 382.7 at night.    That's a pretty good estimate I think.   I'm open to a better estimate if someone can think of a way to do it. </p>

CC BY-SA 4.0
2020-02-09T18:14:57.543Z
This diagram with back-radiation is pretty hard to understand for me. Back radiation could only happen for IR capturing gasses (I hate the expression "greenhouse gas" because Earth has nothing to do with a greenhouse with glasses suppressing convection) and since the atmosphere is almost black for these gasses wavelengths (CO2 captures almost 100% 15um in 10-20m from the ground up), no such radiation can cross the atmosphere. This would mean that all the "backradiation" would happen in a thin layer above ground. Quite troubling and in violation with the laws of physics (energy is moving from h
Backradek
CC BY-SA 4.0
2020-02-10T15:07:47.260Z
A physical greenhouse does a few things.   A house suppresses convection but isn't warmer than the outside during the day.  A house with big windows can be warmer by letting light in.    A greenhouse is warmer still for suppressing convection, letting sunlight in and trapping IR radiation.  As for how back radiation works.  I like to think of individual photons.   A photon leaves the Earths surface, it flies generally upwards until it hits a molecule that can absorb it (greenhouse gas), it's absorbed, it makes the molecule dance, and then it's fairly quickly released in a random direction.
userLTK
CC BY-SA 4.0
2020-02-10T15:12:57.000Z
You're thinking of CO2 in the atmosphere as a mirror reflecting back and you're thinking in percentages.   That's kind of what the diagram looks like, but if you think of the photon and molecules that can absorb and release it individually, then it's a lot more like a lot of floating billiard balls moving almost at random and the photon bounces around, shot from molecule to molecule until it escapes into space.   The amount of time it takes the average IR photon to leave the Earth or the average number of collisions is proportional to the amount of heat storage.
userLTK
CC BY-SA 4.0
2020-02-10T15:16:48.670Z
So, more greenhouse gas molecules, more photon storage.   The same thing (kind of) happens inside the sun.  A photon created in the center of the sun by nuclear fusion can take 100,000 years to make it's way to the surface, then just 8 minutes to Earth.   This is, ofcourse, oversimplified as the photons created in the sun are gamma and the photons that leave the sun are visible light and there's some energy conversion on the way and perhaps, splitting a photon into more than one, but the idea is the same.  The photons inside the sun take a long time to get out, so the Sun heats up over time.
userLTK
CC BY-SA 4.0
2020-02-10T15:18:37.960Z
@Backradek The NASA chart just oversimplifies, rather than counting the photons and timing their trip out, it converts to energy in, energy up and energy down, perfectly fine for thermodynamics.   Not quite as good for imagining the precise process.
userLTK
CC BY-SA 4.0
2022-10-19T15:45:20.133Z
@userLTK how could I take this information and find out an answer to the following question? I've been asking myself, if we lowered global power usage by 10GW, distributed uniformly across all energy sources, then what the impact on Earth's temperature would be over 10 years. I'm a mathematician but I know really little about the thermodynamics at play here.
cheater
CC BY-SA 4.0
2022-10-19T22:37:39.983Z
@Cheater This website doesn't like questions to be asked in comments.  You should probably post a new question.   That has the added advantage of having people much smarter than I am see your question as well.     I can say that the heat trapping from the added CO2 far exceeds any heat generated by human energy generation, by about a factor of about 10 to 1.  But that's CO2 added not just this year, but all prior years, and CO2 lasts in the atmosphere a long time, so the math gets tricky because it's not apples to apples.    CO2 added 20 or 30 years ago is still warming the Earth.
userLTK
CC BY-SA 4.0
2020-02-09T11:22:35.053Z
2020-02-09T11:22:35.053Z
userLTK
<p>I am finding lots of information about Earth's energy budget, about how objects emit infrared radiation based on heat, and lots of related things, but nothing helping me figure out how much energy is actually being emitted by the ground at night.</p>

<p>At night, the ground will still be emitting electromagnetic energy. The amount of energy being emitted will vary based on many factors. Information that helps with many different types of land in various conditions would be great, but for my case I'm focusing on a stereotypical grassy plain in an area which has been between 5C-40C (40F-100F) for the past weeks.</p>

<p>Ultimately I'd like to have a rough ballpark figure of watts per square meter.</p>

<p>In response to @Michael's comment:</p>

<blockquote>
  <p>The amount of energy emitted by night depends on how much it receives during the day, water content, length and type of grass and other vegetation, how many "past weeks" have there been, and what was the temperature of the ground when the relevant period began.</p>
</blockquote>

<p>During a sunny day the ground is said to receive approximately 1 kilowatt per square meter.</p>

<p>Assume grass of 10cm-50cm (about a half foot to a couple feet). There have been thousands of past weeks. Since we need to get picky let's say the temperature has been a constant 10C (50F) every hour of every night and 15C (60F) every hour of every day for the last thousand years except for the 1 or 2 hours day/night transition where the temperature adjusts linearly back and forth if that makes the exercise easier. And the average annual rainfall of this location is 100cm (40in), rain was coming regularly at that rate until 4 days ago. In the last 4 days, it has rained once, yesterday afternoon dropping about 1cm (half-inch), and the water table is generally about 5 feet under ground level.</p>

<p>The dirt is completely covered and appears as you would expect if you Googled an image of a grassy meadow. For example, the picture I've added at the bottom of the question.</p>

<p>In my opinion, however, the best answer would not just say "The ground emits 10 watts per square meter under those exact conditions!" (even that would be useful though) but would instead say "The type of grass actually makes a significant difference that can throw off even a ballpark estimate, as the difference between grass type A and grass type B produces a whopping half-order-of-magnitude difference in final results due to the way their outer material insulates them! And the difference between wet grass and dry grass is actually 2 orders of magnitude! If I were to simply pick a random, reasonable scenario of ABC it would be within the order of magnitude of 0.1 to 1 watts per square meter, and that's on the high end."</p>

<p>For anything else that matters feel free to simply pick a reasonable value and mention (even if only in a few words) why it matters and if your pick produces a high, low, or mid-range estimate.</p>

<p><a href="https://i.sstatic.net/4WZi6.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4WZi6.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2020-02-05T17:11:38.933Z
You'll have to be much more specific than that. The amount of energy emitted by night depends on how much it receives during the day, water content, length and type of grass and other vegetation, how many "past weeks" have there been, and what was the temperature of the ground when the relevant period began.
Michael Walsby
CC BY-SA 4.0
2020-02-05T17:48:50.917Z
This article might be interesting for you: https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1002/2014JD022216
samcarter_is_at_topanswers.xyz
CC BY-SA 4.0
2020-02-05T17:55:00.633Z
To get some rough starting point maybe take the temperatures from https://neo.sci.gsfc.nasa.gov/view.php?datasetId=MOD_LSTN_M and assume a black body spectrum?
samcarter_is_at_topanswers.xyz
CC BY-SA 4.0
2020-02-05T18:00:45.470Z
Based on @MichaelWalsby comment, if those thing all really matter so much, I'm wondering if it would be better to start with separate questions like "How much does vegetation water content affect nighttime IR energy output? All other thing being equal, how wide a range does this variable encompass? What are some reasonable estimates to use for high/low/mid range values for this variable?" This could get out of hand though with a bunch of "How does A affect X?" "How does B affect X?" etc..
Aaron
CC BY-SA 4.0
2020-02-05T20:25:05.310Z
As a rough global ballpark figure, it's equal to incoming radiation (ignoring the 0.1 W/m² difference that is climate change)
gerrit
CC BY-SA 4.0
2020-02-05T21:01:06.620Z
To clarify: do you mean radiation emitted by the Earth *surface*, or radiation as observed at the top of the atmosphere (TOA) which is what actually matters for the energy budget?  Most radiation emitted by the Earth surface is immediately absorbed by the atmosphere, then re-radiated by layers colder than the surface; this is exactly what the greenhouse effect is.  Earth surface radiation is impossible to measure from space, but not immediately relevant for the energy budget.  Do you want *surface* radiation or *top of atmosphere* radiation?
gerrit
CC BY-SA 4.0
2020-02-05T21:21:08.270Z
@gerrit Emitter at surface. My specific use-case interest involves very low altitude, as you would see it facing the ground from no higher than an average building or tree, possibly even closer.
Aaron
CC BY-SA 4.0
2020-02-05T21:26:24.403Z
@gerrit "Not immediately relevant for the energy budget." I only mentioned the energy budget because that's the topic that keeps coming up in the research I'm attempting. Personally, and for my specific purposes, the energy budget itself is irrelevant. That just seemed the right way to ask about it based on what little I've found. Basically, what I'm after is if you could hold a 1-meter square "IR meter" above the ground what would the power reading be. That was the end-goal I had in mind when asking this question.
Aaron
CC BY-SA 4.0
2020-02-05T21:45:43.637Z
I had misunderstood the question, and I have deleted my answer, which was for TOA and not for surface.  I don't know the answer for the surface, I hope you'll get a good answer.
gerrit
CC BY-SA 4.0
2020-02-05T21:58:43.297Z
As long as it's not too difficult to measure I don't mind getting my hands dirty, so I've posted a related question on engineering SE about how to take the measurement myself: https://engineering.stackexchange.com/questions/33095/measuring-em-radiation-power-over-surface-area
Aaron
CC BY-SA 4.0
2020-02-05T22:51:48.487Z
In UK the same square metre re-radiates a lot more heat in June than it does in January, but not a lot more on June 6th than on June 3rd. That also raises the chance of 3 sunny days in a row, so if you keep the timescale short there'd be no variation worth talking about.  Otherwise you can only have an average, and averages can be very misleading. I'm also sure that land covered in elephant grass would give a different measurement from pasture close cropped by sheep or other gazing animals.
Michael Walsby
CC BY-SA 4.0
2020-02-06T17:00:10.083Z
@MichaelWalsby Interesting. By "a lot more" and "different" do you mean plus-or-minus 50% kind of "a lot more" like 10w to 4w, or do you mean an order-of-magnitude-significant difference such as 1w to 20w or more?
Aaron
CC BY-SA 4.0
2020-02-06T21:04:32.137Z
I've never actually measured the night time emissions of my back garden lawn in January but it gets no direct sunlight, though in June it does. Never measured it in June either, but it must get at least 50 percent more, probably 150 percent on a hot day.. It can only re-radiate by night what it receives by day, so I know for a certainty my lawn emits far more on a June night, particularly after a sunny day, than it does in January. In a hard winter the lawn is sometimes frosty or snow-covered all day.
Michael Walsby
CC BY-SA 4.0
2020-02-06T21:11:51.880Z
@Aaron PS I've never measured the emissions from elephant grass either, but apart from anything else it absorbs more energy for photosynthesis than short grass would (elephant grass is about 10 feet high).
Michael Walsby
CC BY-SA 4.0
2020-02-06T22:25:03.170Z
@MichaelWalsby Ok, thanks for the insights. I was under the impression that it was mostly dominated by temperature, which is influenced by the sunlight but is smoothed out by a lot of factors, so I was assuming it would not be that different. I'm certainly no expert though, I'm only wildly guessing.
Aaron
CC BY-SA 4.0
2020-02-06T23:48:31.233Z
As I said, the figures for a few days in June would be much less variable, especially if they were all sunny days. You might even get a whole week of  sunny days in June.
Michael Walsby
CC BY-SA 4.0
2024-01-11T11:29:42.527Z
75W/m² from rooftops, and over 100W/m² from ground, if you're looking for some numbers to work with. Here's qr code of link to a nocturnal radiation test.
[![Radiation test](https://i.sstatic.net/Kvx0Y.jpg)](https://i.sstatic.net/Kvx0Y.jpg)
Zimba
<energy-balance>
How much energy does Earth surface radiate at night per square meter?
Aaron
# How much energy does Earth surface radiate at night per square meter?

(Asked by: username_0 on 2020-02-05)

I am finding lots of information about Earth's energy budget, about how objects emit infrared radiation based on heat, and lots of related things, but nothing helping me figure out how much energy is actually being emitted by the ground at night.

At night, the ground will still be emitting electromagnetic energy. The amount of energy being emitted will vary based on many factors. Information that helps with many different types of land in various conditions would be great, but for my case I'm focusing on a stereotypical grassy plain in an area which has been between 5C-40C (40F-100F) for the past weeks.

Ultimately I'd like to have a rough ballpark figure of watts per square meter.

In response to @username_1's comment:

> The amount of energy emitted by night depends on how much it receives during the day, water content, length and type of grass and other vegetation, how many "past weeks" have there been, and what was the temperature of the ground when the relevant period began.

During a sunny day the ground is said to receive approximately 1 kilowatt per square meter.

Assume grass of 10cm-50cm (about a half foot to a couple feet). There have been thousands of past weeks. Since we need to get picky let's say the temperature has been a constant 10C (50F) every hour of every night and 15C (60F) every hour of every day for the last thousand years except for the 1 or 2 hours day/night transition where the temperature adjusts linearly back and forth if that makes the exercise easier. And the average annual rainfall of this location is 100cm (40in), rain was coming regularly at that rate until 4 days ago. In the last 4 days, it has rained once, yesterday afternoon dropping about 1cm (half-inch), and the water table is generally about 5 feet under ground level.

The dirt is completely covered and appears as you would expect if you Googled an image of a grassy meadow. For example, the picture I've added at the bottom of the question.

In my opinion, however, the best answer would not just say "The ground emits 10 watts per square meter under those exact conditions!" (even that would be useful though) but would instead say "The type of grass actually makes a significant difference that can throw off even a ballpark estimate, as the difference between grass type A and grass type B produces a whopping half-order-of-magnitude difference in final results due to the way their outer material insulates them! And the difference between wet grass and dry grass is actually 2 orders of magnitude! If I were to simply pick a random, reasonable scenario of ABC it would be within the order of magnitude of 0.1 to 1 watts per square meter, and that's on the high end."

For anything else that matters feel free to simply pick a reasonable value and mention (even if only in a few words) why it matters and if your pick produces a high, low, or mid-range estimate.

[![enter image description here](https://i.sstatic.net/4WZi6.png)](https://i.sstatic.net/4WZi6.png)

**Tags:** energy-balance

> **Comments:**
> **username_1**: You'll have to be much more specific than that. The amount of energy emitted by night depends on how much it receives during the day, water content, length and type of grass and other vegetation, how many "past weeks" have there been, and what was the temperature of the ground when the relevant period began.  
> **username_2**: This article might be interesting for you: https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1002/2014JD022216  
> **username_2**: To get some rough starting point maybe take the temperatures from https://neo.sci.gsfc.nasa.gov/view.php?datasetId=MOD_LSTN_M and assume a black body spectrum?  
> **username_0**: Based on @username_1 comment, if those thing all really matter so much, I'm wondering if it would be better to start with separate questions like "How much does vegetation water content affect nighttime IR energy output? All other thing being equal, how wide a range does this variable encompass? What are some reasonable estimates to use for high/low/mid range values for this variable?" This could get out of hand though with a bunch of "How does A affect X?" "How does B affect X?" etc..  
> **username_3**: As a rough global ballpark figure, it's equal to incoming radiation (ignoring the 0.1 W/m² difference that is climate change)  
> **username_3**: To clarify: do you mean radiation emitted by the Earth *surface*, or radiation as observed at the top of the atmosphere (TOA) which is what actually matters for the energy budget?  Most radiation emitted by the Earth surface is immediately absorbed by the atmosphere, then re-radiated by layers colder than the surface; this is exactly what the greenhouse effect is.  Earth surface radiation is impossible to measure from space, but not immediately relevant for the energy budget.  Do you want *surface* radiation or *top of atmosphere* radiation?  
> **username_0**: @username_3 Emitter at surface. My specific use-case interest involves very low altitude, as you would see it facing the ground from no higher than an average building or tree, possibly even closer.  
> **username_0**: @username_3 "Not immediately relevant for the energy budget." I only mentioned the energy budget because that's the topic that keeps coming up in the research I'm attempting. Personally, and for my specific purposes, the energy budget itself is irrelevant. That just seemed the right way to ask about it based on what little I've found. Basically, what I'm after is if you could hold a 1-meter square "IR meter" above the ground what would the power reading be. That was the end-goal I had in mind when asking this question.  
> **username_3**: I had misunderstood the question, and I have deleted my answer, which was for TOA and not for surface.  I don't know the answer for the surface, I hope you'll get a good answer.  
> **username_0**: As long as it's not too difficult to measure I don't mind getting my hands dirty, so I've posted a related question on engineering SE about how to take the measurement myself: https://engineering.stackexchange.com/questions/33095/measuring-em-radiation-power-over-surface-area  
> **username_1**: In UK the same square metre re-radiates a lot more heat in June than it does in January, but not a lot more on June 6th than on June 3rd. That also raises the chance of 3 sunny days in a row, so if you keep the timescale short there'd be no variation worth talking about.  Otherwise you can only have an average, and averages can be very misleading. I'm also sure that land covered in elephant grass would give a different measurement from pasture close cropped by sheep or other gazing animals.  
> **username_0**: @username_1 Interesting. By "a lot more" and "different" do you mean plus-or-minus 50% kind of "a lot more" like 10w to 4w, or do you mean an order-of-magnitude-significant difference such as 1w to 20w or more?  
> **username_1**: I've never actually measured the night time emissions of my back garden lawn in January but it gets no direct sunlight, though in June it does. Never measured it in June either, but it must get at least 50 percent more, probably 150 percent on a hot day.. It can only re-radiate by night what it receives by day, so I know for a certainty my lawn emits far more on a June night, particularly after a sunny day, than it does in January. In a hard winter the lawn is sometimes frosty or snow-covered all day.  
> **username_1**: @username_0 PS I've never measured the emissions from elephant grass either, but apart from anything else it absorbs more energy for photosynthesis than short grass would (elephant grass is about 10 feet high).  
> **username_0**: @username_1 Ok, thanks for the insights. I was under the impression that it was mostly dominated by temperature, which is influenced by the sunlight but is smoothed out by a lot of factors, so I was assuming it would not be that different. I'm certainly no expert though, I'm only wildly guessing.  
> **username_1**: As I said, the figures for a few days in June would be much less variable, especially if they were all sunny days. You might even get a whole week of  sunny days in June.  
> **username_4**: 75W/m² from rooftops, and over 100W/m² from ground, if you're looking for some numbers to work with. Here's qr code of link to a nocturnal radiation test. [![Radiation test](https://i.sstatic.net/Kvx0Y.jpg)](https://i.sstatic.net/Kvx0Y.jpg)  

---

## 1 answers

### Answer by username_5 on 2020-02-09. Score: 5

I'm not sure how accurate you need, but beginning NASA's global 398.2 watts per square meter estimate from their energy budget a rounding down based on temperature variation should give a pretty good estimate. A lot of people have a problem with that number because it's more energy than we get from the sun, but that's because our atmosphere is a kind of blanket that traps heat, which warms the Earth above it's planetary eqilibrium temperature.

It's also (but less so), higher because Earth is warmed from below as well as above.

[![enter image description here](https://i.sstatic.net/VPbI4.jpg)](https://i.sstatic.net/VPbI4.jpg)

[Diurnal temperature variation](https://en.wikipedia.org/wiki/Diurnal_temperature_variation) is between highest day temperature and coldest night temperature and you're looking for average. Using that number won't give a correct answer, but using, perhaps 1/2 that number might be in the range and that's over land. The diurnal temperature variation over water is much lower.

So, 288 Kelvin Earth, say 12 degree variation day to night, so 294 / 282. Ratio of 1.0425, to the 4th power means 18% more heat is radiated from land during the day than at night. Adjust the temperature if you think mine is off. That's 18% x 29% of the surface area.

For oceans, 71% of the surface area the temperature variation is smaller, say 2.5 degrees. 289.25 / 286.75, ratio of 1.0087. 4th power, means the oceans radiate about 3.5% more energy during the day than at night.

(18% x 29%) + (3.5% x 79%) = about 7.7%

So if Earth's surface radiation global average is about 398.2 w/m^2, then the day/night numbers work out to about 7.7% variation between them, or +/- 15.5 degrees, about 413.7 watts/m^2 during the day and about 382.7 at night. That's a pretty good estimate I think. I'm open to a better estimate if someone can think of a way to do it.

> **Comments:**
> **username_6**: This diagram with back-radiation is pretty hard to understand for me. Back radiation could only happen for IR capturing gasses (I hate the expression "greenhouse gas" because Earth has nothing to do with a greenhouse with glasses suppressing convection) and since the atmosphere is almost black for these gasses wavelengths (CO2 captures almost 100% 15um in 10-20m from the ground up), no such radiation can cross the atmosphere. This would mean that all the "backradiation" would happen in a thin layer above ground. Quite troubling and in violation with the laws of physics (energy is moving from h  
> **username_5**: A physical greenhouse does a few things.   A house suppresses convection but isn't warmer than the outside during the day.  A house with big windows can be warmer by letting light in.    A greenhouse is warmer still for suppressing convection, letting sunlight in and trapping IR radiation.  As for how back radiation works.  I like to think of individual photons.   A photon leaves the Earths surface, it flies generally upwards until it hits a molecule that can absorb it (greenhouse gas), it's absorbed, it makes the molecule dance, and then it's fairly quickly released in a random direction.  
> **username_5**: You're thinking of CO2 in the atmosphere as a mirror reflecting back and you're thinking in percentages.   That's kind of what the diagram looks like, but if you think of the photon and molecules that can absorb and release it individually, then it's a lot more like a lot of floating billiard balls moving almost at random and the photon bounces around, shot from molecule to molecule until it escapes into space.   The amount of time it takes the average IR photon to leave the Earth or the average number of collisions is proportional to the amount of heat storage.  
> **username_5**: So, more greenhouse gas molecules, more photon storage.   The same thing (kind of) happens inside the sun.  A photon created in the center of the sun by nuclear fusion can take 100,000 years to make it's way to the surface, then just 8 minutes to Earth.   This is, ofcourse, oversimplified as the photons created in the sun are gamma and the photons that leave the sun are visible light and there's some energy conversion on the way and perhaps, splitting a photon into more than one, but the idea is the same.  The photons inside the sun take a long time to get out, so the Sun heats up over time.  
> **username_5**: @username_6 The NASA chart just oversimplifies, rather than counting the photons and timing their trip out, it converts to energy in, energy up and energy down, perfectly fine for thermodynamics.   Not quite as good for imagining the precise process.  
> **username_7**: @username_5 how could I take this information and find out an answer to the following question? I've been asking myself, if we lowered global power usage by 10GW, distributed uniformly across all energy sources, then what the impact on Earth's temperature would be over 10 years. I'm a mathematician but I know really little about the thermodynamics at play here.  
> **username_5**: @username_7 This website doesn't like questions to be asked in comments.  You should probably post a new question.   That has the added advantage of having people much smarter than I am see your question as well.     I can say that the heat trapping from the added CO2 far exceeds any heat generated by human energy generation, by about a factor of about 10 to 1.  But that's CO2 added not just this year, but all prior years, and CO2 lasts in the atmosphere a long time, so the math gets tricky because it's not apples to apples.    CO2 added 20 or 30 years ago is still warming the Earth.  

---

