A student is trying to run some experiments for a project to explore how the concentration of CO2 affects the temperature of air. For this, she:

  1. Drops different ammounts of dry ice (between 0 and 2.5g) in a 0.5l conical flask ans seals it.
  2. Waits until the CO2 has volatilze and the temperature has stabilized in a water bath.
  3. Places the flask in front of a IR incandescent lamp.
  4. Records the temperature every 10 seconds during one hour. She does this by placing a thermometer inside the flask.

She is always obtaining a nice heating curve that increases quickly for 10 min and then stabilizes (sorry, I don't have the data).

The problme is that there is no correlation between the amount of CO2 and the final temperature.

There are still a few variables that she needs to control, but I am wondering if this experiment can work at all. Will the ammount of CO2 be enough to make a difference? Would you advice any changes in the setup?

  • 3
    $\begingroup$ "Drops different ammounts of dry ice (between 0 and 2.5g) in a 0.5l conical flask ans seals it." – Don't do that! That is very dangerous. $\endgroup$ – Loong Jul 20 '19 at 11:24
  • $\begingroup$ It's sealed with a rubber cork. My guess is that it will pop out before the glass breaks (it actually happened once). Shouldn't a conical flask be able to withstand a few atm of pressure? $\endgroup$ – jrglez Jul 23 '19 at 8:14

Three auxiliary measurements/computations spring to mind immediately.

  1. Your student should measure how much infrared light passes through the glass of the flask. A discussion of greenhouse gases shouldn't forget how a literal greenhouse works: they're made of glass! If you like the thermal-curve technique, you might stick your thermometer in some matte-black thermal mass and see whether it warms at the same rate (or to the same final temperature) in and out of the shadow of the flask. Several levels of sophistication are possible here.

  2. Your student should compute how much infrared light she expects to be absorbed by a column of CO$_2$ a few centimeters thick. Consider how distant mountains look blue in the air, even when cover covered with green trees. How much air is that? The relevant parameter is the air density multiplied by the distance, with units of kilograms per square meter.

  • $\begingroup$ My calculations give me a volume of a liter, not a cubic meter. Am I doing something wrong? $\endgroup$ – jrglez Jul 23 '19 at 8:17
  • $\begingroup$ @jrglez You're right, and I'm wrong. Sorry! $\endgroup$ – rob Jul 23 '19 at 21:53

I think the way CO2 works is not so much by absorbing the heat itself but by acting as a blanket to prevent infra-red from the sun being radiated back into space. I doubt whether you can model this in s flask.

  • $\begingroup$ Carbon dioxide "acts as a blanket" by absorbing the infrared light emitted by the ground, transferring that heat to the atmosphere rather than to space. Or are you suggesting that the blanketing mechanism is scattering, rather than absorption? $\endgroup$ – rob Jul 20 '19 at 12:20

The problem is that there is no correlation between the amount of $CO_2$ and the final temperature.

This is to be expected.

The final temperature should always tend to match the environmental temperature. This is simply the flask (and it's contents) reaching thermal equilibrium with the environment.

The amount of $CO_2$ makes no difference to this basic principle.

Concentration of $CO_2$ would only affect the rate the material cools and heats. As the rates of heating and cooling are different for the different materials involved, after an hour the "final" temperature should be tending towards the equilibrium temperature, but may not reach the same value in all experiments.

  • 1
    $\begingroup$ You've ignored the lamp. An object illuminated by an IR lamp is in conductive/convective thermal contact with the air around it, but in radiative thermal contact with the lamp, and will equilibrate at a temperature between the temperature of the air and the temperature of the lamp. How effectively the lamp heats the object depends on its absorption coefficient. A perfectly mirrored or perfectly transparent object wouldn't experience any infrared heating; a perfectly black object would end up warmer than the air around it. $\endgroup$ – rob Jul 20 '19 at 12:03

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