19
$\begingroup$

This question already has an answer here:

When I deflate balloons by cutting one end and letting the air out, why does the surface of the balloon feel cooler?

$\endgroup$

marked as duplicate by AccidentalFourierTransform, John Rennie thermodynamics Aug 19 at 11:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ Possible duplicate of Rubber band stretched produces heat and when released absorbs heat.. Why? $\endgroup$ – Sciencisco Aug 18 at 13:43
  • 1
    $\begingroup$ @Sciencisco even if the answer is essentially the same, it isn't a duplicate, because balloons are not elastic bands. (In fact, I am not completely convinced the answer is the same. The contracting rubber will cool down, but there might also be Joule-Thomson cooling as the gas escapes - I don't know which would be the stronger effect.) $\endgroup$ – Nathaniel Aug 18 at 20:45
  • $\begingroup$ @Nathaniel Joule Thomson cools the gas, not the balloon's surface. $\endgroup$ – Sciencisco Aug 19 at 1:46
  • 1
    $\begingroup$ @Sciencisco the gas being in contact with the balloon, so... $\endgroup$ – Nathaniel Aug 19 at 7:58
23
$\begingroup$

A piece of rubber can act as a refrigerator of sorts, as follows:

If you take a piece of rubber and stretch it out, the molecular chains are pulled out in tension and the kinks in them get straightened out. they are now stiffer, and whatever frequency at which they were originally vibrating will go up slightly- and this means the rubber is now warmer by a tiny amount, as if it were a parcel of gas which you have compressed.

If you maintain the stretch while the heat leaves the rubber, eventually the rubber assumes the temperature of its surroundings- like the parcel of compressed gas would, if stored in a thermally conductive container under pressure.

Then, if you suddenly release the tension on the rubber band, its molecules relax back to their slack state and their vibrational frequency drops down a tiny bit- which means its temperature drops below ambient- just as if you released the pressure on that parcel of compressed gas and allowed it to expand.

This effect can be demonstrated with a rubber band, as follows: Hold the band with both hands and press it against your lip to sense its temperature. Now pull the band out tight with your hands and hold it there for 10 seconds. Then quickly bring your hands together so the rubber band goes slack and immediately press it against your lip again. You will detect that it is now cooler.

In the specific case of a deflating rubber balloon, can it be that the decompression of the air inside the balloon is responsible for the drop in temperature of the balloon material? No, and here is why. If one were to put a pressure gauge inside the balloon as it deflates, you'd see that its pressure remains almost constant as the balloon shrinks- which means the air inside the shrinking balloon is not cooling. The gas expansion is happening at the outlet of the balloon's fill neck, not in its interior.

$\endgroup$
  • 3
    $\begingroup$ That effect is very real and well documented. But how do we know for sure the air inside, which undergoes a quasi-isobaric expansion, does not drop also in temperature a bit? $\endgroup$ – Gert Aug 17 at 23:06
  • 2
    $\begingroup$ Because its pressure remains almost constant as the balloon contracts. The air expansion occurs not in the body of the balloon itself, but in the neck of the balloon near its outlet. $\endgroup$ – niels nielsen Aug 18 at 0:26
  • $\begingroup$ Yes, I believe you are correct. Ta $\endgroup$ – Gert Aug 18 at 2:19
  • $\begingroup$ Could you please add this comment in your answer, niels? And thanks @Gert for your comment. I hope it improves the answer. $\endgroup$ – thermomagnetic condensed boson Aug 18 at 11:51
  • $\begingroup$ @thermomagneticcondensedboson, sure. -Niels $\endgroup$ – niels nielsen Aug 18 at 17:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.