Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If you boil water inside a pot the outer rim bubbles first, I imagine because its hotter. Does that same concept apply for the inside of a refrigerator for example. Is the outer rim always more extreme, or at least at first?

share|cite|improve this question

When you are boiling water bubbles typically form first at the bottom of the pot, and then at the walls. As you guessed, this is because the bottom is usually at higher temperature than the walls, since it's in direct contact with the heat source. The reason that bubbles don't appear sooner in the bulk of the water is that nucleation (the initial formation of tiny bubbles) is easier at surfaces and at their irregularities. To check this, you can use any pot with a big scratch, and you'll see bubbles forming at the scratch first.

In a refrigerator, the conditions are quite different. In a typical refrigerator there's a coiled set of pipes inside the freezer, near the top of the unit. This is the evaporator, and it's where the heat is extracted, and this is where you'll have the lowest temperatures. If the leaks are small, you'll have the lowest temperature at the top and the highest temperature at the bottom. Usually, the back of the refrigerator is colder than the front because heat leaks mainly through the door crack.

share|cite|improve this answer
does the cooling air leaves the cooling coil reaches a saturation point?discuss that,and how does that effect upon the the condensation flow rate at the evaporator? – user8405 Mar 31 '12 at 8:16
because heat leaks mainly through the door crack--No, its because we keep opening the fridge every three seconds :P. Seriously though: Great answer, you may want to expand a bit on nucleation. Also there's the issue about heat transfer requiring a temperature gradient, which is another explanation for this (there is a T difference at the metal-water interface, but the water has a small T gradient) – Manishearth Mar 31 '12 at 11:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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