How much of a difference does it make whether you blow on your soup to cool it down while in your dining room versus if you open the outside door in the winter and blow on it that way?

This would help settle the argument I'm having with my wife every time she does this. I'm not good enough at physics/math to calculate the answer myself.

some details:

  • You have 300 ml of hot soup, heated to 90 degrees celsius
  • the air temperature in your dining room is 25 degrees celsius
  • the outside temperature is 5 degrees celsius
  • your body temperature is 36.5 degrees celsius
  • the temperature of your exhaled breath, while you are in the dining room, is about 35°C and 95% humid
  • I'm guessing that when you open the window you are breathing in the air that is half influenced by the room temperature, and half by the outside cold (let's say 15 degrees celsius)
  • the temperature of your exhaled breath, while you are blowing out of an open window, is unknown to me, but can be calculated
  • you open the window and blow on your soup for 40 seconds

How much sense does your action make? :)


1 Answer 1


Because your body temperature isn't dependent on room/outside temperature this has no effect on cooling rate.

But temperature of the surroundings (of the soup), that is outdoor or dining room, has a strong bearing on convection losses. Newton's Law of Cooling basically states that a 'hot' object loses heat at a rate $\dot{Q}$ ($\mathrm{J/s}$) that is directly proportional to the difference in temperature between the 'hot' object and its 'colder' surroundings $\Delta T$ ($^\circ \mathrm{C}$):

$$|\dot{Q}|\propto \Delta T$$

So the soup will cool down faster in the colder outdoors, compared to the warmer dining room.

  • 1
    $\begingroup$ It can be calculated (estimated) but it's rather lengthy, so please have patience. $\endgroup$
    – Gert
    Mar 1, 2021 at 16:18
  • 1
    $\begingroup$ What can be said with certainty is that the heat transfer rate $\dot{Q}$ is about $31\text{ percent}$ higher for the outdoor case. I don't want to go any further because of the difficulty in getting good values for the convective heat transfer coefficient $h$. But the temperature drop over $40\mathrm{s}$ is of course very easily measured and that is what you should do. $\endgroup$
    – Gert
    Mar 1, 2021 at 16:39
  • 1
    $\begingroup$ It wasn't in your lungs for long enough to completely warm up Lungs are an extraordinary heat exchanger, due to the alveoli. Temperature of outgoing breath is very close to body temperature for that reason. $\endgroup$
    – Gert
    Mar 1, 2021 at 16:45
  • 1
    $\begingroup$ See e.g. pubmed.ncbi.nlm.nih.gov/20529457/…). $\endgroup$
    – Gert
    Mar 1, 2021 at 16:48
  • 1
    $\begingroup$ Have you considered putting the bowl of soup in a pan of cold water and stirring gently for a while, remove the afore-mentioned bowl, dry and hey presto, cooler soup! $\endgroup$ Mar 15, 2021 at 23:13

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