Pam Giles asked:
We did an experiment where we had water that was 117 degrees Fahrenheit. We placed two containers inside in the air conditioning at 78 degrees and two containers of the same temperature, same quantity of water outside in the sun and two outside in the shade.
We were shocked to find that the water outside in 90 degree weather in the shade cooled much quicker than the water in the cool classroom.
Someone suggested it was evaporation and so we covered the tops the next day and re-did the experiment and still got the same results.
Weíre wondering why is the water in the hotter air cooling quicker than the water in the cool, air-conditioned classroom?
Dave - Thatís very interesting. The answer I was going to give immediately was evaporation because the maximum amount of heat which water looses is generally due to evaporation and therefore if youíve got wind blowing across the top, all those hot molecules evaporating off the top will get blown away quicker, you can get more evaporation and itís going to lose heat quicker. It could still be: I would have thought the only big effect is going to be the wind and it could just be that even just losing heat to the air around it is a lot faster if wind is blowing past something because otherwise, if the air is stationary, the hot object just heats up the air around it and then that insulates it from the rest of the air. If thereís a wind blowing past, it will keep moving that hot air away and so you can keep losing heat much more quickly. Certainly the only thing I can think of is that itís to do with the wind but possibly due to conduction as well as evaporation.
Scott - I think Dave's hypothesis about the wind is right, and I think that there are some useful experiments that could be run. Pam says that she put the outdoor water in the shade. That suggests to me that there was probably both shaded and unshaded areas nearby. That would likely create a temperature differential between the sunny area and the shady area. The air in the hotter sunny area would rise, sucking in air from the cooler shaded area. That would generate the wind, which disrupts any insulating warm air that might build up immediately around the glass in the shade.
The only possible reason is the wind... Evaporation is negligible because the temperature of the water is not high enough (far from boiling point).
Heat up the cream in the microwave (much faster since it is a small volume you can do that fast), then pour it in right away. Not recommended for powdered creamer. Even if the cream is not heated, add it right away . See http://web.cecs.pdx.edu/~graig/me323/Solutions-Practical-Problems.pdf item 14 for an explanation. mcgregor94086, Wed, 15th Sep 2010
I suspect the humidity of the air in the two locations could also be a factor. I would not assume they would be the same in both cases. Geezer, Wed, 15th Sep 2010
I should precise that you don't heat the cream... Yes, your right, because the cooling rate is proportional to the difference in temperature between the cooling object and its surrounding (temperature gradient). There is other factors like insulation, but in this case, we can assume that the insulation is fixed.
In answer to your question, Right now. Anything cools rapidly at first and it is better to get the cooling effect immediately as it cools more slowly later. Thanks for comments. Joe L. Ogan Joe L. Ogan, Mon, 20th Sep 2010
Take a look here for the cooling effect of wind.