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And I always wondered if drinking a glass (16oz) of 0 degrees water was burning the calories required to heat it where you could drink enough water to offset a donut in a morning. (Forgive the lack of math here, I'm about to pass out in bed.)


Quite simple math, since a calorie is the amount of energy required to heat a mL of water by 1°C. Thus a food Calorie (1000 calories) is the amount of energy necessary to heat a liter of water by 1°C. Human body temperature is 37°C, so you burn 37 calories for every liter of 0°C water ingested. The average donut has 195 Calories. Thus you must drink 5.27 liters of 0°C water to cancel out the calories gained from eating a donut. Not very realistic.


No, but still more significant than I would have predicted!


No, you are off by a factor of about 10. 16 oz ~ 500 ml. Body temperature - 40 celsius. 1ml water needs ~ 1 calorie to go up 1 degree celsius. The product is about 20 kcal. A medium dunkin donut is about 200 kcal.


Berkeley did the maths for you: http://w.astro.berkeley.edu/~gmarcy/thermal/tpteacher/jokes/...

(notice the word 'jokes' in the URL, but it's pretty well done)


Those numbers seem very peculiar:

> Each ounce of beer contains 16 latent calories, but extracts 1,036 calories (6,216 cal. per 6 oz. portion) in the temperature normalizing process.

I think it's muddling food calories (which are really mislabelled kilocalories) and temperature calories (which aren't). So that 16 should be 16000.

This is why you do this in joules!




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