Simple temperature calculations

Although I recommend the use of a thermometer, sometimes it’s convenient to know how you can also manage without. If you mix water at two different (but known) temperatures, you can easily calculate the temperature after mixing. Just multiply the temperature of each part with the relative amount. For example, if you have 3 dL at 100 °C and 7 dL at 10 °C (which happens to be the approximate temperature of my tap water), this gives (3 dL x 100 °C + 7 dL x 10 °C) / 10 dL = 37 °C which is just perfect for dissolving fresh yeast when making bread.

You can also do it the other way around. Let’s say you have boiling water and you know that your tap water is approximately 10 °C. If you want water at approximately 37 °C, you can do as follows:

temperature-calculation.jpg

Start by writing what you have to the left (100 °C and 10 °C) and what you want in the middle (37 °C). Subtract: (100-37) = 63 and (37-10) = 27. And voilá – you need 27 parts water at 100 °C and 63 parts at 10 °C (and 27:63 simplifies to 3:7 which is what we found above). Now of course if you really wanted water at 37 °C, you would simply put your finger in to see if it’s at body temperature…

Are there any practical applications of this? Yes – a simple, but elegant way to prepare fish would be to drop a fish of known weight and temperature (fridge @ 4 °C or freezer @ -18 °C) into water that has been brought to boil. Cover pot and turn off heat. The amount of water would be calculated based on the desired temperature of the fish. We are assuming here that there is no heat loss to the surroundings, which of course isn’t quite true. How fast pot of water will cool depends on how much water you use and on the pot. This can be corrected for, and luckily someone has already done it. More on this in my post on how to cook fish in cooling water.

We can apply the temperature calculation from above to figure out roughly what the temperature will with this cooking method. 800 g of fish from the fridge (4 °C) and 2,4 L of boiling water gives a temperature of (0,8 x 4 °C + 2,4 x 100 °C) / 3,2 = 76 °C. The cooling curves for a pot with 2,5 L of water suggest a temperature loss of 15-20 °C in 30 min which would bring us down to 55-60 °C which – considering that no thermometer is used – is quite good.

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Filed under: molecular gastronomy, recipe, tips & tricks

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