I am puzzling with that myself at the moment. I want to disinfect something made of plastic. 100C would be too much for it.
So since the amount of energy to warm 1 liter of water one degree Celsius is more or less constant between zero and boiling point, you would have to solve:
20 * X + 100 * Y = 70
At least that is what I think.
Tim Cooke wrote:Oh this is getting really hard really quickly. Might I recommend pouring the boiled water from the kettle into a glass jug and letting it cool to the desired temperature. A thermometer might be required, unless you're happy to guess at some "hot but not boiling" temperature.
I do not have a thermometer.
I was counting on you guys!
Jan de Boer was right, well almost:-
100x + 20y = 70 × (x + y)
Make x + y equal to 1000 and solve the equation. It is what was called simultaneous equations when I was twelve.
I can think of another way to solve the mystery,
We must have a warmer climate; ours are corrected to 15°C.
Pete Letkeman wrote:. . . prices in Canada are corrected at four degrees Celsius for most automotive gasoline/fuel.
The simple formulae will give approximate results, and the end result will vary slightly depending on whether you measure the water in a jug or weigh it.
You want to cool the water from 100° to 70°.
That reduces to 5 units of heating and 3 units of cooling.
So you want 5 parts of hot water and 3 parts of cold water. 5 + 3 = 8 So you want ⅝ of one (=625ml) and ⅜ of the other (=375ml).
Campbell Ritchie wrote:That reduces to 5 units of heating and 3 units of cooling.
The 5 deduced from 70-20, and the 3 from 100-70? Okay.
So for example if I want to do my laundry at 60 max, I would mix equal parts of boiling hot and room temperature water. I see camping possibilities here.
I use a similar technique for baking bread: mix equal volumes of cold and boiling water and whisk the sugar and yeast into the water. I think you will find that is a common technique in bread recipes.
Campbell Ritchie wrote:Do you wash clothes when camping? Can you get cold water as warm as 20°? Yes, you probably can, but more like 20°F than 20°C .
20°F would be frozen. That would be another calculation. You would have to calculate the energy to get the ice to zero degrees, and then calculate the melt warmth and add it. As last you would have to do the same calculation as above. That is a new challenge.
Looks like nerd camp meets girls scouts here! ;)
Campbell Ritchie wrote:You think you can get girls to warm up the water for you?
No I will warm up the water, and they will do the laundry! But seriously, girls, girl scouts too probably, like smart boys. Especially when they are a little rebellious, creative and witty. At least that is my high school experience. I will blind them with science!
x = grams cold water
y = grams hot water
x+y = 1000
so far, so good. So the heat added to the cold water has to equal the heat lost by the warm water. The warm water cools 30 calories per gram. The cold water must be heated 50 calories per gram. so...
50x == 30y
so ultimately, x = 375 grams, y = 625 grams.