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In a lot of recipes, there is a step that includes reducing your liquid. In most sauces, this isn't a problem, since you just need to thicken it and that's something you can see and doesn't take a long time.

However, last week I was making a base for a sauce, that included water, vinegar and white wine. The recipe asked me to reduce it till it was about 1/3 of the original amount. I didn't think it would take as long as it did and as a result, everything was ready except the sauce.

Is there a way to calculate how fast something reduces? I can understand it would depend on the ingredients, on the surface area and on the temperature, but assume that I know these variables.

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It seems like to a good approximation it might just depend on the power output of your stove - do you know that? –  Jefromi Feb 28 '12 at 17:13
    
You mean the amount of Watt? Or the setting? Or the temperature? –  Mien Feb 28 '12 at 17:15
    
Watts are units of power, yes. (In some places you might also see values in BTU.) Settings don't mean much in absolute terms. The temperature of the burner is not that useful - it depends on power, but also on the characteristics of what the heat is being transferred to. –  Jefromi Feb 28 '12 at 17:24
    
So what I understand you're saying, is that you made a reduction and then finished the sauce once you had the reduction down to the correct consistency. In future, make your reduction ahead of time and store it in the fridge. When you need it, pull it out and finish your sauce from that point. It's what we do in restaurants when we're doing sauces to order. –  Chef Flambe Mar 6 '12 at 5:49
    
@ChefFlambe, good suggestion, thanks. It was the first time I made this and the recipe did not give any estimate in time. I have the rest in my freezer now. –  Mien Mar 6 '12 at 9:46

4 Answers 4

up vote 4 down vote accepted

If you're boiling something rapidly, and it's not in a terribly deep, narrow pot, then essentially all of the heat output of the burner is going into turning water into steam. The latent heat of vaporization of water is 2260 kJ/kg, so if you want to reduce something by a volume V, and your stove has power P, the time required is:

t = V * (1 g/mL) * (2260 J/g) / P

If V happens to be in mL, and P is in W (J/s):

t (s) = V / P * 2260

This would be modified slightly if you're using a really tall, skinny pot, since the convection within the pot, from the bottom to the top of the liquid, would be less efficient, with more heat transferred to the sides of the pot and out into the air, but I doubt you're actually going to try to reduce something like that. The P here is the effective power; for example, a gas burner wastes a lot of heat out the sides, so the advertised power will be higher. See TFD's answer for approximate efficiencies.

If you don't know the power of your stove, in all honesty, the easiest way to measure it would probably be to just see how long it takes to boil away a given volume of water, and work backwards. To get an accurate result, you should not boil a pot dry - once the water is a thin enough layer, the heat transfer might start working differently, with the pot itself heating up more, and water splattering. So you could, for example, put in a liter of water, boil away at the stove setting you intend to measure until it's substantially reduced in volume, record the time, then pour it out to measure how much you boiled away. At this point, knowing the power output might be overkill, though; you can really just measure the time per volume reduction, and use that, unless you care about the power for other reasons.

Trying to deduce the power of the stove from, say, the temperature of an empty pot or of the burner without a pot on it (assuming it's electric) would be difficult; you'd have to deal with the heat transfer between metal and air, and the convection in the air.

Dependence on ingredients shouldn't be significant - you're still just boiling water, unless there's a substantial amount of alcohol, in which case the latent heat of vaporization will be different. Pure alcohol has a latent heat of vaporization of 841 kJ/kg; I haven't found a good table for mixtures.

For solutions, as I noted in the comments, the latent heat of vaporization should be that of water, plus/minus the heat of solution of the solutes (I forget which direction that's measured in). The most common solutes are probably salt and sugar, which have heats of solution of 70 and 16 J/g, respectively. (I found this table, and converted.) The next most common thing I could think of that might be present in substantial concentrations is citric acid; this paper reports a heat of solution of -57 J/g. In all these cases it's small compared to the latent heat of vaporization of water, so pretending the liquid is water should be a good approximation. It's possible that things change if you're reducing really far: heat of solution does depend on concentration. That is, things are different thermodynamically (statistical mechanically?) in a nearly-saturated sugar syrup than in slightly sweet water.

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Interesting. Do most liquids (wine, vinegar, juice) have about 2260 J/g? Stronger liquors will have slightly less, I assume. –  Mien Feb 28 '12 at 19:48
    
@Mien: If I'm thinking about it right, I think that the difference in latent heat of vaporization should be the same as the heat of solution - the amount of energy given off or absorbed when the solutes are dissolved into water. I suspect that that's fairly small for anything we eat; you don't notice large temperature changes when you dissolve salt or sugar into water. But real data would make me more confident asserting that... –  Jefromi Feb 28 '12 at 20:48
    
That formula isn't quite right—some portion of the water is escaping the pot without turning into water vapor/steam, which is why you can see it (water vapor isn't visible). You'd need to account for the portion escaping as mist. –  derobert Feb 29 '12 at 0:07
    
@derobert: I'm pretty sure that most of what you see coming off of a boiling pot is water vapor which has started to condense a bit, since it's passing up into air which is well below the boiling point. –  Jefromi Feb 29 '12 at 0:26

To use Jefromi's calcs you will need this too

Efficiency for stoves (hobs) at heating plain water in metal pot:

  • Gas 50% ±5%
  • Halogen hob 60% ±5%
  • Immersion 90% ±5%
  • Induction 90% ±5%
  • Resistive (most common electric type) 50% ±5%

It's hard to get over 95% efficient due to loses through pot wall etc.

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Note that if you measure the effective power as I described, by boiling away some water, you don't care about this. I'll add a note referencing your answer, unless you'd like to edit it into mine (you're welcome to). –  Jefromi Feb 28 '12 at 21:50

Theoretically yes, practically no. If you have a lot of experience of reducing sauces with your specific equipment then you can try but it's never a good idea to use this as an excuse to leave it to reduce whilst you do something else.

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Forget all that allow 20 to 30 min. It's then thick enough to use.

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If you are using a small amount of liquid, 20-30 min is way too big a variance. If you are using a tiny amount (reducing a deglazing) or a very large amount of liquid, the reducing time won't fall in that interval at all. –  rumtscho Jul 15 '12 at 11:12

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