What volume of 14 C water do I need to add to 30 L of 24 C water to have a product of water at 20 C?

October 19, 2011 10:15 AM

What volume of 14 C water do I need to add to 30 L of 24 C water to have a product of water at 20 C? Is there an easy calculation for this sort of thing so that I can do this with different levels of water and/or temperature differences?

Really, this isn't for Physics homework, I promise. I have two aquaria at these temperatures and need to get the warmer water down to the colder temperature quickly and easily.

posted by pwb503 to Science & Nature (11 answers total)

Assuming that there is no heat lost or gained from stirring, contact with the surrounding air, etc., the formula is just that volume * temperature = volume * temperature.

Hence, 30 * 24 + Y * 14 = (30 + Y) * 20 -- > 720 + 14Y = 600 + 20Y. --> 6Y = 120 --> Y = 20L. You need 20 L.
posted by willbaude at 10:21 AM on October 19, 2011

Best answer: (Maybe I should spell out that formula a little more. The general formula you want is just to average the temperature of your water (in C!), weighted by the volume of water. Hence, (Volume 1) (Temperature 1) + (Volume 2) (Temperature 2) = (Total Volume) (New Average Temperature).
posted by willbaude at 10:23 AM on October 19, 2011

Response by poster: I don't need to convert anything to Kelvin do I?

This is starting to come back to me... I assume this is slightly different that V1/T1 = V2/T2 since this is a liquid rather than a gas?
posted by pwb503 at 10:29 AM on October 19, 2011

You don't need to convert anything to Kelvin (although if you did the calculations in Kelvin they would still work!). This doesn't involve the Volume-Temperature-Pressure formulas you learned for gases, because a liquid's size doesn't really change with its temperature. It would also be more complicated if you were mixing substances with different specific heat capacities, but as long as it's all water, it's not that complicated.
posted by willbaude at 10:35 AM on October 19, 2011

It's the same rules as for gasses (1/2 f k T). And, yes, convert to Kelvin. Thermal energy of a liquid is proportional to the absolute temperature of the liquid. (A liquid at 0 C doesn't contribute 0 energy to the system.)
posted by introp at 10:37 AM on October 19, 2011

Best answer: (Incidentally, if you want, we can solve the V1T1 + V2T2 formula for problems like your original one-- i.e., I have a bunch of water at one temperature, how much water of a second temperature do I need to add to get to a third temperature?)

V1T1 + V2T2 = (V1+V2)T3
V1T1 - V1T3 = V2T3 - V2T2
V1 (T1 - T3) = V2 (T3 - T2)
V1 (T1 - T3) / (T3 - T2) = V2
posted by willbaude at 10:39 AM on October 19, 2011

I'm assuming that none of the temperatures are going to hit or go below 0C, since then we wouldn't be dealing with water anymore. (Ice has its own complications.)

I admit that I could be missing something, but I don't see why it matters if the temperatures are in Kelvin or Celsius. If you substitute all of the temperatures in my formula with T + 273, the +273 will just drop out of the equation because the volume on both sides is the same.
posted by willbaude at 10:41 AM on October 19, 2011

(Oh, unless you're doing a mass estimate per willbaude's system. Then temperature offsets don't matter.)
posted by introp at 10:42 AM on October 19, 2011

Hee. I keep missing your posts while I'm writing mine. The offsets don't matter. My brain immediately jumped to the minor density differences at temperature and wants to do the thing as an energy balance equation. Of course, the constants would still cancel out in that case, but it's an old habit from modeling cooling in electronics (where radiation can matter): thermodynamics work is done in kelvin. Bad introp! *smacks own hand*
posted by introp at 10:47 AM on October 19, 2011 [1 favorite]

As long as the liquids are the same, as long as you don't need to put in heat capacities, you can do it by relative difference and it doesn't matter if you plug temperatures in with Centigrade or Kelvin scales.

Ice works ever better, in my experience. You get the latent heat of melting as well as the heat capacity terms and your starting temerature is usually well defined (0 C/273 K). We cool our 300 L test tanks this way.
posted by bonehead at 10:59 AM on October 19, 2011 [1 favorite]

You're basically just doing a weighted average (i.e., weighting temperature by volume) if you're assuming a closed system.
posted by supercres at 12:41 PM on October 19, 2011