Let's say we want to dilute a 90 w% H2SO4 solution to a 60 w% solution. Regarding the heat released, does it matter if we do it with the help of water or with e.g. a 10 w% H2SO4 solution? If yes, in which case will more heat be released?
Thanks

Posted 02 July 2019  07:15 AM
Let's say we want to dilute a 90 w% H2SO4 solution to a 60 w% solution. Regarding the heat released, does it matter if we do it with the help of water or with e.g. a 10 w% H2SO4 solution? If yes, in which case will more heat be released?
Thanks
Posted 02 July 2019  07:31 AM
Hi,
Let you study the pamphlets attached .
Hope it helps you and others
Breizh
Edited by breizh, 02 July 2019  07:33 AM.
Posted 02 July 2019  04:33 PM
Thank you!
So the heat calculation methods let me conclude that it doesn't matter if we dilute with water or with a diluted H2SO4 solution, but it would be nice to hear a confirmation of my conclusion.
Posted 02 July 2019  08:19 PM
hi ,
If you perform a mass balance you should be able to evaluate how much Sulfuric will be used for both cases . To me the diluted acid will save about 5% of the total quantity you need . All depends if you have sufficient enough diluted acid available .
Recycling diluted acid is a bit more challenging and need more equipment to match the mass balance.
In other words , no much difference in term of heat released .
Breizh
Posted 03 July 2019  05:55 AM
So the heat calculation methods let me conclude that it doesn't matter if we dilute with water or with a diluted H2SO4 solutionThen you must have done it not quite accurate.
Assuming all the starting fluids have the same temperature whether diluting with water or diluting with 10w%, diluting with water will result in a higher product temperature than diluting with 10w% as more heat will be released.
The difference is not big, but in a calculation clearly visible and in actual practice clearly measurable.
Posted 03 July 2019  07:00 AM
In order to get a 60 w% solution, we need to add more 10 w% solution than water, so it's clear that in case of water the product temperature will probably be higher, because the amount of product solution is smaller which absorbs the released heat. However my question is not about the product temperature, but the heat released during the dilution process.
Posted 03 July 2019  09:50 AM
In order to get a 60 w% solution, we need to add more 10 w% solution than water, so it's clear that in case of water the product temperature will probably be higher, because the amount of product solution is smaller which absorbs the released heat.
That is not correct.
Amount of product solution is not smaller, because it is the product solution you want to make.
If you want to obtain say 100 kg 60 w% product then you need to select the amount of 90w% and the amount of water or 10w% such that you end up with 100 kg 60 w% product.
However my question is not about the product temperature, but the heat released during the dilution process.That is the same thing: the warmer the resulting 60w% product the more heat must be released during the mixing.
You can read both from a sulfuric acid  water enthalpy diagram like in the links provided by breizh, or many other sources on internet or in textbooks.
It seems that you did not really try to calculate the effect of the mixing.
Edited by PingPong, 03 July 2019  09:55 AM.
Posted 03 July 2019  11:48 AM
Ok, there are some misunderstandings...
In order to get a 60 w% solution, we need to add more 10 w% solution than water, so it's clear that in case of water the product temperature will probably be higher, because the amount of product solution is smaller which absorbs the released heat.That is not correct.
Amount of product solution is not smaller, because it is the product solution you want to make.
However my question is not about the product temperature, but the heat released during the dilution process.That is the same thing: the warmer the resulting 60w% product the more heat must be released during the mixing.You can read both from a sulfuric acid  water enthalpy diagram like in the links provided by breizh, or many other sources on internet or in textbooks.
It seems that you did not really try to calculate the effect of the mixing.
By "the heat released" I mean e.g. how many Joules will be produced.
If during mixing Q [J] heat is produced, then the product's temperature will change according to Q = c•m•ΔT, where c is the heat capacity. So the product's temperature depends on its mass (m), but I only want to know if the Q [J] is the same in case of water and in the case of a 10 w% solution.
So if my conclusion (based on the diagram) is correct, then:
if we have a given amount of 90 w% H2SO4 solution, and we dilute it to a 60 w% solution, the same Q heat is produced, it doesn't matter if we use water, or a diluted H2SO4 solution. In the latter case we just have to add more diluting agent, (so the amount of the product will be bigger).
Edited by Sulfuricacid, 03 July 2019  11:56 AM.
Posted 03 July 2019  01:41 PM Best Answer
if we have a given amount of 90 w% H2SO4 solution, and we dilute it to a 60 w% solution, the same Q heat is produced,
That is in general not correct.
Diluting a given amount of 90 w% acid with water to obtain a resulting amount of 60 w% will produce more heat than diluting it with 10w% acid.
For example:
100 lb of 90 w% acid at 70 ^{o}F plus 50 lb of water at 70 ^{o}F gives 150 lb of 60 w% acid at about 228 ^{o}F and 29 BTU/lb
100 lb of 90 w% acid at 70 ^{o}F plus 60 lb of 10 w% at 70 ^{o}F gives 160 lb of 60 w% acid at about 210 ^{o}F and 38 BTU/lb
Enthalpy of 60 w% acid at 70 ^{o}F is 115 BTU/lb
So heat released by diluting with water is 150 * (29  (115)) = 12900 BTU
and heat released by diluting with 10 w% acid is 160 * (38  (115)) = 12320 BTU
The difference is about 5 %, not big but still noticeable.
enthalpy diagram for sulfuric acid  water mixtures.jpg 356.02KB 4 downloads
Edited by PingPong, 03 July 2019  01:43 PM.
Posted 04 July 2019  05:07 AM
Well, this diagram also considers the temperature of the solution to be diluted and the diluting agent, which seems reasonable, so this might be the best way to calculate the released heat. Thank you!
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