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# why does the water become cold when adding more ammonium nitrate? Answered

(explanation)

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4 elements
air
earth
water
fire
water is opposite of fire .
fire is hot
water is cold

endothermic reaction

+1

I concur we like to help solve specific problems but prefer not to school readily available physical behavior on Google or youtube...

But it is soo much faster and easier to register here, post a useless question and never come back.
Imagine how long it would take you to find all that info on youtube or Google...
At least here you get an answer that you can copy and paste for your teacher to be proud of you....

Why worry ? Jack will post a text book answer for them.

Jack's elegant prose are recognizable by teachers every where on most English continents especially the Lone Star one...

In general, a change in enthalpy (i.e. heat taken or heat given) can occur, on dissolving any substance in water.

https://en.wikipedia.org/wiki/Enthalpy_change_of_s...

As you can see from the table, in the article linked above, the deltaH for some substances is positive, for some it is negative. Actually, ammonium nitrate is included in this table, and they say it has a deltaH of solution, of +25.69 kJ/mol.

By the way, deltaH is the heat required for a reaction.

Positive delta H, i.e. deltaH > 0, corresponds to heat absorbed; i.e. solution gets colder.

Negative delta H, i.e. deltaH < 0, corresponds to heat released; i.e. solution gets hotter.

You might wonder why the difference: Why do some solids release heat when dissolved in water? Why do some solids require heat to dissolve in water, and get colder by spontaneously stealing that heat from the surroundings?

I think the usual answer to that question comes from the, uh... what is the name for this equation?

deltaG = deltaH - T*deltaS

I don't recall the name for this equation, but it is mentioned in the Wikipedia article titled, "Spontaneous process"

https://en.wikipedia.org/wiki/Spontaneous_process

There is also a Wikipedia article for "Gibbs free energy", but reading that makes me think the above equation is a special case, for constant volume. That is to say there is a longer version of the equation, with a P*deltaV term in it, but that goes to zero in the case of constant volume.

Anyway, where I was going with this story, is that only processes with negative deltaG are spontaneous, and deltaG being negative depends on delaH and deltaS.

So for example if you have a process with positive deltaH (i.e. the solution gets colder) , the only way that can happen. I mean the only way deltaG can be negative, is if

T*deltaS > deltaH

which implies large, positive, deltaS. In other words, it is the big increase in entropy that makes the possible.

Actually you could consider maybe two extreme cases for spontaneous processes. (1)Those for which large negative deltaG is made possible by large negative deltaH, i.e. precesses which are bigly exothermic, and call those processes, "enthalpy driven", and (2) those for which large negative deltaG is due to a large positive deltaS, and call those "entropy driven"