Question about dissolving Ammonium Chloride in Water

[vwishndaetr]

This is a question from a lab previously done for my chem class.

If you had used 40 mL of water and 6 g of ammonium chloride, rather than the 20 mL and 3g in the experiment, would you expect to get a larger, smaller, or identical temperature change?

In the experiment I dissolved 3g in 20 mL and the temperature dropped 10.7*. Now, I understand there is an endothermic reaction and it is caused by the lattice energy from the breaking of ionic bonds, but I'm not sure how it is affected when BOTH are doubled. Being both water and ammonium chloride.

Seems to me though more water would result in a smaller temp drop, and more ammonium chloride would result in a larger temp drop, so being that both are involved, the temp change would remain the same?

Though that's what my intuition tells me, I am not convinced since I really have no "details" to back up what I think.

Any input??

Thanks.

 

[Borek]

Your intuition is right, although you may need more systematic approach.

If you use twice the mass of NH₄Cl, how many times more heat do you need for dissolution?

Do you know how to calculate amount of heat involved for a given mass and temperature change of water?

 

[vwishndaetr]

If you're referring to

$Q = m\ c\ \Delta t$

Then maybe? If this is the equation, then yes, double the mass would result in double the heat. But does double the heat really result in a same temp change for double the water?

I mean, if both water and heat are doubled, delta t would in fact stay equivalent to the first change no?

Feels like I'm walking in circles.

 

[Borek]

You are on the right track - if you feel like you are walking in circles, that's because in fact you do, but that's the correct route!

You doubled the mass of the salt - so you have twice the heat. That makes it 2Q.

You doubled the mass of the water and salt - that makes mass of solution 2m.

So you have

$Q = m\ c\ \Delta t$

and

$2Q = 2m\ c\ \Delta t$

for the first and second case. Calculate Δt for each.

Note: m is now mass of the solution (and earlier I asked about mass of water) - but it doesn't change the outcome not the logic behind these calculations.

 

[vwishndaetr]

So the 2's pretty much cancel and I am left with the same temp change.

 

[Borek]

Exactly - temperature change is a function of RATIO, if both quantities change but ratio is constant, temperature change doesn't change.

 

[vwishndaetr]

Nice. Thanks for that. Even though I submitted with a different response, at least I know what the real deal is.