Currently I'm measuring the rate of completion of a hydration reaction by putting a temperature probe in the middle of a cup full of reagents and measuring the temperature in an isolated box. The main reaction going on in the cup is:

$$\ce{CaSO4\cdot 0.5H2O + 3/2 H2O -> CaSO4*2H2O}$$

Now I'm planning to change from one type of cup to another kind of cup with a different volume and a different shape (that will also be completely filled). Will this influence the temperature profile of the reaction?

Cup 1 (with a volume of 192 ml):

Cup 1

Cup 2 (with a volume of 250 ml):

Cup 2

  • $\begingroup$ Hello and welcome to Chemistry.SE. If you have any questions about the site please feel free to visit the help center. Best of luck! $\endgroup$ – airhuff Mar 22 '17 at 7:04
  • $\begingroup$ @Fl.pf. Do you have something to back your answer up? Because I'm not gonna believe you if you simply say yes. $\endgroup$ – Michthan Mar 23 '17 at 14:06
  • $\begingroup$ @Fl.pf. I can see that the generated heat would be different for a different volume, but that heat would also be spread over a different volume. Wouldn't this cancel out then? $\endgroup$ – Michthan Mar 23 '17 at 14:14
  • $\begingroup$ @Michthan Over volume maybe. But if you change the shape aswell? Also you didn't state if you used the same amount of material for the different volume of cup. Maybe I don't understand exactly what you mean, but I don't think your question is written clearly enough (too much room for interpretation of what you really want to know). $\endgroup$ – Fl.pf. Mar 23 '17 at 14:32
  • $\begingroup$ @Fl.pf. I'm leaving work now, but I'll add more specifications tomorrow to make the question more clear. $\endgroup$ – Michthan Mar 23 '17 at 14:47

I don't believe there would be a temperature difference unless the concentration of reagents changed too.

If per unit area the same amount of product is created, then energy created by said conversion would be the same, regardless of the total volume.

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