[Student99]

How many CaO and Water do I need to heat 10 grams of water from 25 degrees Celsius to 40 degrees Celcius?

Obs: Water must be heated indirectly by primary water outside of the beaker.




Taking 18 mL of water and 56 g of CaO:

CaO(s) + H2O(l) ---> Ca(OH)2(s)     Enthalpy ΔH= 64 KJ.mol^-1

Water: 1,0 cal/gºC
64 KJ/Mol = 15296 cal/Mol.  Ca(OH)2(s)=74g/Mol.  15296/74 = 206,7 cal/g. (Is it correct?)

206,7 cal/g=18 x 1,0 cal/gºC x ΔT
ΔT = 11,48 ºC


This is all I know until this moment. How can I determine the mass of CaO and water required?
Any Idea?
Thanks!

[magician4]

this problem can't be solved with the data given, and for several reasons at this:

- you'd need to know the mass of the water you're dissolving the CaO in, as this water also needs heating

- for more precise measurement, you'd need the heat capacity of CaO (i.e. Ca(OH)₂ )- solutions , both with respect to concentration and to the respective temperatures  (both functions are next to linear with small conc. / small temperature differences, if memory serves, but you'll need them anyway)

- the solubility of Ca(OH)₂ being low, you'll need a certain minimum ratio of "water, inside" and "water, outside" (additionally, dependant on the temperature difference you wish to perform inside )

- you'll need the heat capacity of your apparatus

- an idea of solvation enthalpy of Ca(OH)₂ would be nice, too, as you don't just "make" it, but dissolve it ( at least partially) , too

... and you must have made damned sure that next to no heat got lost ( better: know the heat loss of you apparatus with respect to temperature difference inside-outside, time, beforehand)

hence, with the data given, we can't help you

regards

Ingo