While attempting to make Schweizer's reagent in household conditions, using $\ce{CuSO4 + NaOH + 25\% NH4OH}$ route (all at least 95% pure, either technical or lab grade purity), I've stumbled upon a problem. While trying to precipitate $\ce{Cu(OH)2}$, the temperature of the solutions rose to about 50 Celsius, and instead of expected blue precipitate, I got an almost black (with a hint of violet) one instead, which I assumed is $\ce{CuO}$. However, I was able to dissolve some of the precipitate in $\ce{NH4OH}$ slightly (a suspension was made from about 10 g of precipitate in 200 ml ammonia), resulting in dark blue solution with only traces of suspended black particles ($\ce{CuO}$?), appearing very like the expected Schweizer's reagent itself.
The resulting substance was able to dissolve regular cotton & cellulose samples overnight, however the dissolution was quite slow and unimpressive, with some of the sample material remaining almost intact; a chalk-coated paper was quite resistant to it (although it also dissolved somewhat eventually).
I don't have prior experience with Schweizer's reagent, so I don't know if the rate of dissolution I got here was the proper, expected behaviour here or a sign of dud.
Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?
As side questions:
was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?
how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\% NH4OH}$ route) in dissolving various celluloses in practice?