sorry, 4.8 calories in a joule. Nvm On Sun, Dec 28, 2014 at 2:51 PM, Blaze Spinnaker <blazespinna...@gmail.com> wrote:
> If it takes 540 calories are needed to turn 1 gram (at 100 degrees > Celsius) of water to steam, than that's 540 * 1200 * .23 = 149000 joules, > right? > > On Sun, Dec 28, 2014 at 2:45 PM, Blaze Spinnaker <blazespinna...@gmail.com > > wrote: > >> Also... is 2712000 the right amount for vaporization of the water? >> Doesn't it have to be dry steam to reach that energy amount? Looking at his >> set up, it seems like it would be pretty wet. >> >> On Sun, Dec 28, 2014 at 2:26 PM, Blaze Spinnaker < >> blazespinna...@gmail.com> wrote: >> >>> >>> https://docs.google.com/document/d/1Y3Bxr_aE2iosEKpGFUZiQgAcuT8AFN78RFCAlR-JqNw/edit >>> >>> On Sun, Dec 28, 2014 at 2:26 PM, Blaze Spinnaker < >>> blazespinna...@gmail.com> wrote: >>> >>>> Note by translator Stoyan Sarg: The initial heating power and >>>> temperature before reaching 10000C is not shown in the plot of slide >>>> #16 (does he mean 17?). Is it taken into account for the accumulated >>>> energy? If not a much longer test is needed for estimation of the COP. >>>> >>> >>> >> >
