Welcome to our forum.
As for the if it's linear or logarithmic, I believe it's linear - water needs a certain amount of energy to raise 1 degree regardless of temp.
Chas and Vinny,Based upon your responses, I didn't make the scenario very clear though Vinny addressed more of what I was looking. Taking the same spa with the same usage trends under the same conditions (understanding that a poorly insulated tub/cover would amplify any energy needs), what would the difference in energy consumption be if operated at an outside temp of 70 and a tub temp of 97 and again at 104? Starting at 97 as a baseline, would the difference at 104 be 7 times greater than the difference at 98 or would it be greater or less than 7 times?
Saturday morning I too was having an early morning soak in 103-degree water, feeling the fall chill in the air, when a bug of some kind landed in the spa. As I watched closely, he struggled between life and death in the hot water and I wondered if he was more frightened by the hot water or the fact that he was drowning.... Suddenly I had what most of you would call an "epiphany" and I knew clearly what I had to do........STOP DRINKING BEFORE NOON!
Ah. OK, with those parameters set, then my answer becomes much easier to arrive at: I don't know.But I will look into it. I may even ask one of my brainy kids - and get back to you. This is a very good question, and very apropos to this forum!
I didn't do well with word problems in school. Shouldn't one of these scenarios involve a train going east??
Windsurfdog,If you want an exact answer then it's going to take some time to research. Those books are long gone from my possession - replaced by hot tub brochures. ASSUMING ALL THINGS ARE EQUAL IN THE TUBFor heat loss > it's generally an equation of temp difference and insulation factor (I believe that someone posted the formula on this site a while back). All things being equal the closer the two temps are the slower the heat loss. For heating water > it's generally an equation of (temp desired - actual temp ) x kw/ degree (F or C). Example: if the water temp is 50 F and you want 97 F and ASSUME each degree of temp rise takes 13kw then (97-50) = 47 x 13kw = 611kw, using the same info except the desired water temp is now 104 (104 - 50) = 54 x 13 = 702 kw. Now once the water gets to the desired temp, the other equation takes place to keep the water warm. Actually the heat loss calculation comes in while heating the water and as the water's temp moves further away from the outside air temp the tub loses more heat (I believe Chas mentioned that before)These are over simplifications and the actual calculations are probably more complex (I forget what they are).Windsurfdog, if you really want the calculations, I will try to come up with them but and this is a BIG BUT - YOU will start turning into a Engineering Techno GEEK and won't be happy with the simpler pleasures in life. We are an elite group of people that will ask questions like "How do you really know that your pump is putting out the correct HP" and "how many lumens does that 7 LED light put out vs a halogen bulb vs the 22 LED bulb". You will buy a UV light meter so that you know how much output your ozonator bulb is producing. Friends might shun you but you'll have a wealth of useless knowledge! Please take my advice ponder on more important things in life - Beer or Wine in the tub!Anyway, thanks for the good word to the spatopian goddess!