Can a Body of Water's Temperature Rise Linearly with Constant Heating?

[DaveC426913]

I'm trying to guess when my hot tub will be ready for use.

If an ideal body of water is insulated and has a constant source of heating applied to it, can its temperature be expected to rise linearly?

(No. Even as I write this I see it can't be true. As the water temp rises, it will approach the temp of the heater, reducing the difference, meaning the rate of heat transfer will decrease.)

Well, my next question is moot then.

I was wondering if I saw a leveling off of temp rise over the next 25 hours, that would indicate inefficient insulation. i.e. as the hot tub heated, more and more heat would be lost because heat transfer is affected by temp difference. But nevermind, that's confounded by the above.

Anyway, it's risen from 56F to 63F in 4h20m. If it increased linearly, it would reach a max temp of 104F in just over 25h. (7 degrees in 300 minutes is ~ 1d/37m ... x 41d = 1517m = 25h17m)

 

[Bystander]

(No. Even as I write this I see it can't be true. As the water temp rises, it will approach the temp of the heater, reducing the difference, meaning the rate of heat transfer will decrease.)

Do not confuse/confabulate rates; you are running a "fixed" wattage (infinite T).
Edit: Think Zeno's paradoxes.

 

[DaveC426913]

Do not confuse/confabulate rates; you are running a "fixed" wattage (infinite T).
Edit: Think Zeno's paradoxes.

Elaborate?

Do you mean that, with a fixed wattage, the temp differential won't shrink?

 

[Bystander]

the temp differential won't shrink.

Just so.

 

[Nugatory]

You're assuming that the heating element is maintained at a fixed temperature, so that the heat transfer rate will decrease as the water temperature asymptotically and non-linearly approaches the temperature of the heater.

However, a constant energy transfer rate is a better model for how most heating elements behave. You will eventually find a limit on how hot the water gets, and that limit will come from what you first thought: heat loss to environment is equal to heat input from the heating element.

[edit: @Bystander beat me to it]

 

[Nugatory]

Elaborate?

Do you mean that, with a fixed wattage, the temp differential won't shrink?

Yes. Consider that if there were no water at all, the heating element would glow red hot and burn out - its temperature is limited only by the water's ability to carry heat away. So if the input is some number of joules per second, the element will heat up until the temperature differential is sufficient to transfer the same number of joules to the surrounding water.

 

[DaveC426913]

So my original premise is correct?
A leveling of rise in temp will be a result of heat loss via poor insulation (+ gaps, etc.)?

 

[Chestermiller]

So my original premise is correct?
A leveling of rise in temp will be a result of heat loss via poor insulation (+ gaps, etc.)?

Yes.

 

[sophiecentaur]

So my original premise is correct?
A leveling of rise in temp will be a result of heat loss via poor insulation (+ gaps, etc.)?

= Thermal Equilibrium
If you want to be able to predict how long it will take to warm up your bath water , you can plot a 'cooling curve', which will show the rate of heat loss at a particular temperature (heater turned off) and you can subtract that loss from the electrical power supplied to give you the power left over for increasing the temperature. Ideally, it would call for a bit of calculus but you could use a 'piecewise linear' approach (for which Excel is excellent) to plot a heating curve which will allow you to see how long it will take to reach a target temperature. The slope of that curve will approach horizontal, as you say.
There is no 'time' for that sort of thing in modern School Science curricula but it was the sort of thing that we used to do for A level Physics, in 1960. Basic experimental method is a thing of the past.
PS I will expect a fully written-up experiment from you with quantitate results and a 'conclusion'.

 

[DaveC426913]

So far it's tracking pretty well.
Then again, I don't expect any leveling off until it gets well into the 90s.

 

[Chestermiller]

If there were significant heat loss, you would already be seeing some curvature in the plot. So it looks like the curvature doesn't set in until well above 104 F, and your estimate of 25 hours is pretty accurate. To get a much better handle on all this, turn off the heater, and measure the temperature vs time during cool-down.

 

[sophiecentaur]

measure the temperature vs time during cool-down.

Yes. A cooling curve. Post #9.
I would think that the cooling curve over the range of temperatures so far will probably be pretty straight or the heating curve would be visibly curved. Unfortunately (for the experiment) the timescale involved will mean that the ambient temperature will probably vary so you would really need to be plotting a few other temperatures around the water tank. If one knows the Heater Power and the water mass, one can deduce the heat loss (at a given temperature or within a temperature range) by comparing the heating and cooling curve gradients.
Whatever happens, I think Dave has to commit himself to quite a long term experiment. A simple data logger, such as they use in schools, could take away some of the pain.

 

[Chestermiller]

Yes. A cooling curve. Post #9.
I would think that the cooling curve over the range of temperatures so far will probably be pretty straight or the heating curve would be visibly curved. Unfortunately (for the experiment) the timescale involved will mean that the ambient temperature will probably vary so you would really need to be plotting a few other temperatures around the water tank. If one knows the Heater Power and the water mass, one can deduce the heat loss (at a given temperature or within a temperature range) by comparing the heating and cooling curve gradients.
Whatever happens, I think Dave has to commit himself to quite a long term experiment. A simple data logger, such as they use in schools, could take away some of the pain.

Oops missed that.

 

[DaveC426913]

Tracking pretty much perfectly it seems.

The resolution of the temp scale is integer, so it's not surprising that it can vary from the ideal curve by as much as a degree.

I'm a bit surprised though to see it rise after short period since last reading. I'd have thought that, if anything were going to set the temp back, it would be opening the cover to check the temp - like checking roast in the oven too often or a steak on the BBQ.

 

[DaveC426913]

Well that was disappointing in its ... shortjevity.

Two days have passed and a leak somewhere has drained so much water that it's just blowing air.
The worst part is if I just unplug it, it will freeze.

 

[DaveC426913]

New gasket, another two hours to fill, and I'm back in busin...

Oh great. Now the pump seal has gone.

And that's not just a gasket. That's a new pump.



First world problems...

 

[Bystander]

Do you have enough DIY skills to feel safe rebuilding the pump on your own? The "after-market" on/for sub-assemblies is shameless.

 

[DaveC426913]

Do you have enough DIY skills to feel safe rebuilding the pump on your own? The "after-market" on/for sub-assemblies is shameless.

I've pulled pumps apart and gotten down to the nitty gritty, but ultimately it always comes down to badly worn bearings. This one has rusted out.
Getting the parts and replacing the bearings is ultimately not cost-effective in time and effort. For pros, let alone for little o' me, and without the right tools. And (my) DIY is never as good as the pros.

A new (or refurbished) $250 pump every few years is an acceptable price. (Repairing things is not my hobby. It takes too much time away from PF.)

What kills me is the delay. The first night the temp drops below freezing, the tub is in danger of becoming a much bigger problem.

 

[DaveC426913]

You can't just LIKE and walk away! You've got to argue!

"Ten for that you must be mad!"

Oy! Brutus! I got a guy here who refuses to argue!

 

[Bystander]

Now, THAT I likeLIIKKKE.