Question I was pondering ...

Smoggy

I make a bath for my baby each night and use to do 2 fully boiled kettles and 3 cold water kettles. Now just because I could I did 3 part boiled kettles and 2 cold water kettles. The temperature ended up the same.

Yet it seemed quicker doing it the part boiled method, is there a calculation that will prove me right one way or the other ?

Thanks
Smoggy.

Podge_irl

A stopwatch?

Smoggy

I guess that would be the easiest way.

I will try again tomorrow, this time keeping an eye on the clock.

SOL

I'd say the party boil method is definitely faster, as there are atleast two factors in it's favour I can think of off hand,

Firstly, as you heat the kettle up more than the surroundings it will lose heat to them, the higher you heat it the faster it starts losing heat, and the less the difference between the heating element so the slower it gains heat...

Secondly and probably much more importantly, as you start to get towards the boiling point a lot of heat gets lost into making steam as well as having to undissolve all the dissolved gasses, so you have to put much more heat in...


That said I have no rigorous calculations to offer, so if anyone else does I'd love to see them...

Arciphel

It may do it quicker, as SOL said as you try and increase the water temperature more then the delta T increases and therefore it uses more energy, or in a system where energy input is constant it will take more time.

Another interesting variation would be to alter the ordering of the hot and cold kettles. Try one cold kettle - one hot kettle - one cold kettle - one hot kettle - one cold kettle. This way you are using the residual heat in the element to partially heat the cold water.

Or you could just turn on the immersion! :pac:

FISMA

Smoggy wrote: »

I make a bath for my baby each night and use to do 2 fully boiled kettles and 3 cold water kettles. Now just because I could I did 3 part boiled kettles and 2 cold water kettles. The temperature ended up the same.

Yet it seemed quicker doing it the part boiled method, is there a calculation that will prove me right one way or the other ?

Thanks
Smoggy.

Smoggy,
Definitely check the final temperature with something more accurate than your hand.

If I am reading this correctly, then there has to be a difference in temperature.

You have case 1 (2 boiled and 3 cold) and case 2 (3 boiled and 2 cold). Case two has to have a higher final temperature, it's just energy conservation.

Heat is just energy in motion and energy must be conserved.

In case 2, you have a greater mass of hot water which yields a higher average temperature for the five gallons.

Again, you're conserving energy. When you mix the liquids, one gets hot (heat gained) and the other gets cooler (heat lost). But what one gains, the other loses - conserved.

Heat gained or lost is represented by Q. Since you are talking about water, we'll say that there are no phase changes: melting, vapor, condensation. That makes things easy.

You'll need to measure a few things: mass of the water (or volume), initial temperatures, and final temperatures.

Then how you want to prove it is up to you.

Simple case, get the baby's thermometer and stick it in the final bath of water.

Question: are you using a steel sink? That could be causing a result like the one you are talking about.

Also, are you waiting of the kettles to boil or do you have 2 or 3 available on demand?

Slan

Arciphel

FISMA wrote: »

Smoggy,


You have case 1 (2 boiled and 3 cold) and case 2 (3 boiled and 2 cold). Case two has to have a higher final temperature, it's just energy conservation.

Devil is in the detail - in the case where he has the 3 boiled, he states that they are only part boiled, not fully boiled.

FISMA

Arciphel wrote: »

Devil is in the detail - in the case where he has the 3 boiled, he states that they are only part boiled, not fully boiled.

Ah ha!

When I read that, I thought the OP meant it as in 2 parts boiling and three parts cold...