[Closed] Help with temperature calculation...

I'm not a clever man, but having pondered it for a minute, I'm going with never.

Surely heat won't transfer well in that scenario and even then for the small amounts that may do, wouldn't the room then drop from 20 as it loses heat energy to the water?

(EDIT: I've not assumed a constantly heated room for the 'drop' part of theory)

Interesting. I'd also say never (or at least 'a very long time')
edit: but I'd like to know why.

I'm no Einstein but I think you would need to know a few more details such as material of bucket & thickness, bucket dimensions to work out surface area or something and more.

The system will reach an equilibrium temperature, the bucket will warm and the room will cool. It depends on what's heating the room. Is it always at 20 deg (some external heat source). If so heat will flow from the hot body to the cold body. To work out how long you have to specify how much water, what's the bucket made from etc.

If this is some cunning plan to turn a cold bucket of water into a warm bucket of water to wash your bike with upon returning from your ride?

Too many variables I would think. As has been said the cooler water will try to reduce the temp in the room. How is the ambient temp maintained? Is it allowed to drop? What material is the uninsulated bucket? Metal / plastic which will have effect on the heat transfer. How large is the room? The smaller volume will potentially drop in temp more than a larger volume - if the ambient temp is not maintained.

I'm not a physicist but I think you would have to get the answer by practical experiment - empirical?

Thanks guys, this doesn't need to be exact.

I'm thinking of a real world scenario where an average sized bedroom is kept at a relatively steady 20 degrees and a plastic bucket (say 2mm thick) with 16 degree water is placed in it.

At what sort of rate would the temperature in the bucket increase to reach room temp (or thereabouts)?

Nottingham Uni now do Physics videos as well as the periodic ones. maybe something there to help with entropy?

http://www.bbc.co.uk/schools/gcsebitesize/science/aqa/heatingandcooling/buildingsrev3.shtml

There's an exponential decay in the rate of change of temperate, as the temperature difference between the bucket and the room changes.

The new temperature is equal to the original temperature multiplied by e raised to the power of (some constant multiplied by time). The constant depends upon the surface area of the bucket and its insulating properties.

That is a fantastically difficult question to answer as you have all sorts of variables. e.g. the mass of water, the area available for heat transfer, the heat transfer co-efficients of the air and the water (assuming the bucket is made of metal this will be irrelevant), how is the water mixed (assuming that it is at all) how is the air mixed.

Basically other than "quite some time" you don't have a hope of getting anything close to an accurate answer.

I'm thinking of a real world scenario where an average sized bedroom is kept at a relatively steady 20 degrees and a plastic bucket (say 2mm thick) with 16 degree water is placed in it.

Is this something to do with penis beakers?

If you've got form for sleepwalking it'll probably be nearer 36.9 degrees before you know it.

You can work out exactly how much energy needs to be transferred from the room to the water, but working out how quickly this will occur is, as pointed out, very hard. The easiest way would be to actually do it, measure the temperature over time and plot a graph empirically. Even if you were to build a computer model of the system, most of the input for that would need to be from various empirical experiments anyway.

Basically other than "quite some time" you don't have a hope of getting anything close to an accurate answer.

I don't need accurate. For the water to rise by a few degrees, approaching the room temp, are we talking, very roughly:

0-12 hours
12-24 hours
multiple days

I suspect I'll just give it a bash and monitor the temp a few times (i'll post my findings)!

in that case then, less than 12.

I may be wrong, but...

time(secs) = mass water (kg) x Specific Heat Capacity water (j/kg.°C) / Overall Heat Transfer Coefficient (W/m2.°C) / Area of bucket (m2) x natural log ((Temp air - starting water temp)/(Temp air - final water temp))

Assume 10kg water
OHTC value of 15W/m2.°C (similar to household convection radiator)
Area = 1m2

17°C = 800s
18°C = 1931s
19°C = 3863s
19.9°C = 10279s
20°C = to infinity (and beyond)

I THINK!!!!

EDIT, although the bucket is probably more like 0.5m2, and this method doesn't allow for any heat held up in the mass of the bucket. And it assumes that the bucket is metal!

Sorry for the hijack but I have a piece of string...

You can increase the rate of heat transfer by increasing the surface area. Throw your bucket of water on the floor for faster results.

It will never reach 20°C. Evaporation will keep the water below ambient temperature until there is no water left in the bucket.

Well your very basic numbers are 1 calorie per gram of water per °C. You've got 20,000g of water all needing a temperature increase of 4°C. So that's 80,000 calories required. You just need to figure out how long it will take for that amount of heat transfer to take place between air & water, air & plastic and plastic & water.

Edukator - git, I wanted to make that point 🙁

Don't forget that, as the bucket is full of fermenting homebrew, it'll generate its own heat

Have no idea how to do the calculation but I'm dying to know why you need to work this out.

Ormancheep, you've also assumed that there is perfect thermal mixing of the water within the bucket which won't be the case in reality.

You left out a pile of variables which makes the calculations impossible. Come back with at least the dimensions of the bucket. Someone else can then do the maths as my thermo is way too rusty. Edukator's spot on. The temperature will tend to approach 20C, but never, ever actually get there: even without evaporation, the graph just tails off towards 20.

Don't forget that, as the bucket is full of fermenting homebrew, it'll generate its own heat

Aye. Mine is kept in a temp controlled chamber, but you've got the gist of my OP 🙂

Basically, there's a number who suggest that in the absence of proper temp control, you're better pitching cool so as to mitigate the effects of the exothermic fermentation. ie, if your room is 20 degrees, pitch at 16 (despite most yeast manufacturers recommending pitching at 20ish).

It doesn't make sense to me to pitch cool, since there's a lag until fermentation kicks in, so surely the temp of the wort will have risen to almost room temp anyway before it starts? So it's irrelevant whether you pitch at 16 or 20, the wort will be pretty much the same temp by the time active fermentation starts?

You left out a pile of variables which makes the calculations impossible.

If I knew what variables would be needed, I'm sure I could have worked it out myself with a bit of help from google 🙂

This thread needs a conveyor belt

[i]Sorry for the hijack but I have a piece of string... [/i]

measure the distance from one end to the middle. Double it.

hth.

Edukator - Troll

It will never reach 20°C. Evaporation will keep the water below ambient temperature until there is no water left in the bucket.

Edukator - git, I wanted to make that point

Don't worry, he's wrong, the room will reach 100% RH (assuming the walls of the room are well insulated to prevent condensation and there's no airflow). Then no more water will evaporate as the vapour will be in equilibrium with the water.

Asside from that, OrmanCheaps method is probably close as long as the heat transfer from the air to the bucket is an order of magntude slower than the heat transfer through the bucket and into the water, there's a lid on it and the room is of average humidity. Stick a kettle in the room and it'll heat up quicker as the cold bucket will cause condensation on the outside which is a much better method for heat transfer than relying on conduction from air.

Is the cow spherical and in a vacumn?

You can work out exactly how much energy needs to be transferred from the room to the water

Quite some time. I do this with bottles of fishtank water.

Why not just cut out the guesswork and stick a probe in it?

or just pour a bit of hot out of a kettle in there?