Thermal Dynamics question for you techno science folk

NeilF

NeilF

NeilF

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OK, you have a beaker of hot water and a small cup of water at room temperature.

You want the water in the beaker to be as cool as possible in ten minutes time.

Do you:-
a) Add the colder water straight away and leave for ten mins?
b) Leave it for ten mins and add the colder water at the end?
 
B

Why let the cold water from the small cup battle against 50 degrees when you can wait for it to work less by cooling a mug at 30 degrees in 10 mins.
 
Quick guess would be add it later. Heat transfer is proportional to the difference in temperature. So really hot to really cold transfers heat quicker than room temp to really cold. If you don't mix the water then the cold water will stay the same temperature but the beaker will cool down quicker.

That said, with a small cup it doesn't sound like there will be enough water in it to cool it down significantly.
 
I reckon B. Heat loss through radiation in the beaker with be directly proportional to temperature of the beaker, so the longer you leave it at the higher temperature, the greater the heat loss through radiation.

Rgds
 
The heat from the warm water will transfer to the colder water in the same room to balance the temperatures by heating the surrounding air and raising the ambient temperature.

So it doesn't matter when you add the colder water:
A) add colder water straight away, overall temperature will average out over the two after 10mins
B) leave them separate for 10mins, the warmer water will transfer it's heat to the colder water as mentioned above so when you mix them 10mins later, the overall temperature will still be the same.

PROBLEM?
 
Six6siX said:
The heat from the warm water will transfer to the colder water in the same room to balance the temperatures by heating the surrounding air and raising the ambient temperature.

So it doesn't matter when you add the colder water:
A) add colder water straight away, overall temperature will average out over the two after 10mins
B) leave them separate for 10mins, the warmer water will transfer it's heat to the colder water as mentioned above so when you mix them 10mins later, the overall temperature will still be the same.

PROBLEM?
Click to expand...

Sorry - that's not right; while the warm water will transfer heat / energy to the room, the room won't transfer all of this energy to the cold water. The energy is transferred to the surrounding air and surface the beaker is sitting on via convection / conduction. Therefore energy will be lost to the surroundings, the question is is it more or less depending on whether you add the colder water sooner or later?

As radderfire says the heat loss is proportional to the temperature difference between the hot water and the surrounding room temperature as the hot water will be at a higher temperature before the colder water is added the rate of energy loss will be greater - therefore colder when the two are mixed 10mins later.
 
B.

You are losing heat energy proportional to the difference in temperature between the hot water and the air (... this leads to an exponential decay in temperature over time, by simple A-level standard maths). By leaving the hot water on its own, you're keeping this temperature differential high, and so keeping the rate of heat energy loss high.

The room-temperature water does not gain or lose heat energy, so nothing happens to it (in terms of energy change) while the hot water is cooling. There is no counter-balance to the effect described above...



edit: see here for mathematical proof, subject to reasonable assumptions about the shape of the pots.
 
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hmm wouldnt you add the cold water first and let it sit for 10 mins?
as the thermal heat transfer in water is far better than in air?
so it should cool down quicker that way.
 
I have just proved mathematically that there is no difference using the following:

* assume mixing is instantaneous when mixing beaker's water and cup's water

* Temp(time) = T_initial . exp(-time / tau) + T_environment

* tau is some constant
 
NeilFawcett said:
Wouldn't affect the answer though would it :)
Click to expand...

Well it depends on the function for heat loss. I'd imagine it's proportional to mass, surface are, ambient temp and the temp of the mass. The answer would depend on the increase in rate based on increase in SA in relation to the reduction in rate due to the lower temp of the mass once mixed. It's been a while since I've studied thermodynamics, so I can't be certain lol.
 
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Isnt the rate of heat loss proportional to the 4th power of the temp difference?

As small container is already at room temp then it may gain a (very) little temp from any rise in room temp from large beaker as it cools down.

Big beaker is much hotter than ambient and so leave it to cool for 10 minutes first to maximise temp difference and thus rate of heat loss and then add cooler water.

Real world the differnce between each option will not be that great.
 
I've always put my milk in my tea asap to help it stay hotter for longer:D

Isnt the rate of heat loss proportional to the 4th power of the temp difference?
This is correct (i.e. proportional to t1^4-t2^4)
 
jak731 said:
I've always put my milk in my tea asap to help it stay hotter for longer:D

Isnt the rate of heat loss proportional to the 4th power of the temp difference?
This is correct (i.e. proportional to t1^4-t2^4)
Click to expand...

you're mixing up some principles there.

t^4 is stefan-boltzmann for black body radiation. It is not affected by local temperature.

and "4th power of temp difference" is (t1 - t2)^4, not what you wrote.
 
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