1
$\begingroup$

A question on how to apply thermodynamics principles to figure out how much work is needed to hold 300ml water in room temperature at 3°C.

So far I have:

1Cal for each degree per g of water. We have 22 degrees delta, and 300g.

25->3°C = 22Cal * 300g = 6.6kcal ~ 30kJ

then; W = J / s

W = 30000 / 3600 (hour)

8.3 W/h

Plus the heat that is absorbed from the surrounding air (which will later become the maintenance-work for the device).

If the above is not correct, what are some pointers for things I should be studying better? if the above is somewhat accurate, how do I figure out the next step, i.e. the heat gained from surrounding air? Most material I've found only deals with heat loss from evaporation in a heated system...


some random specs in case they help

mass = 300ml/300g water at sea level

ambient temperature = 20~30 Celsius. Anything in that range that makes the calculations easier. I'm set at 25 for no reason.

water area in contact with ambient air: 125mm2

container: cylindrical, r=40mm h=200mm (think a tall drinking cup), base metalic (Al) in contact with heat transfer element, top open. Walls thick ceramic or glass (can I consider it as ideal insulation since the open top would lead the heat loss/gain calculation anyway?)

$\endgroup$

closed as off-topic by David Z Apr 20 '15 at 4:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

This might be better on the engineering SE site but here is some physics to consider:

You are right the the heat you need to remove is the mass of water times the temperature difference times the specific heat capacity.

Peltier devices and other heat pumps typically have a parameter called a COP - coefficient of performance. This compares their efficiency with that of a perfect heat pump (Carnot engine). Efficiency is a function of temperature difference - the greater the difference, the harder you have to work to get heat across. Peltier devices usually have a maximum temperature difference beyond which they cannot go.

This means that in order to answer your question you need to give thought to the heat sink of your Peltier - if the "output" becomes hotter than room temperature the device becomes less efficient.

And finally, the ability of a surface to absorb or give off heat to the environment depends a LOT on air flow. Stagnant air acts as an insulator while flowing air carries heat efficiently.

So there are a few things you need to know more about before you can solve your problem. These are just some pointers for further thought.

$\endgroup$
  • $\begingroup$ thanks for confirming the 1st part @floris. I want to keep this to the physics of the work required, because the device parameters (peltier or not) can change depending of that. you can consider the peltier/heatsink to be ideal for this problem. in fact, even the cheapest peltier module can handle way over 60C deltas. the 25C delta required here is trivial for their efficiency range, i hope. Good point on the stagnant air. so i think the cylinder being much deeper than the water volume will help. i can probably consider most of that volume to be air insulation. $\endgroup$ – gcb Apr 20 '15 at 1:22
2
$\begingroup$

You are sloppy with units, but the result is correct. To go from 25C to 3C is 22 cal/g. When you multiply by 300 g you have cal and your conversion to kJ is correct. Converting to W-hr is silly, but that is the unit of energy, not W/hr. You have 8.3 W-hr you want to remove. That chills the water assuming no new heat is added, so insulate the water. Otherwise you need to worry about the rate of heat input vs. the cooling capability of your Peltier device. Your radiator will rise above ambient temperature, so you need not worry about heat absorbed, you need to worry about heat rejection capability. Your calculations show clearly that the Peltier device must reject its heat somewhere that does not heat your system. If any of your heat rejection comes back into the system you are toast.

$\endgroup$
2
$\begingroup$

Q = mc(t1-t2), Now, m = (density)(volume), Specific heat of water, c(in joule/gramCelsius) = 4.186,

Hence, you can find the energy it would require for this conversion. . And the work you do can be a bit more pertaining to your efficiency.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.