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)
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?)