A 10-Pound Bag of Ice and Cooling Room Temperature

[Pengwuino]

Our air conditioner has broken so i was bored and started thinken about something.

I calculated that a 10 pound bag of ice will cool my room (which i figured is about 4500 ft^3... yah... i love my room lol) from about 86 freaken degrees! to about 72 degrees fahrenheit. In my calculations i used the specific heat of dry air by the way :-/... but this is more of a excercise then hoping for a real world solution. Anyhow, does that 14 degree drop seem reasonable?? Not looking for an exact figure or anything because i know there's too many things going on for that that i couldn't account for... just wondering if it sounds reasonable.

 

[FredGarvin]

That delta T seems very high to me, especially without any forced convection. A 10 pound bag of ice is only a couple of bucks. Why not do an experiment?

 

[ramollari]

Looks resonable! $$L_{ice} >> c_{air}$$

 

[ramollari]

That delta T seems very high to me, especially without any forced convection. A 10 pound bag of ice is only a couple of bucks. Why not do an experiment?

It wouldn't work in practice because the ice needs time to melt and meanwhile hot air and radiation come continuously from outside.

 

[Pengwuino]

Yah there's a lot that wasnt taken into account... it was just a thought experiment to say the least. I might put a big block right next to me though and see if it'll just cool me down lol

 

[russ_watters]

It wouldn't work in practice because the ice needs time to melt and meanwhile hot air and radiation come continuously from outside.

True...

However, the calculation itself looks sound: it just takes into account the static situation the ice absorbs heat from the air - in a perfectly insulated room.

Just to make sure, you did add both the heat absorbed by melting the ice and the heat absorbed by heating the water from 32F to 72F, right?

Little random, useless tidbit - the cooling capacity of air conditioners is measured in "tons" which is actually referring to tons of ice, used in springhouses. Someone (maybe Fred) said recently that that's the rate of cooling provided by melting a ton of ice in a day, but I thought it was a longer timeframe. I guess I could calculate it if I wasn't so damn lazy...

 

[Pengwuino]

Yah i added in the melting water along with it heating up to ambient... air conditioners worken now i think so I am happy though :D

 

[FredGarvin]

One ton = 288,000 btu/day which is approximately the heat of fusion of 1 ton of ice at 32°F

 

[quark]

Fred is right about the definition of Ton(actually TR-Ton Refrigeration).

You can extract about 184 btu per pound of ice(144btu/pound latent and 40btu/pound sensible heat). So this will aproximately reduce the room temperature by 22.71F(184*10 = 0.075*4500*0.24*dT)

But the average heat load for a residential building will be 1ton/450 sq.ft so considering a 10ft. height room, you have to remove heat at the rate of 12000btu/hr.
The melting ice can provide you comfort for approximately (1840/12000)*60 = 9.2 minutes.

 

[Q_Goest]

The calculation sounds about right. The problem with actually doing this in a room is the large amount of thermal mass in the room. All the furniture, walls, everything in the room will also reject heat to the air. Because of the huge surface area, that heat rejection is relatively efficient - the heat from the room/objects in the room can overcome the cooling of the ice very easily.

Note: The thermal mass of the room/objects is a transient which is different from the steady state heat transfer. You won't get much heat transfer from the environment until you cool down the thermal mass of the room/objects.