# Cooling a room with 2 liters of ice: calculation.

Lately, while I've been waiting for my new AC unit to arrive, I've been freezing a 2L bottle of water and putting it in front of my window fan to cool down my living room in the evening.

Anecdotally, it seems to work... but I've been wondering if it's actually working or if there's something else going on.

Maybe it's just the cooler evening air combined with my idleness in the evenings that's cooling me down. Fortunately, the calculations are quick and relatively painless. Like most (ok, maybe not most, but a lot) of engineering, the difficulty is not in the arithmetic but rather the set-up of the problem, and knowing which assumptions to make and what formulae to use.

**The Set-Up**

We have a 2L bottle of water. Two liters of water has a mass of 2 kg. The whole bottle gets frozen and then placed in front of a window fan. Lets assume that the ice, as soon as it's placed in front of my fan, is -10 C (14 F). We'll assume ambient temperate is 25C (77F) -- but this ends up not mattering because I never let the water bottle warm to room temperature anyway. Instead, let's assume that the bottle warms to only about 10 C (50 F) before I remove it from the window. Assume that the efficiency of heat transfer from the bottle to the room is 100%.

One final assumption: let's assume that we don't care that the kitchen will heat up due to extra work the refrigerator has to do to freeze the water.

The room itself measures (and I'm mixing units here, I know) 20' by 12' by 8'--we'll need the mass of that volume of air though, so let's just ask Wolfram Alpha. The site reports that my living room has 72 kg of air in it.

So the 2kg bottle of ice starts at -10 C and warms to +10 C in a room of 72 kg air. The question we're looking to answer is: how much can that block of ice cool the room? As a corollary to that (since we know that the following won't be terribly accurate, and it won't apply to *your *living room either), we can more generally ask: **will a 2L ice block noticeably cool a room, or would it only change the temperature by a silly amount, like a tenth of a degree or less?**

**Heat Capacity**

Thermodynamics has a property called "heat capacity." Heat capacity is the amount of energy required to change the temperature of a substance by one degree (in whatever scale you're using). Slightly more useful is the "specific heat capacity", or "specific heat" for short, which is the amount of energy required to change the temperature *of one unit mass* of a substance by one degree. That distinction is important: heat capacity is how much energy you need to change a whole block of a substance, but substances come in different shapes and sizes, so it's more useful to use the specific heat capacity and then calculate the "overall" heat capacity if you know the mass of the thing you're looking at.

Anyways, we can look up the specific heat of ice and find out how much energy it takes to bring our 2kg of ice from -10 C to 0 C. We need to do this in steps because *melting* ice also has its own energy requirement (called latent heat). Because we have that phase change in the middle, our ice block actually does three different things in sequence:

- warms from -10 C to 0 C (as ice the whole time),
- melts completely while remaining at 0 C (as a mix of ice and water the whole time),
- and then warms from 0 C to 10 C (as liquid water the whole time).

The specific heat of ice at -10 C is 2.1 kJ / kg * K (from Wikipedia). Those units are "kilojoules per kilogram-degree Kelvin"--and again, that's the amount of energy (kilojoules) it takes to warm a kilogram of ice one degree Kelvin (since we're only concerned with relative temperatures, we can freely substitute Kelvin with Celsius. When I write out units or equations I'll use K, but when I'm describing the problem scenario I'll use C in this post).

To calculate how much energy goes into warming the ice to 0 C, we'll just multiply our specific heat by the temperature difference and the mass. Here's the general formula:

ΔE = m * Cp * ΔT

Plugging in values, we get:

( 2.1 kJ / kg * K ) * ( 2 kg ) * ( 10 C ) = **42 kJ**

(Note that the equation above is a little rearranged--I put down the numbers in this order: Cp * m * ΔT = ΔE)

Warming our ice from -10 C to 0 C requires 42,000 Joules of energy. That energy comes from the air in the room, and when the air in the room loses that energy, it cools down.

Let's see how much energy it takes to warm the water from 0 C to 10 C. The specific heat of water at that temperature is roughly 4.2 kJ / kg * K:

( 4.2 kJ / kg * K ) * ( 2 kg ) * ( 10 C ) = **84 kJ**

Finally, we need to figure out how much energy goes into converting our ice to water. This property is called the "latent heat" (different from heat capacity), and the term "latent" is used there because the temperature of a substance stays the same during a phase transition. Phase transitions (like ice to water, or water to steam) stay at a constant temperature, and also usually require a lot more energy than raising the temperature. I wouldn't be surprised if the energy required to melt the ice is several times larger than the two values above.

The "latent heat of fusion" (fusion is the ice -> water or water -> ice transition) for water is 334 kJ / kg. Note that there's no degree Kelvin dimension there; that's because the temperature won't be changing during this calculation. Therefore, the energy required to melt our ice is:

ΔE = Lf * m

Or, plugging in numbers (again, rearranged to have ΔE on the right side):

( 334 kJ / kg ) * ( 2 kg ) =** 668 kJ**

As we guessed earlier, that figure is much larger than the two above. Almost all the heat that goes into warming our ice is taken at the phase transition, and for that reason many engineers will simply ignore the two other steps in the sequence because they're an order of magnitude less important. We'll keep them in, just to be thorough.

Add up the energies required by the three steps above, and we get the total energy required to warm our ice:

668 kJ + 84 kJ + 42 kJ = **794 kJ**.

Let's just call it 800 kJ.

Finally, let's see how much our 800 kJ will change the temperature of 72 kg of air. The specific heat of air is 1.0 kJ / kg * K. We have 72 kg of the stuff in our room, so all we have to do is rearrange our first formula above to solve for ΔT:

ΔT = ΔE / ( m * Cp )

Plugging it in:

ΔT = 800 kJ / { ( 72 kg ) * ( 1.0 kJ / kg * K) } = **11 C**

(Again, Celsius and Kelvin are interchangeable here).

What a result! The math says that a single 2L block of ice is capable of cooling a room down by 11 C (20 F)!

Now, there are other things going on that are stopping my room from turning into a refrigerator. For one, my living room isn't an air-tight box, and we're probably talking more along the lines of 4,000 cubic feet of air (from the hallways, attached kitchen, etc) that need to be cooled. Additionally, this process certainly isn't 100% efficient. I don't even have a good guess towards the efficiency of the effective heat transfer from the bottle to the room--it could be anywhere from 50% to 80% (surely lots of cool air is just bouncing right back out of the window, or is cooling the back of my couch). Assuming 50% efficiency already cuts us down to a 5 C (9 F) difference, and doubling the volume of air cuts us down further to 2.5 C (5 F).

But still, a 2.5 C (5 F) cool-down in the middle of the summer without an AC unit is absolutely worth the minimal effort it takes to freeze a 2L bottle of water during the day and stick it in front of the fan at night. It absolutely works, and you just saw the math that proves it! Even if the calculations above have an 100% error margin, it's still a significant result--because the real goal of this exercise was to see if the frozen bottle would cool a room something significant like 1-10 degrees, or something insignificant like 0.01 degree.

So I guess I should re-word my conclusion: the above order-of-magnitude calculation shows that a frozen 2L bottle can noticeably cool a small or moderately sized room.

**Edit**: a friend asked if I had considered the heat given off by a human body. Several Google sources say that a human gives off ~ 100 W (one Watt = one Joule per second). If you sit in a room for an hour, you'll provide 6,000 J energy to the room -- or a temperature increase of only 0.1 C! This leads us to another interesting conclusion: turning off the lights (if you only have one in the room) doesn't reaaaaallly make it any cooler. If you have 10 bulbs, that's a different story... but the 60 or 100 W bulb doesn't make much of a difference in a large room.