Source
r/theydidthemath•
CalicoFox

[Request] How much heat does my laptop produce in a 12-hour period and by how much does this raise the temperature of an 11'x11' room?

Okay, it's been hot here, lately, and my laptop is also hot (even though I've got a cooling pad underneath). I was wondering how much heat my laptop gives off in a given day. (I'm averaging 12 hours, though I may be leaving it on longer than that.) I was also wondering in an enclosed bedroom that's about 11' by 11' by how much does the laptop heat up the room? In other words, would I be noticeably cooler if I turned the laptop off during the day?

Thanks in advance! :)

Posted Aug 1, 2015, 7:08 PM UTC

916
hiworldtomv1✓•

We would need to know the watt rating of your charger to give any accurate answer, also if your windows are covered/double glazed and if your house has insulation.

Also keep in mind that your body releases 100 watts of heat so that's going to have a large effect as well.

Bottom line, yes, it would make a difference of maybe 1-5 degrees C, at a guess.

7
CalicoFoxOP•

watt rating of your charger

Where do I find this?

There are curtains on my windows, if that's what you mean by covered, and they're definitely not double-glazed. Also, I have no insulation.

0
hiworldtomv1✓•

In the middle of the cord that runs from the plug to your computer will be a small black box that has numbers on it. What are the numbers.

2
CalicoFoxOP•

There are a lot of numbers on it, but none of them say watts. It does say that the input is between 100 to 240V, ~50 - 60 Hz, and 1.5A.

1
hiworldtomv1✓•

I'm assuming you live in America so you're on 110volts if memory serves. The watt rating is simply amps x volts, so your laptop gives off approximately 160 watts of heat. Compare that to a toaster, which is rated at maybe 1000 watts or more.

So your little laptop gives off not much heat really, most of it would be absorbed by the walls, and realistically it would raise the temp in your room a degree or two.

Why do you want to know?

2
CompellingProtagonis1✓•

A simple answer depends on two things: the average power usage of your laptop and how well-insulated the room is. If we assume the room is perfectly insulated it becomes a heck of a lot easier. I'm also going to assume that your laptop converts 100% of the energy used to power it to heat.

Let's say your laptop is consumes power at a rate of 50 watts on average.

Then the total amount of energy being converted to heat by your laptop is:

50W *12h * 3600s/h / 1000 kW/W = 2160 kJ

This is the total amount of energy being introduced into the air by the laptop in a 12h period. Now the question is: what is extent to which this influx of energy will change the temperature of the room?

To start with this: I'm going to make some other assumptions, the first, most important one, being that energy is distributed evenly throughout he room. The second one is that there is only air in the room. The third is that you have 8 foot ceilings. The fourth one is that your room is at 25 degrees celsius, and the 5th and final one is that the airs at 0% humidity.

The first step here is to figure out how much energy is being put into each unit-mass of air (kg in this case, because that is the units my old thermo book uses.

The Volume of the room is:

11ft * 11ft * 8ft * (12/39ft/m)^3 = ~38.75 m^3 of air

(I'm as proud as a 'merican as the next guy, but for doing stuff like this SI is just way better)

at 25 C, the density of air is: 1.184 kg/m^3, so in your room you have:

1.184 kg/m^3 * 38.75 m^3 = 45.88kg of air

So, going back to the second assumption, that the energy is distributed evenly (which is the natural end-state for the system), your computer is introducing:

2160 kJ / 45.88 kg = 47.1 kJ of energy to each kg of air

Now the final step is to figure out exactly how much this raises the temperature by (it's going to be a lot, our assumptions turned your room into a slow-cooker).

At 25C, or 298K, the specific heat of air is ~1.005 kJ/kg*K. We already have the energy introduced per unit mass, so to figure out the change in temperature is very simple:

Temp change = Heat Added per kg(kJ/kg) / specific heat (kJ/kg*K) Temp change = 47.1 / 1.005 = ~46.8K

It's actually a bit below this, because as the temperature of air increases, so does it's specific heat, so let's say 46K

So, given the assumptions I have outlined, your computer is introducing enough energy to raise the temperature of your room by 46C over the course of 12 hours! This seems like a lot, but when you consider that there really isn't much air in your room, I took out ventilation, and I took out heat sinks (walls, bed, plants, the actual body of the computer, etc), it's not too surprising.

Just remember to turn your fan on and open a window!

3
AnAppleSnail•

The walls cool the air.

2
CompellingProtagonis1✓•

"Perfectly Insulated". The extent to which the walls cool the air is extremely variable. I calculated the answer based upon the minimum amount of assumptions possible. OP didn't mention how high the ceilings are, and you expect me to assume building material and wall thickness? Even if I were to go that route, without $10,000 FEA software, which is the only way I would realistically solve a problem that approaches the complexity required if I add in walls, windows, glass, etc, it would take many hours, assuming I could do it at all.

2
AnAppleSnail•

I find this problem with a lot of models. Calling something a simple assumption does not make it a good one. "No heat leaves the system" is a terrible assumption.

Make the model and compare it to reality. If it doesn't match, change the model.

There is simply no way for a laptop (150W) to heat up a normal room by 46C. Giving this answer with a straight face means you ignored problems with your model. Here is a linear approximation for heat flow out of a room:

Tinside - Toutside x 2 W/m^2.

So a room 3x4x2 has 60 square meters, which can transport about 120W per 1C temperature difference.

So a normal laptop will manage about 1C rise. A very powerful computer may manage a 15C rise.

2
CompellingProtagonis1✓•

That's a fair point, but the overall purpose of presenting the model is to answer OP's question, which is: "will the laptop heat the room appreciably over a time period of 12 hours?"

I think that the model demonstrates that there is a significant amount of heat being introduced to the room, thus it is sufficient for the purposes of the problem. There is a difference between presenting a model as reality, and presenting a model as an edge case.

I will grant that I should have explicitly stated that this was an edge case, but I really don't think it is unreasonable of me to not want to attempt an exhaustive solution to the problem, which is the implication I am getting from your statement.

2
CalicoFoxOP•

You were right when you assumed the ceiling was about 8 feet. The walls are thin, not insulated at all, and I don't know what materials they're made out of... The average weather for where I live shows that the humidity is typically between 38% and 90% throughout the year. (I thought it was drier here, but I was wrong.) This summer has been unusually humid, too, which is uncomfortable.

I had thought that there would be a lot of variables to consider, but 46℃ does seem like too much, especially since my room isn't insulated. With the assumptions you gave, it makes more sense.

(Edit: Fact checks.)

1
CalicoFoxOP•

✓ Thank you for taking the time to answer my question. I realize that there are more variables to consider for a more accurate answer, but given the assumptions you put forth, it makes sense.

For the record:

  • The ceiling is about 8 feet tall. (You guessed right.)
  • I have four windows.
  • The walls are thin and not insulated.
  • There is a bed in my bedroom. ;)
1
TDTMBotMODBeep. Boop. •

Confirmed: 1 request point awarded to u/CompellingProtagonis. ^[History]

^View ^My ^Code | [^Rules ^of ^Request ^Points] (http://www.reddit.com/r/theydidthemath/w/RPRules)

2