Estimate a room's temperature based solely on outdoor measurement, and a well known average CPU temp. fluctuation?

Question

I'm trying to wrap my dyscalcic amoeba mind around the following:

1. There's a room, in a house.

2. In that room there's a PC, with CPU, GPU and PSU, running a very invariable and 100% spending task 24/7. If sampled at no external influences, the/their temperature would always report to be the same; The temperature the cooling fan(s) were able to provide. It is running at it's reins at all times, so there's no fluctuation on it's part.

3. for all intents and purposes, (and as to not explode my puny brain); Disregard time of year, sun position, barometric pressure, air moisture % etc. .. That one PC is THE ONE AND ONLY source of "external" heating influencing the room's ability to rize above outside temperature. Nothing else but perhaps night and day temp would be worthy of accounted for as deviation between the two.

The outside temperature can be sampled quite accurately at pretty much any desired frequency. The CPU temp. sensor of the PC inside the indoor room can also be sampled at any given time and rate.

How might I, if even possible; Estimate (if even ball-parking it) the temperature of the room where the PC is situated, based solely on it's providing temp. report, and the outdoor's fluctuating temperature perhaps lowering it?

Help my hangover, please dear Wizards and God 'o Mathematics; I got a headache just attempting to formulate my problem trough my dyscalculia, to you guys.. :(

Answer 1

The room looses energy through heat transfer at walls and windows. In a very simplified way, the energy loss per time unit depends on several fixed factors (total area of wall, wall thickness, material properties of the wall) and then is proportional to the temperature difference between inside and outside temperature.

The room gains energy from your PC. Essentially, precisely the electric power the device consumes will end up in room heating. In equilibrium, the energy loss equals the energy gain, so that the temperature difference room-outside turns out to be proportional to the electric power fed to the PC. Add about 100 W if you are also in the room.