How much heat does an individual PC or PC server add to the room temperature for an average size room?

Answer 1

From the specific heat equation: c = Q / (m * T), the density equation: d = m / V and the power equation: P = Q / t we have: T = P * t / (V * d * c) where: T = temperature delta, in °C, P = heat power dissipation, in watts, t = time, in seconds, V = volume, in m³, d = density, in kg/m³, c = specific heat capacity, in J/(kg*K). Considering the air at 20°C we get: T = 8,257e-4 * P * t / V. Supposing a room of 50m³ and a heat dissipation of 100W we'd have a raise of 5,94ºC per hour. If we double the power, we double the increase in temperature. Doubling the room volume cuts the increase by one half. All that ignores some small changes in air properties which vary with temperature and the dissipation across the walls and windows in the room. The later is particularly important, as the calculations assume the room does not leak any heat. To get the right number one needs to know also exactly what amount of heat the computer generates. (It's safe to assume all power consumption from a computer is converted into heat.)