Heat absorbed by a room

How can one calculate the amount of heat absorbed inside a room in a single day using the temperature readings of the day?

If the temperature has to be reduced from 30C to 20C how much heat should be removed?

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migrated from theoreticalphysics.stackexchange.comMar 11 '12 at 17:26

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The second part of the question is straightforward to answer. Assuming a room filled with air, closed so that no mass could exchange with the outside, the energy required to be removed can be simply estimated as:

$$Q = \rho_{air} V_{room} C_{p,air} (T_f - T_i)$$

With $\rho_{air}=1.23~kg/m^3$, $C_{p,air}=1.00~J/gK$, all you need is an estimation of the room's volume to calculate the energy to be removed.

However, the first part of the question is a little more tricky to consider. A real room has heat being lost (or gained on a hot day!) through the windows, generated or lost by heating or cooling devices respectively and so on. Further, heat can also be transferred through exchange of mass, e.g. if the windows are open.

The amount of heat absorbed by the room is therefore subject to interpretation. Do you mean the difference in heat from the room's initial state as compared with its final state? Or do you mean the absolute amount of heat absorbed, accounting for the amount also lost through other means? The former can be easily calculated by the formula above.

The latter, however, requires consideration of all of the energy sources and sinks, and as such; is much more complicated to calculate.

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