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As I know, heat capacity of normal air is ~1kJ/kg and heat capacity of water is ~4.1kJ/kg.

I am trying to cool down room (around 6m³) from 30C to 20C just by cold water.

Let's say I have water with temperature of 5C in insulated tank and I am pumping it trough radiator what picking up heat from this room.

I just know that I need 4times less water than air, but I dont know how to add temperatures to this equation :D

How to calc how much of this cold water I need to drop 10C in that room? (this means that at the end, room will have 20C and water 20C or less)

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    $\begingroup$ See Newton's Law of Cooling $\endgroup$
    – Sam
    Commented Feb 14, 2020 at 14:11
  • $\begingroup$ The practical point of really doing this is that you are missing the energy stored in walls, ceiling, etc. Cooling down that volume of air (<1 kg?) is cheap, but having it rise its temperature also, and the building will keep radiating. So your air just couples your radiator to your real problem, the building. $\endgroup$
    – Andrestand
    Commented Jul 3, 2020 at 20:38

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Firstly it has to be understood that, given enough time, both water and air will end up at the same temperature, call it $T_e$.

Now assuming there are no heat losses to the rest of the universe, with the air initially at $T_{air}$ and the water initially at $T_{water}$ then the air 'loses' heat energy according to:

$$\Delta Q=m_{air}c_{air}(T_{air}-T_e)$$

and the water 'gains' heat energy according to: $$\Delta Q=m_{water}c_{water}(T_e-T_{water})$$

where the $c$ are specific heat capacities.

So we have:

$$\boxed{m_{air}c_{air}(T_{air}-T_e)=m_{water}c_{water}(T_e-T_{water})}$$

from which the unknown $T_e$ can be extricated easily. If $T_e$ is known then another unknown can be calculated.

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