# Cooling down room with just a water

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)

• – Sam
Commented Feb 14, 2020 at 14:11
• 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. Commented Jul 3, 2020 at 20:38

## 1 Answer

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.