# Water Heat Loss

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1. Oct 6, 2014

### gerry7

Hi all, I have a (probably very simple) problem that I need some help with. If I have a body of water in which the temperature of the water, ambient air temperature and mass of the water is known; how do I calculate the rate of heat loss from the water?

Using Q=m*cp*(Twater-Tair) I can find out the total heat lost when the water drops to ambient temperature, but how do i calculate the rate at which heat is lost? I should also note that the mass of the water is also dropping due to evaporation and liquid water loss. Can I integrate the above equation knowing the two mass values to get the rate of heat loss?

Or am I going about this completely wrong and need to take into account the surface area of water. (Heat is assumed to be lost equally in all directions)

Thanks,
Gerald

2. Oct 7, 2014

### sanka

The equation Q=mCpdeltaT will give you the energy lost from the water. This will be in Joules or kJ. If you can record the time it takes the mass of water to reach ambient temp, divide Q by this time to get the rate of energy lost. Units will be Joules/seconds to give Watts.
Regarding the evaporation aspect, if the water is cooling why is this happening? I don't understand this. In any case, if there is not a significant amount of water lost to evaporation, the calculation outlined above will be valid.

3. Oct 8, 2014

### gerry7

Hey thanks for the reply, yeah that's what I thought. I forgot to mention that I have hot air flowing through the water which would be the cause of evaporation.