Calculating Rate of Heat Loss from Water

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SUMMARY

The discussion focuses on calculating the rate of heat loss from water using the equation Q=m*cp*(Twater-Tair). The key insight is that to determine the rate of heat loss, one must divide the total heat lost (Q) by the time taken for the water to reach ambient temperature, resulting in a rate measured in Watts (Joules/seconds). Additionally, the impact of evaporation is acknowledged, particularly when hot air flows over the water, which can affect the mass of water and thus the calculations. The importance of considering surface area for accurate heat loss calculations is also highlighted.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with the specific heat capacity (cp) of water.
  • Knowledge of basic calculus for integrating mass loss over time.
  • Experience with measuring temperature and time in experimental setups.
NEXT STEPS
  • Research the effects of surface area on heat loss in fluids.
  • Learn about the specific heat capacity of various liquids.
  • Explore the principles of evaporation and its impact on thermal dynamics.
  • Investigate methods for measuring heat transfer rates in real-time experiments.
USEFUL FOR

Students and professionals in physics, engineering, and environmental science, particularly those interested in thermodynamics and heat transfer calculations.

gerry7
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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
 
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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.
 
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.
 

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