# Mass flow rate of evaporating water in a vacuum

I'm melting ice in a vacuum using a heating plate, and attempting to condense any water vapor evaporated from the melted water onto a cold plate. Is there a way to describe the mass flow rate of the water vapor in terms of the power output of the heating pad and temperature of the cold plate (which remains constant through constant piping of liquid nitrogen)? Thanks for your help!

## 1 Answer

The rate of evaporation of water is related to the thermal power input and to the specific latent heat of water, which is the thermal energy absorbed by a unit mass of water in order to undergo a phase transition from liquid to gas. At standard pressure: $$L_{water-SP}=2264.7 kJ/kg$$.

Thus the mass evaporation rate would be $$\dot{m}=\dfrac{P_{input}}{L_{water}}$$, where $$P_{input}$$ is the thermal energy absorbed by the water per unit time.

In your case you need to find $$L_{water}$$ at zero pressure (since you are in vacuum). You also need to be aware that $$P_{input}$$, the rate of heat absorption by evaporating water from the surroundings, may or may not be the same as the power output from the heating pad in case your system is not closed.

• What if the ice is sublimating in the vacuum? From ice at ~200k to gas at T>200k? is there a way to relate the thermal energy absorbed by the ice per unit time to the vapor pressure of the sublimated water vapor? – Peter Connors Sep 22 '18 at 23:29
• Hmm I can see a clear way. The pressure of water vapor depends largely on the details of the system and the efficiency of your condensation mechanism. – Tofi Sep 22 '18 at 23:55
• I can't* see a way – Tofi Sep 23 '18 at 10:27