There's no free lunch, and heat transfer / thermodynamics is no different.
The short term result is that you would increase the amount of ice while increasing the overall temperature of the polar regions.
specific heat of air at polar temperatures (250K) is 1.006 kJ/kg*K with a density of 1.3947 kg/m^3
specific heat of water at 273K is 4.217kJ/kg*K with a density of 1000kg/m^3 the latent heat of fusion for water is 334.7kJ/kg (energy to change from water to ice).
specific heat of ice is 2.040kJ/kg*K with a density of 920kg/m^3, because you won't be stopping heat transfer until temperatures come close to equalizing.
So the water is 1000 times more dense and takes over 4 times as much energy as air to change the temperature by 1 degree K, not to mention the energy required to change it from a liquid to a solid.
Assuming some nice round numbers:
Polar air temp = 250K (-23C)
Freshwater incoming temp = 275K (2C)
final equilibrium temp of ice and air = 260K (-13C)
For every cubic meter of water you turn to ice in this manner, you raise the air temperature of 26195m^3 of air by 10 degrees C. If the amount of ice were to be significant on a global scale, I would think a temperature change of that magnitude for that much air would be catastrophic over a short (at least in a climate sense) timeframe.
those are just some 'back of the envelope' calculations, but for ballparking, you see what you're asking.
edit: if you find some actual air/ice/water temps, let me know, I can easily plug them in and find a more accurate result.
The short term result is that you would increase the amount of ice while increasing the overall temperature of the polar regions.
specific heat of air at polar temperatures (250K) is 1.006 kJ/kg*K with a density of 1.3947 kg/m^3
specific heat of water at 273K is 4.217kJ/kg*K with a density of 1000kg/m^3 the latent heat of fusion for water is 334.7kJ/kg (energy to change from water to ice).
specific heat of ice is 2.040kJ/kg*K with a density of 920kg/m^3, because you won't be stopping heat transfer until temperatures come close to equalizing.
So the water is 1000 times more dense and takes over 4 times as much energy as air to change the temperature by 1 degree K, not to mention the energy required to change it from a liquid to a solid.
Assuming some nice round numbers:
Polar air temp = 250K (-23C)
Freshwater incoming temp = 275K (2C)
final equilibrium temp of ice and air = 260K (-13C)
For every cubic meter of water you turn to ice in this manner, you raise the air temperature of 26195m^3 of air by 10 degrees C. If the amount of ice were to be significant on a global scale, I would think a temperature change of that magnitude for that much air would be catastrophic over a short (at least in a climate sense) timeframe.
those are just some 'back of the envelope' calculations, but for ballparking, you see what you're asking.
edit: if you find some actual air/ice/water temps, let me know, I can easily plug them in and find a more accurate result.