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Here's a phase diagram of water from here.

If you find Mars's atmosphere of 600Pa on the Y-axis and track right you'll see that H20 can't be water on Mars. It can only be ice or vapor.

In your example where you have a dome of Earth like atmosphere protecting your Martian pool that blows away somehow then you'll either have a skating rink or an empty hole but no water. Going by the average Martian temperature you'd have a skating rink but once the temperature exceeds the sublimation point all your H2O would turn to vapor and that'd be the end of it.

Edit:

The other answer asserts that it would first boil and perhaps there'd be some H20 left as ice after some of it boiled off. I suppose this depends on the exact circumstances of losing the dome. The way OP asks the question, it seems that first heat is lost and then the dome goes away. If the water has already cooled to below the sublimation point before the pressure drop occurs then there's no need for it to boil. If everything is working fine and then the dome flies away while the ambient temperature is above boiling temperature then clearly it would boil off. However, if the ambient temperature is below sublimation when the dome is lost I don't think it is clear that the water would boil rather than freeze. The very act of decompressing would cool the water so we can't simply say that the pressure can drop faster than the water cools because the pressure dropping speeds the cooling.

Here's a phase diagram of water from here.

If you find Mars's atmosphere of 600Pa on the Y-axis and track right you'll see that H20 can't be water on Mars. It can only be ice or vapor.

In your example where you have a dome of Earth like atmosphere protecting your Martian pool that blows away somehow then you'll either have a skating rink or an empty hole but no water. Going by the average Martian temperature you'd have a skating rink but once the temperature exceeds the sublimation point all your H2O would turn to vapor and that'd be the end of it.

Edit:

The other answer asserts that it would first boil and perhaps there'd be some H20 left as ice after some of it boiled off. I suppose this depends on the exact circumstances of losing the dome. The way OP asks the question, it seems that first heat is lost and then the dome goes away. If the water has already cooled to below the sublimation point before the pressure drop occurs then there's no need for it to boil. If everything is working fine and then the dome flies away while the ambient temperature is above boiling temperature then clearly it would boil off. However, if the ambient temperature is below sublimation when the dome is lost I don't think it is clear that the water would boil rather than freeze. The very act of decompressing would cool the water so we can't simply say that the pressure can drop faster than the water cools because the pressure dropping speeds the cooling.

Here's a phase diagram of water.

If you find Mars's atmosphere of 600Pa on the Y-axis and track right you'll see that H20 can't be water on Mars. It can only be ice or vapor.

In your example where you have a dome of Earth like atmosphere protecting your Martian pool that blows away somehow then you'll either have a skating rink or an empty hole but no water. Going by the average Martian temperature you'd have a skating rink but once the temperature exceeds the sublimation point all your H2O would turn to vapor and that'd be the end of it.

Edit:

The other answer asserts that it would first boil and perhaps there'd be some H20 left as ice after some of it boiled off. I suppose this depends on the exact circumstances of losing the dome. The way OP asks the question, it seems that first heat is lost and then the dome goes away. If the water has already cooled to below the sublimation point before the pressure drop occurs then there's no need for it to boil. If everything is working fine and then the dome flies away while the ambient temperature is above boiling temperature then clearly it would boil off. However, if the ambient temperature is below sublimation when the dome is lost I don't think it is clear that the water would boil rather than freeze. The very act of decompressing would cool the water so we can't simply say that the pressure can drop faster than the water cools because the pressure dropping speeds the cooling.

Here's a phase diagram of water from here.

If you find Mars's atmosphere of 600Pa on the Y-axis and track right you'll see that H20 can't be water on Mars. It can only be ice or vapor.

In your example where you have a dome of Earth like atmosphere protecting your Martian pool that blows away somehow then you'll either have a skating rink or an empty hole but no water. Going by the average Martian temperature you'd have a skating rink but once the temperature exceeds the sublimation point all your H2O would turn to vapor and that'd be the end of it.

Edit:

The other answer asserts that it would first boil and perhaps there'd be some H20 left as ice after some of it boiled off. I suppose this depends on the exact circumstances of losing the dome. The way OP asks the question, it seems that first heat is lost and then the dome goes away. If the water has already cooled to below the sublimation point before the pressure drop occurs then there's no need for it to boil. If everything is working fine and then the dome flies away while the ambient temperature is above boiling temperature then clearly it would boil off. However, if the ambient temperature is below sublimation when the dome is lost I don't think it is clear that the water would boil rather than freeze. The very act of decompressing would cool the water so we can't simply say that the pressure can drop faster than the water cools because the pressure dropping speeds the cooling.

Here's a phase diagram of water.

If you find Mars's atmosphere of 600Pa on the Y-axis and track right you'll see that H20 can't be water on Mars. It can only be ice or vapor.

In your example where you have a dome of Earth like atmosphere protecting your Martian pool that blows away somehow then you'll either have a skating rink or an empty hole but no water. Going by the average Martian temperature you'd have a skating rink but once the temperature exceeds the sublimation point all your H2O would turn to vapor and that'd be the end of it.

Here's a phase diagram of water.

If you find Mars's atmosphere of 600Pa on the Y-axis and track right you'll see that H20 can't be water on Mars. It can only be ice or vapor.

In your example where you have a dome of Earth like atmosphere protecting your Martian pool that blows away somehow then you'll either have a skating rink or an empty hole but no water. Going by the average Martian temperature you'd have a skating rink but once the temperature exceeds the sublimation point all your H2O would turn to vapor and that'd be the end of it.

Edit:

The other answer asserts that it would first boil and perhaps there'd be some H20 left as ice after some of it boiled off. I suppose this depends on the exact circumstances of losing the dome. The way OP asks the question, it seems that first heat is lost and then the dome goes away. If the water has already cooled to below the sublimation point before the pressure drop occurs then there's no need for it to boil. If everything is working fine and then the dome flies away while the ambient temperature is above boiling temperature then clearly it would boil off. However, if the ambient temperature is below sublimation when the dome is lost I don't think it is clear that the water would boil rather than freeze. The very act of decompressing would cool the water so we can't simply say that the pressure can drop faster than the water cools because the pressure dropping speeds the cooling.