Why Does Water Freeze from the Top?

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Water freezes from the top because ice is less dense than liquid water, causing it to float. As water cools to 0°C, the colder, denser water remains at the bottom while the warmer water rises to the top, where it eventually freezes. This phenomenon is crucial for aquatic life, as it allows ponds and lakes to maintain a liquid layer beneath the ice, providing a habitat for organisms. Water reaches its maximum density at approximately 3.98°C, which influences its freezing behavior. The process exemplifies nature's adaptability and efficiency in supporting life.
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Homework Statement



hi all, i just have a small quistion that why always the water is down and ice starting from the up or the top?
i mean for example if we try to freezing a water the ice will start from the top of the bottle and the water will be downsizing in the bottom of the buttle until it become all solid

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The Attempt at a Solution

 
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Hi looody, welcome to PF. At 0°C, liquid water is denser (heavier) than ice.
 
Does it happen that way? I guess I've never watched water freeze for long enough :-p

Here"s what I think is happening: The density of water at nearly freezing point is lower than that of water at room temperature and the density if ice is lower than water (that's why ice blocks float in your cool drink). So, the colder water rises and eventually freezes at the top of the other water.
 
In fact water has its maximum density at a temperature of 3.98 Degrees. Imagine ponds ,lakes etc froze from the bottom up,what would that do to the pond life?Isn't nature clever?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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