system
Given the following post, respond as if you're commenting on Reddit:
human
Does the amount of energy to freeze water scale linearly?

If I was to freeze 1 ml of water does that take 1/1000th the amount of energy as freezing 1L of water? If not, how does it scale?
gpt
Technically, you're actually taking heat energy from the water, instead of using energy. That being said, yes, the energy taken does scale linearly with the mass/volume of the body of water, and if you were to take 1ml, it would lose 1/1000th the amount of heat than when you freeze 1L **as long as the initial and final temperatures of both the bodies of water taken are equal.**

EDIT: In case anyone is interested in the scientific basis of this, first you need to know two terms:

Latent Heat of Fusion (L) - It is the heat energy used to convert unit mass of a body in liquid state to solid state, or the heat lost to convert liquid into solid. During this process the temperature of the body does not change. So when you freeze water at 0℃ to ice at 0℃, the temperature does not change till the water has been completely turned to ice, but energy is used. For example, the latent heat of water is about 80 cal/g.

Specific Heat Capacity - It is the heat energy used to increase the temperature of a body by a unit, per unit mass, or energy lost to decrease its temperature. This is used when there is no change in state, like when cooling water from 80℃ to 20℃. For example the specific heat of water is 1 cal/℃/g and that of ice is about 0.5 cal/℃/g

In OP’s case, assuming water taken is not at 0℃ we must use a combination of the two. Let the temperature of the body initially be T₁ ℃ and the mass taken be M (in grams).

First we must cool the body to 0℃. So we use specific heat of water. The energy lost is:

M x c x T₁ = M x 1 x T₁ = MT₁

Now we must convert the water to ice so we use latent heat. Energy lost is:

M x L = ML = 80M

Finally we cool the ice to a temperature T₂ below 0 (assuming that you do not stop immediately after water is converted to ice).Again we use specific heat (but of ice). Energy lost is:

M x c x T₂ = M x 0.5 x T₂ = 0.5MT₂

Hence total energy used is:

MT₁ + 80M + 0.5MT₂ = M (T₁ + 80 + 0.5T₂)

As you see, the energy used is directly proportional to the mass taken. OP has asked about volume. Since density of water is constant at same initial temperature, the mass taken can be written as V x d (where V is volume and d density).

Hence the equation can be written as:

Vd ( T₁ + 80 + 0.5T₂ ).

Hence the energy lost is directly proportional to the volume of the substance taken. Hope this helped.