# Calculating temperature of water in the freezer

Assuming water volume ($V$), initial water temperature ($T_0$) and environment temperature ($T_e$) are known, what is the easiest way to calculate temperature of water in given time ($T$)?

For the sake of question, let's assume water container is so thin it doesn't matter. Also, water container is airtight (for example, soda can or sparkling wine bottle). Environment temperature is constant and air is not moving remarkably.

I already experimented with sparkling wine bottle (filled with water and temperature logger). Blue line is for fridge, red for freezer and greenish for ice water (without salt). But it's not very practical to repeat same experiments with different and different containers.

Why? I thought it would be nice to make a simple mobile application for giving alarms when your drinks are chilled to whatever temperature you want.

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Related or even duplicate: physics.stackexchange.com/questions/24013 –  Pygmalion Apr 26 '12 at 13:49

You have exact equations for the solution in the related question Time it takes for Temperature Change. Here I would add a few comments.

It is actually easier if container is thick! Then suppose that all water is at same temperature $T$ and all the air in the frizer is at the same temperature $T_e$. $T_e$ is constant.

If that is so, you can use only Fourier's law to describe how heat $Q$ leaves the container

$$\frac{\text{d}Q}{\text{d}t} = \frac{\lambda A}{d} (T-T_e).$$

$d$ is thickness, $A$ area and $\lambda$ thermal conductivity of the container.

Knowing that water cools as heat is leaving the container

$$\text{d}Q = m c \text{d}T,$$

where $m$ is mass and $c$ is specific heat capacity of the liquid, you get rather simple differential equation

$$m c \frac{\text{d}T}{\text{d}t} = \frac{\lambda A}{d} (T-T_e),$$

$$\frac{\text{d}T}{(T-T_e)} = \frac{\lambda A}{d m c } \text{d}t = K \text{d}t,$$

which has exponential solution:

$$K t = \ln \left(\frac{T-T_e}{T_0-T_e}\right),$$

$$T = T_e + (T_0-T_e) e^{-Kt}.$$

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