5. Ice and water Why does ice float on water? Because the density of ice is lowerl In fact, water is the only common liquid whose maximal density occurs above its freezing temperature, and this has important biological implications, e.g. stratification of water in lakes by temperature, and preventing ice from forming at the bottoms of lakes and them freezing solid. Between $$ 0^{\circ} \mathrm{C} $$ and $$ 30^{\circ} \mathrm{C} $$ the volume $$ V $$ (in cubic centimetres) of 1 kg of water at a temperature $$ T $$ is given by the formula, $$ V=999.87-0.06426 T+0.0085043 T^{2}-0.0000679 T^{3} $$ Find the temperature at which water has its maximum density. $$ { }^{1} $$

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We are given that the volume (in cubic centimetres) of 1 kg of water at temperature $$ T $$ (in $$ ^\circ\mathrm{C} $$) is

$$
V = 999.87 - 0.06426T + 0.0085043T^2 - 0.0000679T^3.
$$

Since the mass is constant, water is densest when its volume is minimum. To find the temperature at which this occurs, we need to find the critical points by setting the derivative $$ \frac{dV}{dT} $$ equal to zero.

1. **Differentiate the volume function:**

$$
   \frac{dV}{dT} = -0.06426 + 2(0.0085043)T - 3(0.0000679)T^2.
   $$
   
   Simplify the coefficients:

$$
   \frac{dV}{dT} = -0.06426 + 0.0170086\,T - 0.0002037\,T^2.
   $$

2. **Set the derivative equal to zero:**

$$
   -0.06426 + 0.0170086\,T - 0.0002037\,T^2 = 0.
   $$

It is convenient to multiply the whole equation by $$-1$$:

$$
   0.06426 - 0.0170086\,T + 0.0002037\,T^2 = 0.
   $$

Rearranging gives:

$$
   0.0002037\,T^2 - 0.0170086\,T + 0.06426 = 0.
   $$

3. **Solve the quadratic equation using the quadratic formula:**

The quadratic formula for an equation $$ aT^2 + bT + c = 0 $$ is

$$
   T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
   $$

Here, 
   $$
   a = 0.0002037,\quad b = -0.0170086,\quad c = 0.06426.
   $$

Compute the discriminant $$\Delta$$:

$$
   \Delta = b^2 - 4ac = (-0.0170086)^2 - 4(0.0002037)(0.06426).
   $$
   
   First, calculate $$ (-0.0170086)^2 $$:

$$
   (-0.0170086)^2 \approx 0.00028929.
   $$

Next, calculate $$ 4(0.0002037)(0.06426) $$:

$$
   4(0.0002037)(0.06426) \approx 0.00005236.
   $$

So,

$$
   \Delta \approx 0.00028929 - 0.00005236 = 0.00023693.
   $$

The square root of the discriminant is:

$$
   \sqrt{\Delta} \approx \sqrt{0.00023693} \approx 0.015391.
   $$

Now substitute back into the quadratic formula:

$$
   T = \frac{-(-0.0170086) \pm 0.015391}{2(0.0002037)} = \frac{0.0170086 \pm 0.015391}{0.0004074}.
   $$

4. **Determine the physically relevant solution:**

We obtain two solutions:

$$
   T_1 = \frac{0.0170086 - 0.015391}{0.0004074} \quad \text{and} \quad T_2 = \frac{0.0170086 + 0.015391}{0.0004074}.
   $$

First, calculate $$ T_1 $$:

$$
   T_1 = \frac{0.0016176}{0.0004074} \approx 3.97^\circ\mathrm{C}.
   $$

Next, $$ T_2 $$ is:

$$
   T_2 = \frac{0.0323996}{0.0004074} \approx 79.53^\circ\mathrm{C}.
   $$

Since the formula is valid for temperatures between $$ 0^\circ\mathrm{C} $$ and $$ 30^\circ\mathrm{C} $$, the only acceptable solution is

$$
   T \approx 4^\circ\mathrm{C}.
   $$

Thus, water has its maximum density at approximately $$ 4^\circ\mathrm{C} $$.
Water has its maximum density at approximately $$ 4^\circ\mathrm{C} $$.

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