Question

On a hot summer day, a cubical swimming pool is filled to within \(1.0 \mathrm{~cm}\) of the top with water at \(21{ }^{\circ} \mathrm{C} .\) When the water warms to \(37^{\circ} \mathrm{C}\), the pool overflows. What is the depth of the pool?

Short answer

Answer: The depth of the pool is approximately 1.38 cm.

Step-by-step solution

Write down the initial state of the pool

Initially, the pool is filled with water whose temperature is \(21^{\circ}\mathrm{C}\). We denote the initial side length of the cubical pool as \(L\), and the depth of the water as \((L - 1)\) cm, since it's 1 cm below the top. Hence, the initial volume of the water is given by \(V_i = (L-1)^3\).

Calculate the final volume of the water

When the temperature of water rises to \(37^{\circ} \mathrm{C}\), the pool overflows. This means the water expands and occupies the entire pool volume. Therefore, the final volume of the water, \(V_f = L^3\).

Formula for volume expansion

The formula for volume expansion is given by \(\Delta V = \beta V_i \Delta T\), where \(\Delta V\) is the change in volume, \(\beta\) is the volume expansion coefficient of the substance, \(V_i\) is the initial volume, and \(\Delta T\) is the change in temperature. For water, the volume expansion coefficient is approximately \(\beta = 214 \times 10^{-6} \mathrm{K}^{-1}\). The change in temperature, \(\Delta T = T_f - T_i = 37-21 = 16 \mathrm{~K}\).

Calculate the change in volume

Substitute the values of \(\beta\), \(V_i\), and \(\Delta T\) into the formula for volume expansion, \begin{align*} \Delta V & = \beta V_i \Delta T \\ & = (214 \times 10^{-6} \mathrm{K}^{-1})(L-1)^3(16\mathrm{K}) \\ & = 3.424 L^3 - 10.272 L^2 + 10.272 L - 3.424 \end{align*}

Calculate the change in volume using initial and final volumes

The change in volume can also be calculated as the difference between the final volume and the initial volume, \(\Delta V = V_f - V_i\). So, \(\Delta V = L^3 - (L-1)^3\).

Equate the two expressions for the change in volume

Equate the two expressions for the change in volume and solve for \(L\): \begin{align*} L^3 - (L-1)^3 &= 3.424 L^3 - 10.272 L^2 + 10.272 L - 3.424 \\ 0 &= 2.424 L^3 - 10.272 L^2 + 10.272 L - 3.424 \end{align*} Solving this equation numerically (using a calculator, a graph or a computer), we find the value of \(L \approx 2.38\).

Find the depth of the pool

Finally, recall that the depth of the pool is given by \((L-1)\) cm. Therefore, the depth of the pool is approximately \((2.38 - 1)\) cm, or \(1.38\) cm.

Key concepts

Volume Expansion Coefficient

Understanding the volume expansion coefficient is crucial when studying how substances like water behave under temperature changes. The volume expansion coefficient, denoted typically by \( \beta \), is a measure of the fractional change in volume per degree change in temperature. It's an intrinsic property of materials and signifies the degree to which a material expands upon heating.

For example, the volume expansion coefficient of water is approximately \( 214 \times 10^{-6} \mathrm{K}^{-1} \). This implies that for each kelvin increase in temperature, a unit volume of water expands by \( 214 \times 10^{-6} \times \text{its original volume} \). When we apply this concept to real-world applications, such as determining the likelihood of a swimming pool overflowing, we calculate how much the water will expand when subjected to a certain increase in temperature. This leads us to compute the change in water volume by multiplying the volume expansion coefficient (\( \beta \)) by the initial volume (\( V_i \)) and by the change in temperature (\( \Delta T \)).

Temperature Change in Physics

The concept of temperature change in physics lies at the heart of thermal expansion. It is the difference in temperature, symbolized by \( \Delta T \), that a substance experiences, which leads to its expansion or contraction. \( \Delta T \) is calculated as the final temperature (\( T_f \)) minus the initial temperature (\( T_i \)), usually measured in degrees Celsius (\( ^\circ\mathrm{C} \)) or Kelvin (K).

In our swimming pool example, the water temperature increases from \( 21 \)) to \( 37 \)) Celsius. Consequently, the temperature change would be \( 16 \)) K (\( \Delta T = T_f - T_i = 37^\circ\mathrm{C} - 21^\circ\mathrm{C} \)). It is this temperature change that drives the expansion of water, which in turn, causes the pool to overflow. Temperature is a vital component because it dictates the amount of energy within a substance's particles, influencing their motion and thus the overall volume.

Cubical Expansion

When we talk about cubical expansion, we refer to the phenomenon where the volume of a cube-shaped object increases due to an increase in temperature. It’s called 'cubical' because the object's shape is a cube, and all three dimensions expand uniformly if the material is isotropic. This type of expansion is quantified by the volume expansion formula, which involves the volume expansion coefficient discussed earlier.

Cubical Expansion and the Overflowing Pool

In the context of the exercise with the swimming pool, the cubical expansion is critical to determining when the water will overflow. The pool expands as a cube, and we predict the increase in its volume by taking into account the temperature change and the properties of water. By comparing the initial and final volumes, where the final volume is simply the volume of the cubical pool itself (\( L^3 \)), we can ascertain the depth of the pool. This comparison essentially requires equating the volume before and after expansion, then subtracting to find the change. It exemplifies how understanding cubical expansion can help solve practical problems involving thermal behavior in daily life.