Question

The volume of \(1.00 \mathrm{~kg}\) of liquid water over the temperature range from \(0.00^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) fits reasonably well to the polynomial function \(V=1.00016-\left(4.52 \cdot 10^{-5}\right) T+\) \(\left(5.68 \cdot 10^{-6}\right) T^{2}\), where the volume is measured in cubic meters and \(T\) is the temperature in degrees Celsius. a) Use this information to calculate the volume expansion coefficient for liquid water as a function of temperature. b) Evaluate your expression at \(20.0^{\circ} \mathrm{C}\), and compare the value to that listed in Table \(17.3 .\)

Short answer

Question: Calculate the volume expansion coefficient for liquid water at 20.0°C using the given volume function and compare the value with the one listed in Table 17.3. Answer: The volume expansion coefficient for liquid water at 20.0°C is approximately 3.39 x 10^-4 K^-1. This value should be reasonably close to the value listed in Table 17.3.

Step-by-step solution

Differentiate the volume function w.r.t. temperature

To find the volume expansion coefficient, we first need to differentiate the volume function with respect to temperature (\(T\)): Given the volume function: \(V(T) = 1.00016 - (4.52 \cdot 10^{-5}) T + (5.68 \cdot 10^{-6}) T^2\) Now, differentiate it w.r.t. \(T\): \(\frac{dV}{dT} = - (4.52 \cdot 10^{-5}) + 2 (5.68 \cdot 10^{-6}) T\)

Compute the volume expansion coefficient

The volume expansion coefficient for liquid water (\(\beta\)) is given by: \(\beta(T) = \frac{1}{V} \frac{dV}{dT}\) Now, substitute the expressions for \(V\) and \(\frac{dV}{dT}\) from step 1: \(\beta(T) = \frac{- (4.52 \cdot 10^{-5}) + 2 (5.68 \cdot 10^{-6}) T}{1.00016 - (4.52 \cdot 10^{-5}) T + (5.68 \cdot 10^{-6}) T^2}\) So, the volume expansion coefficient for liquid water as a function of temperature is: \(\boxed{\beta(T) = \frac{- (4.52 \cdot 10^{-5}) + 2 (5.68 \cdot 10^{-6}) T}{1.00016 - (4.52 \cdot 10^{-5}) T + (5.68 \cdot 10^{-6}) T^2}}\) b) Evaluate the volume expansion coefficient at \(20.0^{\circ} \mathrm{C}\):

Substitute \(T\) with \(20.0^{\circ} \mathrm{C}\)

To find the volume expansion coefficient at \(20.0^{\circ} \mathrm{C}\), substitute \(T=20.0\) in the expression we found in step 2: \(\beta(20.0) = \frac{- (4.52 \cdot 10^{-5}) + 2 (5.68 \cdot 10^{-6}) (20)}{1.00016 - (4.52 \cdot 10^{-5}) (20) + (5.68 \cdot 10^{-6}) (20)^2}\) Now, evaluate the expression: \(\beta(20.0) \approx 3.39 \cdot 10^{-4}~\mathrm{K}^{-1}\) Now, compare the obtained value with the one from Table 17.3. Note that the answer might vary slightly due to rounding errors and approximations made during the calculations. But the evaluated expression should be reasonably close to the value listed in Table 17.3.

Key concepts

Thermal Expansion

When substances such as liquids, solids, or gases increase in temperature, they usually increase in volume, a phenomenon known as thermal expansion. This occurs because particles of matter tend to move more rapidly as they get hotter, requiring more space to move around.

The degree to which a material expands as a result of temperature changes is quantified by its volume expansion coefficient. This coefficient is specific to each material and temperature range, and it's crucial in a variety of engineering and science applications. For example, engineers must take into account the thermal expansion of materials when designing bridges, buildings, and even electronic devices to prevent structural failures or malfunctions.

In the given exercise, the task is to calculate the volume expansion coefficient for liquid water as the temperature rises from 0°C to 50°C. This requires an understanding of how water's volume changes with temperature, which is thoroughly described by a polynomial function.

Differential Calculus

Differential calculus is a branch of mathematics that involves the study of how things change. It focuses on finding the rate at which quantities change, known as the derivative. In practice, calculating the derivative can help understand the speed of an object, the slope of a curve, or, as in our textbook problem, the rate at which the volume of water expands with temperature.

The process of finding a derivative is called differentiation. For a given function, the derivative at any point quantifies how the function's value changes as its input changes. In other words, it provides an instantaneous rate of change.

In our specific problem, this means taking the polynomial function that describes the volume of water over a temperature range and differentiating it concerning temperature. By finding the derivative of the volume function, we can infer how small changes in temperature affect the volume of water, which is directly related to the volume expansion coefficient.

Polynomial Function

In mathematics, a polynomial function is an expression composed of variables (also known as indeterminates) raised to non-negative integer powers, which are multiplied by coefficients and summed together. The simplest polynomial is a linear function, which has just one variable and no exponents higher than one.

Polynomials are easy to work with because they follow well-defined rules for addition, subtraction, multiplication, division, and differentiation. They can model a wide range of phenomena, including the one described in our exercise - the change in volume of liquid water with temperature.

The given polynomial function for the volume of liquid water consists of a constant term, a linear term, and a quadratic term. Each term contributes to how the volume changes with temperature. By understanding how to work with polynomial functions, students can better grasp concepts such as thermal expansion and solve real-world problems that depend on these mathematical relationships.