Question

The volume expansivity \(\beta\) values of copper at \(300 \mathrm{K}\) and \(500 \mathrm{K}\) are \(49.2 \times 10^{-6} \mathrm{K}^{-1}\) and \(54.2 \times 10^{-6} \mathrm{K}^{-1},\) respectively, and \(\beta\) varies almost linearly in this temperature range. Determine the percent change in the volume of a copper block as it is heated from \(300 \mathrm{K}\) to \(500 \mathrm{K}\) at atmospheric pressure.

Short answer

Answer: The percentage change in the volume of the copper block as it is heated from 300 K to 500 K at atmospheric pressure is approximately 10.34%.

Step-by-step solution

Calculate the average volume expansivity

To determine the average volume expansivity, we need to find the average of the volume expansivity values given at 300 K and 500 K. Average volume expansivity, \(\beta_{avg} = \frac{\beta_{300K} + \beta_{500K}}{2}\) \(\beta_{300K} = 49.2 \times 10^{-6} \mathrm{K}^{-1}\) \(\beta_{500K} = 54.2 \times 10^{-6} \mathrm{K}^{-1}\) \(\beta_{avg} = \frac{49.2 \times 10^{-6} + 54.2 \times 10^{-6}}{2}\) \(\beta_{avg} = \frac{103.4 \times 10^{-6}}{2} \mathrm{K}^{-1}\) \(\beta_{avg} = 51.7 \times 10^{-6} \mathrm{K}^{-1}\)

Calculate the temperature difference

We need to find the temperature difference between initial and final temperature to determine the change in volume. \(\Delta T = T_2 - T_1\) \(T_1 = 300 \thinspace \mathrm{K}\) \(T_2 = 500 \thinspace \mathrm{K}\) \(\Delta T = 500 - 300\) \(\Delta T = 200 \thinspace \mathrm{K}\)

Calculate the volume change

We can use the average volume expansivity and temperature difference to calculate the volume change. \(\Delta V = V_0 \cdot \beta_{avg} \cdot \Delta T\) Here, \(\Delta V\) is the volume change, and \(V_0\) is the initial volume. Since we need to find the percentage change in volume, we can divide the volume change by the initial volume and multiply by 100. \(\% \thinspace \Delta V \thinspace = \thinspace \frac{\Delta V}{V_0} \times 100\) \(\% \thinspace \Delta V \thinspace = \thinspace \frac{V_0 \cdot \beta_{avg} \cdot \Delta T}{V_0} \times 100\) The initial volume \(V_0\) cancels out, and the percentage change in volume becomes: \(\% \thinspace \Delta V \thinspace = \thinspace \beta_{avg} \cdot \Delta T \times 100\) \(\% \thinspace \Delta V \thinspace = \thinspace (51.7 \times 10^{-6}) \cdot (200) \times 100\) \(\% \thinspace \Delta V \thinspace = \thinspace 10.34 \thinspace \%\) The percent change in the volume of the copper block as it is heated from 300 K to 500 K at atmospheric pressure is approximately 10.34%.