Question

How does the density of copper that is just above its melting temperature of \(1356 \mathrm{~K}\) compare to that of copper at room temperature?

Short answer

Answer: The density of copper just above its melting temperature (8.86 g/cm³) is slightly lower than its density at room temperature (8.92 g/cm³).

Step-by-step solution

Review the formula for density changes with temperature

The formula for estimating the change in density of a given substance due to a change in temperature is given by: Density at temperature \(T_2\) = Density at temperature \(T_1\) × \((1 - \beta \times \Delta T)\) Here, \(\beta\) is the coefficient of volume expansion, and \(\Delta T = T_2 - T_1\) is the difference in temperature.

Gather the necessary values

To apply the formula, we need the following values: 1. Density of copper at room temperature: Given as \(\rho_1 = 8.92 g/cm^3\) 2. Coefficient of volume expansion of copper: \(\beta = 5 \times 10^{-5} K^{-1}\) 3. Room temperature: \(T_1 = 298K\) (approximately 25 °C) 4. Copper's melting temperature: \(T_2 = 1356 K\)

Calculate the change in density

Using the values above, we can calculate the density of copper just above its melting temperature (\(\rho_2\)): \(\Delta T = T_2 - T_1 = 1356K - 298K = 1058K\) \(\rho_2 = \rho_1 × (1 - \beta \times \Delta T)\) \(\rho_2 = 8.92 g/cm^3 × (1 - (5 × 10^{-5} K^{-1}) × (1058K))\)

Find the density of copper just above its melting temperature

Now, we can calculate the value of \(\rho_2\): \(\rho_2 = 8.92 g/cm^3 × (1 - (5 × 10^{-5} K^{-1}) × (1058K)) \approx 8.86 g/cm^3\)

Compare the densities

Comparing the two densities, we observe that the density of copper just above its melting temperature (\(\rho_2 = 8.86 g/cm^3\)) is slightly lower than its density at room temperature (\(\rho_1 = 8.92 g/cm^3\)). This is expected as substances typically expand when heated, causing a decrease in density.