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Chapter 17: Problem 30

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

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

Step by step solution

01

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 \(T_1\) 脳 \((1 - \beta \times \Delta T)\) Here, \(\beta\) is the coefficient of volume expansion, and \(\Delta T = T_2 - T_1\) is the difference in temperature.
02

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 掳C) 4. Copper's melting temperature: \(T_2 = 1356 K\)
03

Calculate the change in density

Using the values above, we can calculate the density of copper just above its melting temperature (\(\rho_2\)): \(\Delta T = T_2 - T_1 = 1356K - 298K = 1058K\) \(\rho_2 = \rho_1 脳 (1 - \beta \times \Delta T)\) \(\rho_2 = 8.92 g/cm^3 脳 (1 - (5 脳 10^{-5} K^{-1}) 脳 (1058K))\)
04

Find the density of copper just above its melting temperature

Now, we can calculate the value of \(\rho_2\): \(\rho_2 = 8.92 g/cm^3 脳 (1 - (5 脳 10^{-5} K^{-1}) 脳 (1058K)) \approx 8.86 g/cm^3\)
05

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.

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Most popular questions from this chapter

A plastic-epoxy sheet has uniform holes of radius \(1.99 \mathrm{~cm}\). The holes are intended to allow solid ball bear- ings with an outer radius of \(2.00 \mathrm{~cm}\) to just go through. Over what temperature rise must the plastic-epoxy sheet be heated so that the ball bearings will go through the holes? The linear expansion coefficient of plastic-epoxy is about \(1.3 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

Two solid objects are made of different materials. Their volumes and volume expansion coefficients are \(V_{1}\) and \(V_{2}\) and \(\beta_{1}\) and \(\beta_{2}\), respectively. It is observed that during a temperature change of \(\Delta T\), the volume of each object changes by the same amount. If \(V_{1}=2 V_{2}\) what is the ratio of the volume expansion coefficients?

The Rankine temperature scale is an absolute temperature scale that uses Fahrenheit degrees; that is, temperatures are measured in Fahrenheit degrees, starting at absolute zero. Find the relationships between temperature values on the Rankine scale and those on the Fahrenheit, Kelvin, and Celsius scales.

When a 50.0 -m-long metal pipe is heated from \(10.0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\), it lengthens by \(2.85 \mathrm{~cm}\). a) Determine the linear expansion coefficient. b) What type of metal is the pipe made of?

Which object has the higher temperature after being left outside for an entire winter night: a metal door knob or a rug? a) The metal door knob has the higher temperature. b) The rug has the higher temperature. c) Both have the same temperature. d) It depends on the outside temperature.
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