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Chapter 12: Problem 47

The normal melting point of copper is \(1357 \mathrm{K}\), and \(\Delta \mathrm{H}_{\text {fus }}\) of \(\mathrm{Cu}\) is \(13.05 \mathrm{kJ} \mathrm{mol}^{-1}\). (a) How much heat, in kilojoules, is evolved when a \(3.78 \mathrm{kg}\) sample of molten Cu freezes? (b) How much heat, in kilojoules, must be absorbed at 1357 K to melt a bar of copper that is \(75 \mathrm{cm} \times\) \(15 \mathrm{cm} \times 12 \mathrm{cm} ?\) (Assume \(d=8.92 \mathrm{g} / \mathrm{cm}^{3}\) for \(\mathrm{Cu}\).)

Short Answer

Expert verified
The heat evolved in part (a) when a 3.78 kg sample of molten Cu freezes is -776.23 kJ and the heat absorbed in part (b) at 1357 K to melt a bar of copper that is 75 cm x 15 cm x 12 cm is 24735.89 kJ.

Step by step solution

01

Convert the mass of copper into moles for part (a)

To convert mass into moles, the formula used is Moles = mass ÷ molar mass. The mass of copper is given as 3.78 kg, which is 3780g (since 1kg = 1000g). The molar mass of copper (Cu) is 63.55g. Moles of Cu = 3780g ÷ 63.55g/mol = 59.5 mol.
02

Calculate the heat evolved when the molten Cu freezes

We can find the heat evolved when copper freezes using the enthalpy of fusion formula i.e. q = nΔH_fus. Therefore, q = 59.5 mol × -13.05 kJ/mol= -776.23 kJ. The answer is negative because energy is being released (evolved) when Cu changes from liquid to solid.
03

Calculate the volume of copper in part (b)

The volume of the copper bar can be calculated using the formula for the volume of a rectangular parallelepiped which is volume = length × breadth × height. Therefore, volume = 75 cm × 15 cm × 12 cm = 13500 cm^3.
04

Calculate the mass of copper in part (b)

The mass of the copper can be calculated by multiplying its volume by the density. Therefore, mass = volume × density = 13500 cm^3 × 8.92 g/cm^3 = 120420 g.
05

Convert the mass of copper into moles for part (b)

Converting mass into moles, we get Moles = mass ÷ molar mass = 120420g ÷ 63.55g/mol = 1895.39 mol.
06

Calculate the heat absorbed when the Cu melts

Use the enthalpy of fusion formula i.e. q = nΔH_fus = 1895.39 mol × 13.05 kJ/mol = 24735.89 kJ. The answer is positive because energy is being absorbed when Cu changes from solid to liquid.

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Most popular questions from this chapter

Argon, copper, sodium chloride, and carbon dioxide all crystallize in the fcc structure. How can this be when their physical properties are so different?

In an ionic crystal lattice each cation will be attracted by anions next to it and repulsed by cations near it. Consequently the coulomb potential leading to the lattice energy depends on the type of crystal. To get the total lattice energy you must sum all of the electrostatic interactions on a given ion. The general form of the electrostatic potential is $$V=\frac{Q_{1} Q_{2} e^{2}}{d_{12}}$$ where \(Q_{1}\) and \(Q_{2}\) are the charges on ions 1 and \(2, d_{12}\) is the distance between them in the crystal lattice. and \(e\) is the charge on the electron. (a) Consider the linear "crystal" shown below. The distance between the centers of adjacent spheres is \(R .\) Assume that the blue sphere and the green spheres are cations and that the red spheres are anions. Show that the total electrostatic energy is $$V=-\frac{Q^{2} e^{2}}{d} \times \ln 2$$ (b) In general, the electrostatic potential in a crystal can be written as $$V=-k_{M} \frac{Q^{2} e^{2}}{R}$$ where \(k_{M}\) is a geometric constant, called the Madelung constant, for a particular crystal system under consideration. Now consider the NaCl crystal structure and let \(R\) be the distance between the centers of sodium and chloride ions. Show that by considering three layers of nearest neighbors to a central chloride ion, \(k_{M}\) is given by $$k_{M}=\left(6-\frac{12}{\sqrt{2}}+\frac{8}{\sqrt{3}}-\frac{6}{\sqrt{4}} \cdots\right)$$ (c) Carry out the same calculation for the CsCl structure. Are the Madelung constants the same?

Of the compounds \(\mathrm{HF}, \mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{N}_{2} \mathrm{H}_{4},\) and \(\mathrm{CHCl}_{3},\) hydrogen bonding is an important intermolecular force in (a) none of these; (b) two of these; (c) three of these; (d) all but one of these; (e) all of these.

Would you expect an ionic solid or a network covalent solid to have the higher melting point?

The magnitude of one of the following properties must always increase with temperature; that one is (a) surface tension; (b) density; (c) vapor pressure; (d) \(\Delta H_{\text {vap }}\)
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