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Chapter 5: Problem 100

An aluminum can of a soft drink is placed in a freezer. Later, you find that the can is split open and its contents have frozen. Work was done on the can in splitting it open. Where did the energy for this work come from?

Short Answer

Expert verified
The energy for the work done on the aluminum can, causing it to split open, comes from the heat released during the freezing process of the soft drink. As the liquid inside the can loses heat energy to the surroundings (the freezer) and becomes a solid, the energy released is used to perform mechanical work on the can, causing it to deform and eventually split open.

Step by step solution

01

Understand the freezing process

First, let's understand what happens when the soft drink freezes. When the can is placed in the freezer, the temperature of the liquid inside the can decreases. As the temperature decreases, the liquid molecules slow down and eventually align themselves in an organized pattern. This process causes the liquid to change its phase into a solid (in this case, ice). When water freezes, it expands and requires more volume.
02

Identify what causes the can to split

When the liquid inside the can freezes and expands, it puts a force on the walls of the aluminum can. The walls of the can need to expand to accommodate the increased volume of the solid. If the force exerted by the expanding ice is greater than the can's material strength, the can will deform and eventually split open.
03

Identify the energy conversion

Now, we know that there is a force applied to the walls of the can. This force results in mechanical work being done on the can, causing it to split open. The energy for this work must come from somewhere. During the freezing process, energy is released in the form of heat as the liquid inside the can loses energy to the surroundings (the freezer). This is because the molecules in the liquid slow down and arrange themselves into a solid state, which generally has lower energy than a liquid state.
04

Identify the source of energy

We can now conclude that the energy for the work done on the can, which caused it to split open, comes from the heat released during the freezing process of the soft drink. As the soft drink loses heat energy (exothermic process) and becomes a solid, the energy that is released is used to perform mechanical work on the aluminum can, causing it to deform and eventually split open.

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Most popular questions from this chapter

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{kJ}\) and \(w=-840 \mathrm{J} .\) (b) A system releases 66.1 \(\mathrm{kJ}\) of heat to its surroundings while the surroundings do 44.0 \(\mathrm{kJ}\) of work on the system.

The standard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\) propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are \(+185.4,+20.4,\) and \(-103.8 \mathrm{kJ} / \mathrm{mol}\) , respectively.(a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) (b) Calculate the heat evolved on combustion of 1 \(\mathrm{kg}\) of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$\begin{array}{ll}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-890.3 \mathrm{kJ}} \\ {\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} & {\Delta H^{\circ}=-136.3 \mathrm{kJ}} \\ {2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-571.6 \mathrm{kJ}} \\ {2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-3120.8 \mathrm{kJ}}\end{array}$$

Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that was once used as a surgical anesthetic, has the structure $$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$$ The complete combustion of 1 mol of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{kJ}\) . (a) Write a balanced equation for the combustion of 1 \(\mathrm{mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) .\) (b) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

Two positively charged spheres, each with a charge of \(2.0 \times\) \(10^{-5} \mathrm{C},\) a mass of 1.0 \(\mathrm{kg}\) , and separated by a distance of \(1.0 \mathrm{cm},\) are held in place on a frictionless track. (a) What is the electrostatic potential energy of this system? If the spheres are released, will they move toward or away from each other? (c) What speed will each sphere attain as the distance between the spheres approaches infinity? [Section 5.1\(]\)
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