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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

The complete combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l),\) to form \(\mathrm{H}_{2} \mathrm{O}(g)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases 1235 \(\mathrm{kJ}\) of heat per mole of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and then look up the standard enthalpy of formation for each substance in Appendix C: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g),(\mathbf{b}) \mathrm{CaCO}_{3}(s)\) (c) \(\mathrm{POCl}_{3}(l),(\mathbf{d}) \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\)

The heat of combustion of fructose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) is \(-2812 \mathrm{kJ} / \mathrm{mol} .\) If a fresh golden delicious apple weighing 4.23 oz \((120 \mathrm{g})\) contains 16.0 \(\mathrm{g}\) of fructose, what caloric content does the fructose contribute to the apple?

Using values from Appendix \(\mathrm{C}\) , calculate the standard enthalpy change for each of the following reactions: $$ \begin{array}{l}{\text { (a) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)} \\ {\text { (b) } \mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (c) } \mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)}\end{array} $$

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