/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 An aluminum can of a soft drink ... [FREE SOLUTION] | 91影视

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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

(a) When a 4.25 -g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter (Figure 5.18), the temperature drops from 22.0 to \(16.9^{\circ} \mathrm{C}\) . Calculate \(\Delta H\left(\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) for the solution process: $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)$$ Assume that the specific heat of the solution is the same as that of pure water. (b) Is this process endothermic or exothermic?

Consider the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g) .(\mathbf{a})\) Use the bond enthalpies in Table 5.4 to estimate \(\Delta H\) for this reaction, ignoring the fact that iodine is in the solid state. (b) Without doing a calculation, predict whether your estimate in part (a) is more negative or less negative than the true reaction enthalpy. (c) Use the enthalpies of formation in Appendix \(C\) to determine the true reaction enthalpy.

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?

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.26 L against a constant pressure of 0.857 atm?

Under constant-volume conditions, the heat of combustion of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is 15.57 \(\mathrm{kJ} / \mathrm{g}\) . A 3.500 -g sample of glucose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to \(24.72^{\circ} \mathrm{C}\) (a) What is the total heat capacity of the calorimeter? (b) If the size of the glucose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?
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