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Chapter 5: Problem 91

An aluminum can of a soft drink is placed in a freezer. Later, you find that the can is split open and its contents 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 can in splitting it open came from the heat extracted from the soft drink and the heat absorbed by the freezer. As the soft drink loses heat energy and the freezer absorbs this heat, the freezing process causes the liquid inside to expand and the dissolved gases to exert pressure on the can, leading to its rupture.

Step by step solution

01

Identify the freezing process

When the aluminum can was placed in the freezer, the temperature of its contents started to drop gradually. The soft drink inside the can consists of both liquid and dissolved gases. As the temperature decreases, the liquid in the can begins to freeze, and the gases dissolved in the soft drink expand and try to find a larger volume in which they can spread. When the soft drink freezes, the pressure inside the can builds up due to the increased volume occupied by gases and the expansion of the frozen liquid.
02

Investigate work and energy

In this step, we'll examine the relationship between work and energy. Work is the transfer of energy between two objects or systems in a process in which the objects change their states or conditions. This energy transfer occurs when a force is applied to an object over a certain distance. In the case of the aluminum can, the work performed resulted in the splitting of the can. This work comes from the energy transfer caused by the expansion of the freezing liquid and the released gases, which apply pressure on the walls of the can, leading to its deformation and rupture.
03

Identify the source of energy

Now that we know how work and energy transfers have led to the splitting of the aluminum can, we need to identify the source of the energy. In this case, the energy comes from the heat extracted from the soft drink and the heat absorbed by the freezer. When the soft drink loses heat energy and the freezer absorbs this heat, the result is a transfer of energy to the environment, which is what causes the freezing process. To summarize, the energy that performed the work of splitting the aluminum can came from the heat extracted from the soft drink and the heat absorbed by the freezer.

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Most popular questions from this chapter

Indicate which of the following is independent of the path by which a change occurs: (a) the change in potential energy when a book is transferred from table to shelf, (b) the heat evolved when a cube of sugar is oxidized to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), (c) the work accomplished in burning a gallon of gasoline.

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant- pressure process the system absorbs heat from the surroundings. Does the enthalpy of the system increase or decrease during the process?

From the following data for three prospective fuels, calculate which could provide the most energy per unit volume:$$ \begin{array}{lcc} & \begin{array}{c} \text { Density } \\ \text { at } 20^{\circ} \mathrm{C} \\ \left(\mathrm{g} / \mathrm{cm}^{3}\right) \end{array} & \begin{array}{c} \text { Molar Enthalpy } \\ \text { of Combustion } \\ \text { Fuel } \end{array} & \mathrm{kJ} / \mathrm{mol} \\ \hline \text { Nitroethane, } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}(l) & 1.052 & -1368 \\ \text { Ethanol, } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) & 0.789 & -1367 \\ \text { Methylhydrazine, } \mathrm{CH}_{6} \mathrm{~N}_{2}(l) & 0.874 & -1305 \end{array} $$

Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanest-burning octanes is a compound called \(2,3,4\) -trimethylpentane, which has the following structural formula: The complete combustion of one mole of this compound to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) leads to \(\Delta H^{\circ}=-5064.9 \mathrm{~kJ} / \mathrm{mol}\). (a) Write a balanced equation for the combustion of 1 mol of \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .(\mathrm{b})\) Write a balanced equation for the formation of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) from its elements. (c) By using the information in this problem and data in Table \(5.3\), calculate \(\Delta H_{f}^{\circ}\) for \(2,3,4\) -trimethylpentane.

In what two ways can an object possess energy? How do these two ways differ from one another?
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