/*! 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] | 91Ó°ÊÓ

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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 comes from the energy released during the freezing process of the liquid contents as they turned into a solid.

Step by step solution

01

Identify the Change in State

When the contents of the can freeze, they change from a liquid to a solid state. This transformation is accompanied by a release of energy due to the change in internal energy and the reduction in entropy.
02

Consider the Freezing Process

Freezing is an exothermic process, meaning that it releases energy. In this case, the energy is released from the liquid as it becomes solid, because the molecules are losing energy as they arrange into a lattice structure typical of solid states.
03

Understand the Energy Release

The energy released during freezing is due to changes in the internal energy of the substance. The solid state has lower potential energy compared to the liquid state, so energy must be released during this phase transition.
04

Link Energy Release to Work Done

The energy released from the freezing process contributes to performing work on the can. Specifically, this energy is used to generate enough pressure inside the can that ultimately causes it to split open as the volume of the frozen contents expands.
05

Finalize the Source of Energy

Thus, the energy for performing the work on the can by breaking it open comes from the phase change of the contents freezing and releasing energy as the liquid converts into solid ice.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Change
When a substance undergoes a phase change, it moves from one state of matter to another. In the scenario where the liquid inside an aluminum can freezes, it changes from a liquid to a solid state. This transition is known as a phase change and is a critical concept in thermodynamics.
Throughout a phase change, the temperature of the substance remains constant. All the energy provided or released is used in changing the state. In this case, the can's contents freezing indicates a liquid-to-solid phase change, which involves arranging the molecules into a more structured form typical of solids.
  • Liquid state: Particles move freely and have more energy.
  • Solid state: Particles are tightly packed and have less energy.
Understanding phase changes helps explain the energy transformations which occur during these processes.
Exothermic Process
An exothermic process is one that releases energy, usually in the form of heat. When the soft drink in the can freezes, it is undergoing an exothermic process.
During freezing, the molecules of the liquid lose energy as they shift into a solid configuration. This release of energy is an essential aspect of exothermic reactions. In freezing, the solid formation provides a structured environment where molecules have less kinetic energy.
Key aspects of exothermic processes include:
  • Energy is released, decreasing the overall energy of the system.
  • Results in a temperature drop in the surroundings, which is why squeezing energy out of the liquid helps to create ice.
This kind of process is critical in understanding how energy transfer impacts physical changes, like the splitting of a can due to pressure from expanding ice.
Internal Energy
Internal energy refers to the total energy contained within a substance, encompassing both potential and kinetic energy of particles.
During the phase change from liquid to solid, the internal energy of the substance decreases. This is because the particles in a solid arrange in a fixed position, resulting in lower potential energy compared to their more dynamic arrangement in a liquid.
When the can's contents froze, the substance's internal energy changed:
  • Potential energy decreased as particles became more tightly packed.
  • Kinetic energy decreased leading to less movement of particles in the solid state.
These changes facilitated the release of energy necessary to alter the state of the matter inside the can.
Entropy
Entropy is a measure of disorder or randomness in a system. In thermodynamics, entropy tends to increase over time, signifying more disordered states.
However, during a phase change from liquid to solid, entropy decreases because particles settle into an ordered arrangement. Freezing the soft drink contents results in diminished entropy as the molecules pack closely in a lattice-like arrangement typical of solids.
Key points about entropy include:
  • Entropy reduction is associated with the system becoming more ordered as it transitions from liquid to solid.
  • This orderliness requires energy release, allowing the shift from a higher to lower entropy state.
Understanding entropy helps clarify why energy is released when matter transitions to a more orderly form, illustrating essential dynamics in thermodynamics.

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Most popular questions from this chapter

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}:\) $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) and \((\mathbf{b}) 10.4 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

Consider the following reaction: $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) \quad \Delta H=-1204 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(3.55 \mathrm{~g}\) of \(\mathrm{Mg}(s)\) reacts at constant pressure. (c) How many grams of \(\mathrm{MgO}\) are produced during an enthalpy change of \(-234 \mathrm{~kJ}\) ? (d) How many kilojoules of heat are absorbed when \(40.3 \mathrm{~g}\) of \(\mathrm{MgO}(s)\) is decomposed into \(\mathrm{Mg}(s)\) and \(\mathrm{O}_{2}(g)\) at constant pressure?

When an 18.6 -g sample of solid potassium hydroxide dissolves in \(200.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.18), the temperature rises from 23.7 to \(44.5^{\circ}\) C. (a) Calculate the quantity of heat (in kJ) released in the reaction. (b) Using your result from part (a), calculate \(\Delta H\) (in k]/mol KOH) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

It is estimated that the net amount of carbon dioxide fixed by photosynthesis on the landmass of Earth is \(5.5 \times 10^{16} \mathrm{~g} / \mathrm{yr}\) of \(\mathrm{CO}_{2}\). Assume that all this carbon is converted into glucose. (a) Calculate the energy stored by photosynthesis on land per year, in \(\mathrm{kJ} .\) (b) Calculate the average rate of conversion of solar energy into plant energy in megawatts, MW \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s})\). A large nuclear power plant produces about \(10^{3} \mathrm{MW}\). The energy of how many such nuclear power plants is equivalent to the solar energy conversion?

Assume that 2 moles of water are formed according to the following reaction at constant pressure (101.3 kPa) and constant temperature \((298 \mathrm{~K}):\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the pressure-volume work for this reaction. (b) Calculate \(\Delta E\) for the reaction using your answer to (a).

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