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Chapter 6: Problem 68

If a system has \(48 \mathrm{~J}\) of work done on it and absorbs \(22 \mathrm{~J}\) of heat, what is the value of \(\Delta E\) for this change?

Short Answer

Expert verified
\(\Delta E = 70 \mathrm{~J}\)

Step by step solution

01

Understanding the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in the internal energy of a system, denoted as \(\Delta E\), is equal to the heat, \(Q\), added to the system minus the work, \(W\), done by the system on its surroundings. The equation is \(\Delta E = Q - W\). However, if work is done on the system, we consider \(W\) to be negative, so the equation becomes \(\Delta E = Q + W\).
02

Apply Values to the Formula

Using the given values, \(Q = 22 \mathrm{~J}\) as the heat absorbed and \(W = 48 \mathrm{~J}\) as the work done on the system, we can insert these into the equation \(\Delta E = Q + W\), giving us \(\Delta E = 22 \mathrm{~J} + 48 \mathrm{~J}\).
03

Calculate the Change in Internal Energy

Adding the quantities of heat and work yields the change in internal energy: \(\Delta E = 22 \mathrm{~J} + 48 \mathrm{~J} = 70 \mathrm{~J}\). So the value of \(\Delta E\) for this change is \(70 \mathrm{~J}\).

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Key Concepts

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

Internal Energy Change
The concept of internal energy change is central to understanding the First Law of Thermodynamics. Internal energy refers to the total energy contained within a system, encompassing both the kinetic and potential energies of the particles. When we talk about a change in internal energy, denoted as \(\Delta E\), we are referring to the difference between the total energy of a system at two different states.

For an isolated system, internal energy can change due to two main interactions: heat exchange and work done. Imagine you have a balloon that you are heating while simultaneously compressing it. Heat will increase the internal energy, while compression work also contributes positively to the system's internal energy. When there is an increase in internal energy, it suggests that the system has either absorbed heat, had work done on it, or a combination of both. It's important to visualize internal energy as the cumulative outcome of these interactions rather than a standalone concept.
Heat Absorption
Heat absorption is the process whereby a system takes in heat (\(.Q.\)) from its surroundings. Heat can be thought of as energy in transit, and it only exists when there is a temperature difference between the system and its environment. As heat is absorbed, it increases the internal energy of the system, leading to various possible changes such as an increase in temperature, a change in state (like ice melting to water), or an increase in volume.

Imagine placing a cold metal rod into a cup of hot coffee. The rod absorbs heat from the coffee and gets warmer. This heat transfer continues until the temperature of the metal rod and the coffee equalize, highlighting the direction of heat flow from hot to cold objects. Heat absorption can be measured in joules (J), and it's crucial in calculating the change in the internal energy of a system.
Work Done on System
When we discuss work done on a system in thermodynamics, we are referring to the external force causing a displacement of the system's boundary. Simply put, for work (\(.W.\)) to be done on a system, there must be a force that acts on it through a distance. In a closed system, this work can result in a change in volume, such as compressing a piston in a cylinder.

Take, for instance, a bicycle pump. Each time you push down on the pump, you're doing work on the air inside by decreasing its volume and increasing its internal energy. This type of work, especially when it involves gas compression, is a driving factor in engines and refrigeration cycles. Work is also measured in joules (J), and it's fundamental to understand that when work is done on a system, it is considered positive and adds to the internal energy of the system.
Thermodynamics Equations
Thermodynamics equations are the mathematical representations that describe the principles governing energy exchanges. One of the core equations in thermodynamics is the expression for the First Law, which relates internal energy change, heat, and work. It is given by \(\Delta E = Q + W\), where \(\Delta E\) is the change in internal energy, \(.Q.\) represents the heat absorbed by the system, and \(.W.\) is the work done on the system.

The utility of this equation can be seen in various disciplines—from engineering systems that harness energy to understanding biological processes. Remembering that this equation assumes work done on the system is positive can help avoid common mistakes when applying it to real-life situations. Knowing how to use this equation properly allows one to predict how a system will respond to energy input or output, which is a fundamental aspect of thermodynamics.

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Most popular questions from this chapter

What distinguishes a thermochemical equation from an ordinary chemical equation?

Magnesium burns in air to produce a bright light and is often used in fireworks displays. The combustion of magnesium can be described by the following thermochemical equation: \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) \quad \Delta H^{\circ}=-1203 \mathrm{~kJ}\) How much heat (in kilojoules) is liberated by the combustion of \(6.54 \mathrm{~g}\) of magnesium?

Toluene, \(\mathrm{C}_{7} \mathrm{H}_{8}\), is used in the manufacture of explosives such as TNT (trinitrotoluene). A \(1.500 \mathrm{~g}\) sample of liquid toluene was placed in a bomb calorimeter along with excess oxygen. When the combustion of the toluene was initiated, the temperature of the calorimeter rose from \(25.000^{\circ} \mathrm{C}\) to \(26.413^{\circ} \mathrm{C}\). The products of the combustion were \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l),\) and the heat capacity of the calorimeter was \(45.06 \mathrm{~kJ}^{\circ} \mathrm{C}^{-1}\) (a) Write the balanced chemical equation for the reaction in the calorimeter. (b) How many joules were liberated by the reaction? (c) How many joules would be liberated under similar conditions if 1.000 mol of toluene was burned?

How does the potential energy change (increase, decrease, or no change) for each of the following? (a) Two electrons come closer together. (b) An electron and a proton become farther apart. (c) Two atomic nuclei approach each other. (d) A ball rolls downhill.

Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 \(\left(\mathrm{CFCl}_{3}\right)\) and Freon-12 \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right),\) which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the "ozone hole" that develops at the end of the Antarctic winter. In other parts of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earth's inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, \(\mathrm{O}_{3}\). Possible reactions are $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{ClO}(g) & \Delta H^{\circ}=-126 \mathrm{~kJ} \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) & \Delta H^{\circ}=-268 \mathrm{~kJ} \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) & \Delta H^{\circ}=? \end{array} $$ The \(\mathrm{O}\) atoms in the second equation come from the breaking apart of \(\mathrm{O}_{2}\) molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of \(\Delta H^{\circ}\) (in kilojoules) for the last equation, the net reaction for the removal of \(\mathrm{O}_{3}\) from the atmosphere.
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