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Chapter 17: Problem 36

For a process, \(w=34 \mathrm{~J}\) and \(q=-109 \mathrm{~J} .\) What is \(\Delta E\) for this process?

Short Answer

Expert verified
\( \Delta E = -75 \text{ J} \)

Step by step solution

01

Identify the Formula

To find the change in energy, \( \Delta E \), you need to use the first law of thermodynamics. The formula is \( \Delta E = q + w \), where \( q \) is the heat exchange and \( w \) is the work done by or on the system.
02

Substitute Known Values

Plug the given values into the formula. Here, \( q = -109 \text{ J} \) and \( w = 34 \text{ J} \). Substitute these into the formula: \( \Delta E = -109 + 34 \).
03

Perform the Calculation

Calculate \( \Delta E \) by performing the arithmetic operation. \( -109 + 34 = -75 \).
04

Interpret the Result

The outcome of the calculation is \( \Delta E = -75 \text{ J} \). This negative value indicates that the system has lost 75 joules of energy during the process.

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Key Concepts

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

Energy Change
The concept of energy change, or \( \Delta E \), is a fundamental part of understanding the first law of thermodynamics. According to this law, energy cannot be created or destroyed in an isolated system; it can only change forms.
In practical terms, this means the total energy of a system is the sum of the heat added to the system and the work done on the system.
  • If \( \Delta E \) is positive, the system gains energy.
  • If \( \Delta E \) is negative, the system loses energy.
For instance, given \( q = -109 \text{ J} \) and \( w = 34 \text{ J} \), the energy change is calculated as \( \Delta E = q + w = -109 + 34 = -75 \text{ J} \). This result tells us the system has lost energy. It highlights the fact that the system's total energy has decreased by 75 joules during the process.
Heat Exchange
Heat exchange, denoted by the symbol \( q \), refers to the transfer of thermal energy between a system and its surroundings. It is a crucial component in the analysis of thermodynamic processes.
When a system absorbs heat, \( q \) is positive, indicating energy is entering the system. Conversely, when a system releases heat, \( q \) is negative, denoting energy leaves the system. In the given problem, \( q = -109 \text{ J} \) indicates the system has lost 109 joules to its surroundings.
  • This loss of heat suggests an exothermic process.
  • Heat exchange affects the system's energy balance, influencing the calculated \( \Delta E \).
Understanding heat exchange helps in comprehending how energy flows in and out of a system, allowing for a deeper insight into thermodynamic reactions and processes.
Work Done
Work done, represented by \( w \), is another key element in the study of thermodynamics. It refers to the energy transferred by the system, excluding heat. Work can be done on the system or by the system.
A positive \( w \) means work is done on the system, adding energy to it. A negative \( w \) indicates work is done by the system, resulting in an energy loss. In our scenario, \( w = 34 \text{ J} \) means 34 joules of energy have been added to the system through work done on it.
  • Work can involve processes such as compression or expansion of gases.
  • It is essential to differentiate between heat exchange and work because they represent distinct ways energy can enter or leave a system.
By analyzing work done, students can better understand how physical processes influence the energy dynamics of a system.

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Most popular questions from this chapter

For Exercises 17.87 to 17.102 , assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. State whether increasing temperature increases or decreases the value of the equilibrium constant for the following reactions. (a) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{~g})\) (c) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{HCOOH}(\mathrm{g})\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (e) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\)

What is the sign of the entropy change for each of the following processes? The system is underlined. (a) A plate is dropped on the floor and shatters. (b) A shuffled deck of cards is reordered from aces to deuces. (c) Iron rusts into iron oxide. (d) \(A\) wooden fence rots.

Explain the difference between internal energy and enthalpy.

When most soluble ionic compounds dissolve in water, the enthalpy of solution is positive (the process is endothermic). What conclusion(s) can be made about the entropy change that accompanies the dissolution of these substances?

Calculate the work performed if a balloon contracts from \(12.90 \mathrm{~L}\) to \(788 \mathrm{~mL}\) because of an external pressure of \(3.70 \mathrm{~atm} .\)
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