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Chapter 13: Problem 17

Calculate the energy transforred as work, \(w,\) for a system that releases \(450 \mathrm{J}\) of heat in a process for which the internal energy of the system decreases by 135 J. (Section 13.5 )

Short Answer

Expert verified
The work done on the system is 315 J.

Step by step solution

01

Understanding the Problem

In this problem, we are given that a system releases 450 J of heat, which means the heat transferred, \( q \), is \( -450 \) J. We are also told that the internal energy of the system decreases by 135 J, so the change in internal energy, \( \Delta U \), is \( -135 \) J. We need to calculate the energy transferred as work, \( w \).
02

Using the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat added to the system plus the work done on the system: \( \Delta U = q + w \). In this equation, \( \Delta U \) is the change in internal energy, \( q \) is the heat exchange, and \( w \) is the work. We need to rearrange this equation to solve for \( w \): \( w = \Delta U - q \).
03

Substitute Known Values

Substitute the known values into the rearranged equation \( w = \Delta U - q \). Here, \( \Delta U = -135 \) J and \( q = -450 \) J, so we have: \( w = -135 + 450 \).
04

Calculate the Work

Calculate \( w \) using the substituted values: \( w = -135 + 450 = 315 \). The work done on the system is 315 J.

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Key Concepts

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

Internal Energy
Internal energy is a key concept in thermodynamics. It refers to the total energy stored within a system. This includes both kinetic energy, due to the movement of molecules, and potential energy, from forces between them. Think of internal energy like your body's energy: some comes from how fast you move, and some from the bonds between atoms.
In a thermodynamic process, any change in internal energy (}U) is a result of interactions like heat transfer and work done. This change is crucial because it helps us understand how energy moves and transforms within a system. For example, if a system's internal energy decreases by 135 J, it means more energy has left (through heat or work) than has entered.
Heat Transfer
Heat transfer occurs when there is a temperature difference between two systems or within a part of a system. In the exercise, the system releases 450 J of heat, indicating heat has transferred out.
This outflow of energy is characterized by the term 'q', which is negative when heat is released, as in our example (= -450 J). Heat transfer can happen in three ways: conduction, convection, and radiation, but the exact method isn't specified here.
  • **Conduction**: Direct contact transfer, like a spoon getting hot in a pot of boiling water.
  • **Convection**: Transfer by movement of fluids, like hot air rising.
  • **Radiation**: Transfer through electromagnetic waves, like the Sun warming your skin.
Knowing how heat is transferred helps us understand how energy moves in all sorts of systems, from engines to ecosystems.
Work Calculation
Work in thermodynamics is the process of energy transfer that doesn't involve temperature change directly but instead involves force and movement. When performing work, a system might expand a gas or move an object, contributing to a change in its internal energy.
The first law of thermodynamics helps us calculate the work () done by the system. It states that U = q + w, depicting how changes in the internal energy (U) are due to heat (q) and work (w).
To find , rearrange the formula to  = U - q. Here, the internal energy change (U = -135 J) and the heat transfer (= -450 J) lead to a calculation of:\[w = -135 - (-450)w = 315 J\]This positive work value means the system has done 315 J of work, possibly moving an external component. Understanding this helps clarify how energy within a system can translate into physical processes or mechanical actions.

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Most popular questions from this chapter

The molar heat capacity of water at constant pressure is \(75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) at room temperature. Calculate the mass of methanol that must be burned to heat \(1 \mathrm{dm}^{3}\) of water from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). (Assume the density of water is \(1.0 \mathrm{gcm}^{-3}\) ) (Section \(13.3)\)

Calcite and aragonite are two forms of calcium carbonate. Calculate the enthalpy change for the transition from calcite to aragonite. (AH(calcite) \(=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}\) and \(\Delta \mathrm{H}\) (aragonite) \(=\) \(-1207.1 \mathrm{kJ} \mathrm{mol}^{-1}\), j (Section 13.3)

Classify the following properties as intensive or extensive: density; amount in moles; pressure; length; temperature. (Section \(13.1)\)

A slice of banana weighing \(2.7 \mathrm{g}\) was burned in oxygen in a bomb calorimeter and produced a temperature rise of \(3.05 \mathrm{K}\) In the same calorimeter, the combustion of \(0.316 \mathrm{g}\) of benzoic acid produced a temperature rise of \(3.24 \mathrm{K}\). Ac U for benzoic acid is \(-3251 \mathrm{kJmol}^{-1}\). If the average mass of a banana is \(125 \mathrm{g}\), how much energy in (a) \(\mathrm{kJ}\) and (b) calories (kcal) can be obtained on average from a banana? (1 calorie \(=4.18 \mathrm{kJ}\) ) (Section 13.5 )

Car safoty airbags inflate when the car undergoes a sudden deceleration, setting oft a reaction which produces a large amount of gas. One of the reactions used is \(\mathrm{NaN}_{3}(\mathrm{s}) \rightarrow \mathrm{Na}(\mathrm{s})+1.5 \mathrm{N}_{2}(\mathrm{g}) \quad \Delta, \mathrm{H}^{0}_{29 \mathrm{e}}=-21.7 \mathrm{kJmol}^{-1}\) (Section \(13.3)\) (a) What is the value of \(\Delta_{1} H_{294}\) for \(\mathrm{NaN}_{3} ?\) (b) \(\quad\) An airbag system is inflated by 2.4 mol (around \(60 \mathrm{dm}^{3}\) ) of \(\mathrm{N}_{2}(\mathrm{q})\). Calculate the enthalpy change when this amount of nitrogen is produced at \(298 \mathrm{K}\) (c) Assuming this energy all goes into heating the \(\mathrm{N}_{2}(\mathrm{g}),\) what Will be the final ternperature of the gas? \(\left(C_{p} \text { for } N_{2}(g)=\right.\) \(\left.29.1 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\right)\)
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