/*! 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 35 A closed system of mass \(5 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A closed system of mass \(5 \mathrm{~kg}\) undergoes a process in which there is work of magnitude \(9 \mathrm{~kJ}\) to the system from the surroundings. The elevation of the system increases by \(700 \mathrm{~m}\) during the process. The specific internal energy of the system decreases by \(6 \mathrm{~kJ} / \mathrm{kg}\) and there is no change in kinetic energy of the system. The acceleration of gravity is constant at \(g=9.6\) \(\mathrm{m} / \mathrm{s}^{2}\). Determine the heat transfer, in \(\mathrm{kJ}\).

Short Answer

Expert verified
The heat transfer is \( -3.6 \text{ kJ} \).

Step by step solution

01

Identify known values

List all the given data.- Mass of the system, \( m = 5 \text{ kg} \)- Work done on the system, \( W = 9 \text{ kJ} \)- Change in elevation, \( \text{Δ}z = 700 \text{ m} \)- Change in specific internal energy, \( \text{Δ}u = -6 \text{ kJ/kg} \)- Acceleration due to gravity, \( g = 9.6 \text{ m/s}^2 \)
02

Calculate the change in internal energy

The change in internal energy per unit mass is given by \( \text{Δ}u = -6 \text{ kJ/kg} \). Multiply by the mass to find the total change in internal energy.\( \text{Δ}U = m \times \text{Δ}u = 5 \text{ kg} \times (-6 \text{ kJ/kg}) = -30 \text{ kJ} \)
03

Calculate the change in potential energy

Use the formula for potential energy change: \( \text{Δ}PE = m \times g \times \text{Δ}z \). Plug in the known values.\( \text{Δ}PE = 5 \text{ kg} \times 9.6 \text{ m/s}^2 \times 700 \text{ m} = 33600 \text{ J} \). Convert this to kJ: \( 33600 \text{ J} = 33.6 \text{ kJ} \)
04

Apply the First Law of Thermodynamics

Use the First Law of Thermodynamics which states \( \text{Δ}U = Q - W + \text{Δ}KE + \text{Δ}PE \). Given \( \text{Δ}KE = 0 \), solve for heat transfer, \( Q \).\( -30 \text{ kJ} = Q - 9 \text{ kJ} + 0 \text{ kJ} + 33.6 \text{ kJ} \)Rearrange the equation to solve for \( Q \).\( Q = -30 \text{ kJ} + 9 \text{ kJ} - 33.6 \text{ kJ} = -3.6 \text{ kJ} \)

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle in thermodynamics. It states that energy cannot be created or destroyed in an isolated system. Instead, the total energy of the system is conserved. The equation representing this law is: \[\text{Δ}U = Q - W + \text{Δ}KE + \text{Δ}PE\]Where: \(\text{Δ}U\) is the change in internal energy, \(Q\) is the heat added to the system, \(W\) is the work done by the system, \(\text{Δ}KE\) is the change in kinetic energy, and \( \text{Δ}PE\) is the change in potential energy.The essence of this law is that the energy entering a system must either increase the internal energy, change the kinetic or potential energy, or be used to do work.
Internal Energy
Internal Energy (\text{U}\text{)} refers to the energy contained within a system due to the kinetic and potential energies of its molecules. During a process, changes in internal energy (\text{ΔU}) occur due to heat transfer and work interactions. In our exercise, the specific internal energy of the system decreases by \(-6 \text{ kJ/kg}\). With a mass of \(5 \text{ kg}\), the total change in internal energy is:\[\text{Δ}U = m \times \text{Δ}u = 5 \text{ kg} \times (-6 \text{ kJ/kg}) = -30 \text{ kJ}\]This decrease in internal energy indicates that the system has lost some of its internal energy during the process.
Potential Energy
Potential Energy (\text{PE}\text{)} is the energy possessed by a system due to its position in a gravitational field. For an object at a height \(z\) with mass \(m\), the potential energy is given by \(PE = m \times g \times z \), where \(\text{g}\) is the acceleration due to gravity. In our problem, the change in elevation is \(\text{Δ}z = 700 \text{ m}\). The initial and final potential energies differ by:\[\text{Δ}PE = m \times g \times \text{Δ}z = 5 \text{ kg} \times 9.6 \text{ m/s}^2 \times 700 \text{ m} = 33600 \text{ J} = 33.6 \text{ kJ} \]This positive change in potential energy means the system's elevation increased, adding energy to the system in the form of gravitational potential energy.
Work-Energy Principle
The Work-Energy Principle states that the work done on or by a system results in a change in its energy. Mathematically, it can be written as: \[W = \text{Δ}E \], where \(\text{Δ}E\) is the change in the system's energy. In our exercise, the work done on the system is \(\text{W} = 9 \text{kJ}\). The heat transfer \(Q\) can be found using the First Law of Thermodynamics: \[ \text{Δ}U = Q - W + \text{Δ}PE \]Substituting the known values: \[ -30 \text{kJ} = Q - 9 \text{kJ} + 33.6 \text{kJ} \]Rearranging to solve for \(Q\), the heat transfer is: \[Q = -30 \text{kJ} + 9 \text{kJ} + 33.6 \text{kJ} = 12.6 \text{kJ} \]Thus, a heat transfer of \(12.6 \text{kJ}\) occurs during the process to balance the changes in internal energy, work done, and potential energy change.

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Most popular questions from this chapter

A gas expands from an initial state where \(p_{1}=500 \mathrm{kPa}\) and \(V_{1}=0.1 \mathrm{~m}^{3}\) to a final state where \(p_{2}=100 \mathrm{kPa}\). The relationship between pressure and volume during the process is \(p V=\) constant. Sketch the process on a \(p-V\) diagram and determine the work, in \(\mathrm{kJ}\).

Two objects having different masses fall freely under the influence of gravity from rest and the same initial elevation. Ignoring the effect of air resistance, show that the magnitudes of the velocities of the objects are equal at the moment just before they strike the earth.

A gas is compressed from \(V_{1}=0.3 \mathrm{~m}^{3}, p_{1}=1\) bar to \(V_{2}=0.1 \mathrm{~m}^{3}, p_{2}=3\) bar. Pressure and volume are related linearly during the process. For the gas, find the work, in \(\mathrm{kJ}\).

A household refrigerator with a coefficient of performance of \(2.4\) removes energy from the refrigerated space at a rate of \(200 \mathrm{~W}\). Evaluating electricity at \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), determine the cost of electricity in a month when the refrigerator operates for 360 hours.

A disk-shaped flywheel, of uniform density \(\rho\), outer radius \(R\), and thickness \(w\), rotates with an angular velocity \(\omega\), in \(\mathrm{rad} / \mathrm{s}\). (a) Show that the moment of inertia, \(I=\int_{\text {vol }} \rho r^{2} d V\), can be expressed as \(I=\pi \rho w R^{4} / 2\) and the kinetic energy can be expressed as \(\mathrm{KE}=I \omega^{2} / 2\). (b) For a steel flywheel rotating at 3000 RPM, determine the kinetic energy, in \(\mathrm{N} \cdot \mathrm{m}\), and the mass, in \(\mathrm{kg}\), if \(R=0.38 \mathrm{~m}\) and \(w=0.025 \mathrm{~m}\). (c) Determine the radius, in \(\mathrm{m}\), and the mass, in \(\mathrm{kg}\), of an aluminum flywheel having the same width, angular velocity, and kinetic energy as in part (b).

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