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Does heat generation in a solid violate the first law of thermodynamics, which states that energy cannot be created or destroyed? Explain.

Short Answer

Expert verified
Answer: No, heat generation in a solid does not violate the first law of thermodynamics. The energy in the solid is conserved and simply converted from one form to another without creating or destroying energy.

Step by step solution

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1. Understanding the First Law of Thermodynamics

The first law of thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only converted from one form to another. This means that the total amount of energy in a closed system remains constant throughout any processes occurring within the system.
02

2. Heat Generation in a Solid

Heat generation in a solid occurs due to various processes such as friction, chemical reactions, and other mechanisms that cause the material atoms or molecules to collide and transfer energy within the solid. As a result, the solid's internal energy increases as the heat is generated, leading to an increase in the solid's temperature.
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3. Analyzing the heat generation process

When heat is generated in a solid, energy is being transferred from one part of the material to another. The transfer may convert potential energy to kinetic energy or vice versa, but the total amount of energy within the solid remains constant. The first law of thermodynamics is obeyed as no energy is being created or destroyed during the heat generation process.
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4. Conclusion

Heat generation in a solid does not violate the first law of thermodynamics, as energy is not being created or destroyed in this process. Instead, energy is transferred from one form to another within the solid, ensuring that the total amount of energy within the system remains constant.

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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 of physics, often referred to as the Law of Energy Conservation. This law tells us that energy in a closed system is constant. It cannot just appear or disappear.

When we talk about processes occurring in a system, energy might seem to change, but it's only converting between forms.

This concept is crucial because it helps us understand that all energy transitions in physical processes are balanced. The energy may shift between kinetic (motion), potential (stored energy), thermal (heat), or other forms.
  • If a system gains energy, it must be lost from elsewhere.
  • Similarly, if a system loses energy, it goes somewhere else within the system.
The First Law is the backbone of energy studies and applications in real-life scenarios.
Energy Conservation
Energy Conservation directly follows from the First Law of Thermodynamics. It implies we cannot generate energy from nothing. However, we can manage how it's transformed and transferred.

In everyday activities, energy is being conserved through transformations. When you pedal a bike, your body's stored energy changes to kinetic energy.

Yet conservation also refers to something more practical—saving energy, optimizing its use, and minimizing waste.
  • Energy-efficient technologies aim to conserve energy economically.
  • Limiting unnecessary energy dissipation also achieves this goal.
Understanding this concept allows us to design processes and devices that optimize energy use, reducing costs and environmental impacts.
Internal Energy
Internal Energy is the total energy stored within a system. It is comprised of both kinetic and potential energies at the microscopic level.

In solids, this includes the energy due to atom vibrations and interactions.

The internal energy of a solid increases with heat generation, as atoms collide and transfer energy among themselves.
  • Internal energy is pivotal in understanding temperature changes in materials.
  • Sometimes even slight changes lead to noticeable effects like expansion or phase changes.
This concept is vital in thermodynamics, as it explains how energy, once added to a system, affects it internally.
Energy Transfer
Energy Transfer in a system occurs as energy moves between different parts, causing a change in form but not in quantity.

For example, when heating a metal rod, energy is transferred from the heat source to the rod's atoms, raising their internal energy.

The transfer can involve various forms—thermal, mechanical, or electrical—but always respects the conservation principle.
  • Energy transfer can happen via conduction, convection, or radiation.
  • This process ensures energy distributes throughout a material.
Understanding energy transfer helps in tackling complex processes like engine operations, electricity generation, and even climate systems, elucidating how and why energy moves as it does.

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Most popular questions from this chapter

Consider a small hot metal object of mass \(m\) and specific heat \(c\) that is initially at a temperature of \(T_{i}\). Now the object is allowed to cool in an environment at \(T_{\infty}\) by convection with a heat transfer coefficient of \(h\). The temperature of the metal object is observed to vary uniformly with time during cooling. Writing an energy balance on the entire metal object, derive the differential equation that describes the variation of temperature of the ball with time, \(T(t)\). Assume constant thermal conductivity and no heat generation in the object. Do not solve.

When a long section of a compressed air line passes through the outdoors, it is observed that the moisture in the compressed air freezes in cold weather, disrupting and even completely blocking the air flow in the pipe. To avoid this problem, the outer surface of the pipe is wrapped with electric strip heaters and then insulated. Consider a compressed air pipe of length \(L=6 \mathrm{~m}\), inner radius \(r_{1}=3.7 \mathrm{~cm}\), outer radius \(r_{2}=4.0 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) equipped with a 300 -W strip heater. Air is flowing through the pipe at an average temperature of \(-10^{\circ} \mathrm{C}\), and the average convection heat transfer coefficient on the inner surface is \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming 15 percent of the heat generated in the strip heater is lost through the insulation, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and \((c)\) evaluate the inner and outer surface temperatures of the pipe.

The temperatures at the inner and outer surfaces of a 15 -cm-thick plane wall are measured to be \(40^{\circ} \mathrm{C}\) and \(28^{\circ} \mathrm{C}\), respectively. The expression for steady, one-dimensional variation of temperature in the wall is (a) \(T(x)=28 x+40\) (b) \(T(x)=-40 x+28\) (c) \(T(x)=40 x+28\) (d) \(T(x)=-80 x+40\) (e) \(T(x)=40 x-80\)

Heat is generated in a 10 -cm-diameter spherical radioactive material whose thermal conductivity is \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). If the surface temperature of the material is measured to be \(120^{\circ} \mathrm{C}\), the center temperature of the material during steady operation is (a) \(160^{\circ} \mathrm{C}\) (b) \(205^{\circ} \mathrm{C}\) (c) \(280^{\circ} \mathrm{C}\) (d) \(370^{\circ} \mathrm{C}\) (e) \(495^{\circ} \mathrm{C}\)

Consider a long rectangular bar of length \(a\) in the \(x-\) direction and width \(b\) in the \(y\)-direction that is initially at a uniform temperature of \(T_{i}\). The surfaces of the bar at \(x=0\) and \(y=0\) are insulated, while heat is lost from the other two surfaces by convection to the surrounding medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient two-dimensional heat transfer with no heat generation, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

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