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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
Explain your answer. Answer: No, heat generation in a solid does not violate the first law of thermodynamics. The first law states that energy cannot be created or destroyed, but it can change forms. In the examples of heat generation in solids through chemical reactions, friction, and electrical resistance, energy is transformed from one form to another in accordance with the law of conservation of energy, rather than being created or destroyed.

Step by step solution

01

Understand the First Law of Thermodynamics

The first law of thermodynamics, also known as the law of conservation of energy, states that the total energy within a closed system remains constant, meaning energy cannot be created or destroyed, but it can change forms. In a given process, energy can be transferred or converted from one form to another.
02

Investigate Heat Generation in Solids

Heat generation in a solid typically occurs as a result of various processes, such as chemical reactions, friction, and electrical resistance. In each of these cases, the heat generated is a result of the transformation of other forms of energy, rather than being created or destroyed.
03

Analyze Chemical Reactions

When a chemical reaction occurs in a solid, the energy stored in chemical bonds is transformed into heat energy. This heat energy gets distributed throughout the solid. The energy has not been created or destroyed, but rather, transformed from one form (chemical potential energy) to another (heat energy).
04

Analyze Friction

Similarly, when friction is present between a solid and a surface, the mechanical energy (either kinetic or potential) of the object being moved gets transformed into heat energy due to the resistance between the surfaces. Again, the energy conservation principle holds as energy is transferred from mechanical energy to heat energy.
05

Analyze Electrical Resistance

Finally, electrical resistance in a solid can also generate heat. In this case, electrical energy gets converted into heat energy as electrons in a material encounter resistance while moving through the material. The energy used to overcome this resistance appears as heat, and once again, the energy conservation principle holds - electrical energy is transformed into heat energy without being created or destroyed.
06

Conclusion

Based on the examples of heat generation in solids through chemical reactions, friction, and electrical resistance, it can be concluded that heat generation in a solid does not violate the first law of thermodynamics. In all cases, energy is transformed from one form to another in accordance with the law of conservation of energy, rather than being created or destroyed.

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Most popular questions from this chapter

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at $500 \mathrm{~W} / \mathrm{m}^{2}\( with a surrounding temperature of \)0^{\circ} \mathrm{C}$. The convection heat transfer coefficient at the absorber surface is $5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as $T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\(, and determine net heat flux \)\dot{q}_{0}$ absorbed by the solar collector.

A large steel plate having a thickness of \(L=4\) in, thermal conductivity of \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ} \mathrm{F}\), and an emissivity of \(\varepsilon=0.7\) is lying on the ground. The exposed surface of the plate at \(x=L\) is known to exchange heat by convection with the ambient air at \(T_{\infty}=90^{\circ} \mathrm{F}\) with an average heat transfer coefficient of $h=12 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$ as well as by radiation with the open sky with an equivalent sky temperature of \(T_{\text {sky }}=480 \mathrm{R}\). Also, the temperature of the upper surface of the plate is measured to be $75^{\circ} \mathrm{F}\(. Assuming steady one-dimensional heat transfer, \)(a)$ express the differential equation and the boundary conditions for heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the plate by solving the differential equation, and \((c)\) determine the value of the lower surface temperature of the plate at \(x=0\).

A spherical vessel is filled with chemicals undergoing an exothermic reaction. The reaction provides a uniform heat flux on the inner surface of the vessel. The inner diameter of the vessel is \(5 \mathrm{~m}\) and its inner surface temperature is at \(120^{\circ} \mathrm{C}\). The wall of the vessel has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where $k_{0}=1.01 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0018 \mathrm{~K}^{-1}\(, and \)T\( is in \)\mathrm{K}$. The vessel is situated in a surrounding with an ambient temperature of \(15^{\circ} \mathrm{C}\), and the vessel's outer surface experiences convection heat transfer with a coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2}\). K. To prevent thermal burns to workers who touch the vessel, the outer surface temperature of the vessel should be kept below \(50^{\circ} \mathrm{C}\). Determine the minimum wall thickness of the vessel so that the outer surface temperature is \(50^{\circ} \mathrm{C}\) or lower.

Consider a homogeneous spherical piece of radioactive material of radius \(r_{o}=0.04 \mathrm{~m}\) that is generating heat at a constant rate of \(\dot{e}_{\mathrm{gen}}=4 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The heat generated is dissipated to the environment steadily. The outer surface of the sphere is maintained at a uniform temperature of \(80^{\circ} \mathrm{C}\), and the thermal conductivity of the sphere is $k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(. Assuming steady one-dimensional heat transfer, \)(a)$ express the differential equation and the boundary conditions for heat conduction through the sphere, \((b)\) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and \((c)\) determine the temperature at the center of the sphere.

What is the difference between an ordinary differential equation and a partial differential equation?

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