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Chapter 2: Problem 79

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

01

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.
03

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.
04

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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Most popular questions from this chapter

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

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.

A spherical container, with an inner radius \(r_{1}=1 \mathrm{~m}\) and an outer radius \(r_{2}=1.05 \mathrm{~m}\), has its inner surface subjected to a uniform heat flux of \(\dot{q}_{1}=7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container has a temperature \(T_{2}=25^{\circ} \mathrm{C}\), and the container wall thermal conductivity is \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Show that the variation of temperature in the container wall can be expressed as \(T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 / r_{2}\right)+T_{2}\) and determine the temperature of the inner surface of the container at \(r=r_{1}\).

Heat is generated in a \(3-\mathrm{cm}\)-diameter spherical radioactive material uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). Heat is dissipated to the surrounding medium at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the material in steady operation is (a) \(56^{\circ} \mathrm{C}\) (b) \(84^{\circ} \mathrm{C}\) (c) \(494^{\circ} \mathrm{C}\) (d) \(650^{\circ} \mathrm{C}\) (e) \(108^{\circ} \mathrm{C}\)

Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(180 \mathrm{~W} / \mathrm{cm}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(12.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(13.5 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(64.7 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(180 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(191 \mathrm{~W} / \mathrm{cm}^{2}\)
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