An electric current passes through a metal strip \(0.50 \mathrm{cm}\) by \(5.0 \mathrm{cm}\) by \(0.10 \mathrm{mm},\) heating it at a rate of \(50 \mathrm{W} .\) The strip has emissivity \(e=1\) and its surroundings are at \(300 \mathrm{K}\). What will be the strip's temperature if it's enclosed in (a) a vacuum boule transparent to all radiation and (b) an insulating box with thermal resistance \(R=8.0 \mathrm{K} / \mathrm{W}\) that blocks all radiation?

When an electric current passes through a conductor, such as a metal strip, it generates heat due to the resistance of the material to the flow of current. This process is known as electric current heating or Joule heating. The power of the heating, measured in watts (W), can be calculated by using the formula \( P = I^2 \cdot R \), where \( P \) represents the power dissipated as heat, \( I \) is the current through the conductor, and \( R \) is the resistance.

In the given exercise, the metal strip is heated at a constant rate of 50W. This power is the energy per unit of time that is converted from electrical energy into thermal energy within the strip, causing its temperature to rise. The heating rate also depends on the properties of the material, such as its resistivity and dimensions, which influence how readily it converts the electrical energy into heat. The more resistant the material, the more heat is produced for a given current.

To regulate and control this heating effect, various materials and configurations are used based on their resistance and heat dissipation properties. In practical applications, thermal management solutions are necessary to ensure that the temperature remains within safe limits to avoid damaging the conductor or surrounding components.

Thermal resistance is a property that quantifies the temperature difference across a material when heat is transferred through it. It is analogous to electrical resistance, but instead of hindering electrical current, it impedes the flow of heat. The higher the thermal resistance, the better the material is at insulating and the less heat is lost for a given temperature difference.

In the context of our exercise, the insulating box with a given thermal resistance of \( 8.0 \, \text{K/W} \) dictates how the heat is transferred from the heated strip to its environment. To maintain a steady state, the power of current heating (50W) must equal the power of heat transfer related to the thermal resistance, expressed as \( P = (T - T0) / R \) where \( T \) is the strip's temperature, \( T0 \) is the surrounding temperature, and \( R \) is the thermal resistance. By isolating \( T \) in this equation, we can determine the temperature of the strip while it is enclosed within the insulating box.

The concept of thermal resistance is crucial in designing thermal insulation and managing heat dissipation in electronic devices, buildings, and various industrial processes. It allows engineers to predict how much energy will be needed to maintain certain temperature conditions and how to prevent overheating or inefficiency.

The term emissivity refers to the efficiency with which a surface emits thermal radiation. It is a dimensionless quantity that ranges from 0 to 1, with 1 being a perfect emitter and absorber—a black body—and 0 indicating that the surface does not emit or absorb thermal radiation at all.

In our exercise, the given metal strip has an emissivity (\(e\)) of 1, indicating that it behaves as a perfect black body with respect to thermal radiation. The strip's ability to emit heat by radiation is governed by the Stefan-Boltzmann law, \[ P = e \cdot \sigma \cdot A \cdot (T^4 - T0^4) \], where \(P\) is the radiative power, \(e\) is emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area of the strip, \(T\) is the strip's temperature, and \(T0\) is the surrounding temperature.

The role of emissivity is significant because it affects the rate at which an object will cool down or heat up when exposed to an environment with a different temperature. In the case of the vacuum boule, where the only heat exchange mode is radiation (since no medium exists for conduction or convection), understanding thermal radiation and emissivity becomes essential to predict the strip's temperature. Thus, controlling emissivity is key in thermal engineering to manage the energy efficiency of systems that cool down or heat up through radiative heat transfer.