A copper wire of length \(1.317 \mathrm{~m}\) and diameter \(3.26 \times 10^{-4} \mathrm{~m}\) is blackened and placed along the axis of an evacuated glass tube. The wire is connected to a battery, a rheostat, an ammeter, and a voltmeter, and the current is increased until, at the moment the wire is about to melt, the ammeter reads \(12.8 \mathrm{~A}\) and the voltmeter reads \(20.2 \mathrm{~V}\). Assuming that all the energy supplied was radiated and that the radiation from the glass tube is negligible, calculate the melting temperature of copper.

Ohm's Law is fundamental to understanding electrical circuits. It states that the current passing through a conductor between two points is directly proportional to the voltage across these two points, given a constant temperature. The formula is \( V = IR \) where:

• \( V \) is the voltage (in volts)
• \( I \) is the current (in amperes)
• \( R \) is the resistance (in ohms)

In this exercise, we used Ohm's Law to find the resistance of the copper wire by dividing the voltage (20.2 V) by the current (12.8 A), resulting in a resistance of 1.578 ohms. This step is crucial as it allows us to understand the behavior of the wire under electrical stress.

The Stefan-Boltzmann Law helps us understand how objects radiate energy. The law states that the total energy radiated per unit surface area of a black body is proportional to the fourth power of the black body's temperature. The formula is: \[ P = e \, \sigma \, A \, T^4 \] where:

• \( P \) is the radiation power (in watts)
• \( e \) is the emissivity of the material (ranges from 0 to 1; for blackened surfaces, it's close to 1)
• \( \sigma \) is the Stefan-Boltzmann constant, \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)
• \( A \) is the surface area (in square meters)
• \( T \) is the temperature in Kelvin

In this exercise, the Stefan-Boltzmann Law allowed us to relate the power supplied to the wire to its temperature, taking into account the emissivity and surface area.

To find out how much power is radiated by the wire, we need to know the surface area. For a cylindrical wire, the surface area \(A \) can be calculated using: \( A = \pi \, d \, L \) where:

• \( \pi \) is approximately 3.1416
• \( d \) is the diameter of the wire
• \( L \) is the length of the wire

In the problem, we used the given diameter (\( 3.26 \times 10^{-4} \, \text{m} \)) and length (\( 1.317 \, \text{m} \)) to find the surface area: \( 1.348 \times 10^{-3} \, \text{m}^2 \). This is important as it helps us understand how much of the wire’s surface is radiating energy.

Radiation power is the energy emitted by an object due to its temperature. As per the Stefan-Boltzmann Law, the radiation power \( P \) depends on several factors: emissivity, surface area, and temperature. After calculating the surface area and obtaining the resistance, we could find the power supplied to the wire with: \( P = V \, I \) Substituting the values, we got: \( P = 20.2 \, V \times 12.8 \, A = 258.56 \, W \). This power is what the wire emits as heat. Finally, using the Stefan-Boltzmann equation, we isolated the temperature \( T \) to deduce how hot the wire gets just before melting.

Electrical resistance \( R \) is a measure of how much an object opposes the flow of electric current. The unit for resistance is ohms (\( \Omega \)). The formula, as per Ohm's Law, is: \( R = \frac{V}{I} \) where voltage (\( V \)) and current (\( I \)) are known. In this context, the resistance tells us how much the wire resists the current flowing through it. Higher resistance means more energy is converted to heat. For the copper wire, we computed the resistance to be 1.578 ohms. This information was key in calculating the radiation power and, ultimately, the temperature of the wire. Understanding resistance helps us predict how materials behave in electrical circuits, especially under high loads.