Question

A \(2.50-\mathrm{m}\) -long copper cable is connected across the terminals of a \(12.0-\mathrm{V}\) car battery, Assuming that it is completely insulated from its environment, how long after the connection is made will the copper start to melt? (Useful information: copper has a mass density of \(8960 \mathrm{~kg} / \mathrm{m}^{3}\), a melting point of \(1359 \mathrm{~K},\) and a specific heat of \(386 \mathrm{~J} / \mathrm{kg} / \mathrm{K}\).).

Short answer

Solution: Step 1: Calculate the cross-sectional area of the copper cable \(Area = \frac{Mass}{(2.50 \mathrm{~m})(8960 \mathrm{~kg} / \mathrm{m}^{3})}\) Step 2: Find the resistance of the copper cable \(Resistance = \frac{(1.724 \times 10^{-8} \Omega \cdot \mathrm{m})(2.50 \mathrm{~m})}{Area}\) Step 3: Calculate the power transfer from the car battery to the copper cable \(Power = \frac{(12.0 \mathrm{~V})^2}{Resistance}\) Step 4: Determine the total energy required to increase the temperature of the copper cable till its melting point \(Energy = Mass \times (386 \mathrm{~J} / \mathrm{kg} / \mathrm{K}) \times (1359 \mathrm{~K} - 298 \mathrm{~K})\) Step 5: Calculate the time it takes for the copper cable to start melting \(Time = \frac{Energy}{Power}\)

Step-by-step solution

Calculate the cross-sectional area of the copper cable

We need to find the cross-sectional area of the copper cable to determine its resistance later on. We'll use the mass density formula for that: Density \(= \frac{Mass}{Volume}\) We know that the cable is \(2.50 \mathrm{~m}\) long and its mass density is \(8960 \mathrm{~kg} / \mathrm{m}^{3}\). Hence, the volume of the cable will be: \(Volume = \frac{Mass}{Density} = \frac{Mass}{8960 \mathrm{~kg} / \mathrm{m}^{3}}\) Since the cable is cylindrical, its volume can also be expressed as: \(Volume = Area \times Length\), where the length is \(2.50 \mathrm{~m}\). Now we can find the cross-sectional area of the cable: \(Area = \frac{Mass}{(2.50 \mathrm{~m})(8960 \mathrm{~kg} / \mathrm{m}^{3})}\)

Find the resistance of the copper cable

Next, we need to find the resistance of the cable. The resistance can be found using the following formula: \(Resistance = \frac{Resistivity \times Length}{Area}\) We know the resistivity of copper is \(1.724 \times 10^{-8} \Omega \cdot \mathrm{m}\), the length of the cable is \(2.50 \mathrm{~m}\), and we found its cross-sectional area in step 1. Thus, we can find the resistance: \(Resistance = \frac{(1.724 \times 10^{-8} \Omega \cdot \mathrm{m})(2.50 \mathrm{~m})}{Area}\)

Calculate the power transfer from the car battery to the copper cable

Now we will find the power transfer from the car battery to the copper cable using the following formula: \(Power = \frac{Voltage^2}{Resistance}\) We know the voltage given by the car battery (12 V) and the resistance of the copper cable (from step 2). So we can calculate the power transfer: \(Power = \frac{(12.0 \mathrm{~V})^2}{Resistance}\)

Determine the total energy required to increase the temperature of the copper cable till its melting point

Next, we will find the total energy required to increase the temperature of the copper cable until its melting point. The energy required can be found using the following formula: \(Energy = Mass \times Specific \ Heat \times Temperature \ Change\) We know the specific heat of copper is \(386 \mathrm{~J} / \mathrm{kg} / \mathrm{K}\) and the temperature change required to reach the melting point is \((1359 \mathrm{~K} - 298 \mathrm{~K})\). Thus, we can find the total energy required: \(Energy = Mass \times (386 \mathrm{~J} / \mathrm{kg} / \mathrm{K}) \times (1359 \mathrm{~K} - 298 \mathrm{~K})\)

Calculate the time it takes for the copper cable to start melting

Finally, we will calculate the time it takes for the copper cable to start melting. We will use the following formula for that: \(Time = \frac{Energy}{Power}\) We know the total energy required (from step 4) and the power transfer (from step 3), so we can calculate the time it takes for the copper cable to start melting: \(Time = \frac{Energy}{Power}\)