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When you pass a current through a wire, for a fraction of a second (a time $\Delta t$, say) almost all the electrical work ($I^2 R\Delta t$) goes into internal thermal energy in the wire (that is, as you say, $mc\Delta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat will be lost as infrared radiation. In this case we have $$I^2 R=\alpha\sigma 2\pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $\alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $\alpha=1$ isn't usually too bad an approximation. $\sigma\ (\text{the Stefan-Boltzmann constant})=5.67\times10^{-8}\ \text{Wm}^{-2} \text{K}{^{-4}}$, $2\pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $\alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special ally constantan, you remove one complication. Constantan (aka Eureka) is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $\mu \Omega \ \text m$ and its melting point is 1210 °C = 1483 K. [Using these figures I found that a constantan wire of diameter 0.50 mm would need a current of 13 A to reach its melting point. Not unreasonable?]

When you pass a current through a wire, for a fraction of a second (a time $\Delta t$, say) almost all the electrical work ($I^2 R\Delta t$) goes into internal thermal energy in the wire (that is, as you say, $mc\Delta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat – my limited researches suggest more than 80% for temperatures over 600°C – will be lost as infrared radiation. In this case we have $$I^2 R=\alpha\sigma 2\pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $\alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $\alpha=1$ isn't usually too bad an approximation. $\sigma\ (\text{the Stefan-Boltzmann constant})=5.67\times10^{-8}\ \text{Wm}^{-2} \text{K}{^{-4}}$, $2\pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $\alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special alloy constantan, you remove one complication. Constantan (aka Eureka) is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $\mu \Omega \ \text m$ and its melting point is 1210 °C = 1483 K. [Using these figures I found that a constantan wire of diameter 0.50 mm would need a current of 13 A to reach its melting point. Not unreasonable?]

When you pass a current through a wire, for a fraction of a second (a time $\Delta t$, say) almost all the electrical work ($I^2 R\Delta t$) goes into internal thermal energy in the wire (that is, as you say, $mc\Delta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat will be lost as infrared radiation. In this case we have $$I^2 R=\alpha\sigma 2\pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $\alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $\alpha=1$ isn't usually too bad an approximation. $\sigma\ (\text{the Stefan-Boltzmann constant})=5.67\times10^{-8}\ \text{Wm}^{-2} \text{K}{^{-4}}$, $2\pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $\alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special ally constantan, you remove one complication. Constantan is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $\mu \Omega \ \text m$ and its melting point is 1210 °C = 1483 K.

When you pass a current through a wire, for a fraction of a second (a time $\Delta t$, say) almost all the electrical work ($I^2 R\Delta t$) goes into internal thermal energy in the wire (that is, as you say, $mc\Delta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat will be lost as infrared radiation. In this case we have $$I^2 R=\alpha\sigma 2\pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $\alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $\alpha=1$ isn't usually too bad an approximation. $\sigma\ (\text{the Stefan-Boltzmann constant})=5.67\times10^{-8}\ \text{Wm}^{-2} \text{K}{^{-4}}$, $2\pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $\alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special ally constantan, you remove one complication. Constantan (aka Eureka) is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $\mu \Omega \ \text m$ and its melting point is 1210 °C = 1483 K. [Using these figures I found that a constantan wire of diameter 0.50 mm would need a current of 13 A to reach its melting point. Not unreasonable?]

When you connect a voltage across a wire, for a fraction of a second (a time $\Delta t$, say) almost all the electrical work ($\frac{V^2}{R}\Delta t$) goes into internal thermal energy in the wire (that is, as you say, $mc\Delta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat will be lost as infrared radiation. In this case we have $$\frac{V^2}{R}=\alpha\sigma 2\pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $\alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $\alpha=1$ isn't usually too bad an approximation. $\sigma\ (\text{the Stefan-Boltzmann constant})=5.67\times10^{-8}\ \text{Wm}^{-2} \text{K}{^{-4}}$, $2\pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $\alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special ally constantan, you remove one complication. Constantan is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $\mu \Omega \ \text m$ and its melting point is 1210 °C = 1483 K.

When you pass a current through a wire, for a fraction of a second (a time $\Delta t$, say) almost all the electrical work ($I^2 R\Delta t$) goes into internal thermal energy in the wire (that is, as you say, $mc\Delta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat will be lost as infrared radiation. In this case we have $$I^2 R=\alpha\sigma 2\pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $\alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $\alpha=1$ isn't usually too bad an approximation. $\sigma\ (\text{the Stefan-Boltzmann constant})=5.67\times10^{-8}\ \text{Wm}^{-2} \text{K}{^{-4}}$, $2\pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $\alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special ally constantan, you remove one complication. Constantan is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $\mu \Omega \ \text m$ and its melting point is 1210 °C = 1483 K.