In a wire with a cross-sectional area of $A$, a current $I$ flows during a short circuit. Before the short circuit, the conductor was at a temperature of $T_{0}$. Estimate how long it will take for the wire to melt! Ignore radiation losses! (The specific resistance of the material depends on the temperature: $\varrho=\varrho_{0}\left[1+\alpha\left(T-T_{0}\right)\right]$, where $\varrho_{0}$ is the specific resistance of the conductor at temperature $T_{0}$, and $\alpha$ is the temperature coefficient of the material. The temperature $T$ is given in K. Data: $\alpha=\frac{1}{235}{ }^{\circ} \mathrm{C}^{-1}, \varrho_{0}=1,65 \cdot 10^{-8} \Omega \mathrm{m}, A=2 \mathrm{~mm}^{2}$, $I=120 \mathrm{~A}, T_{0}=313 \mathrm{~K}$
I. solution. On a hollow copper wire of length $l$, the heat

$$
Q=I^{2} R(T) \cdot \Delta t
$$

develops in Joules over time $\Delta t$. $(R(T)$ represents the resistance of the wire at temperature $T$.) From the temperature dependence of the specific resistance:

$$
R(T)=\left(\varrho_{0} l / A\right)\left[1+\alpha\left(T-T_{0}\right)\right]
$$

Over the short time $\Delta t$, the temperature change of the wire is $\Delta T$, so the heat absorbed by it is

$$
Q=c m \Delta T
$$

where $c=385 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$ is the specific heat of copper, which we assume does not depend on temperature; $m$ is the mass of the wire of length $l$,

$$
m=\varrho_{r} A l
$$

$\varrho=8,96 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ is the density of the copper wire.

Since we neglect radiative losses, the Joule heat (1) is equal to the heat absorption (3). Thus, using the relationships (2) and (4), after rearrangement, we get:

$$
\Delta T / \Delta t=(I / A)^{2} \cdot\left(\varrho_{0} / c \varrho_{r}\right)\left[1+\alpha\left(T-T_{0}\right)\right]
$$

Based on equation (5), we numerically calculate the time it takes for the wire to reach the melting point $T_{\text {melt }}$ $\left(T_{\text {melt }}=1356 \mathrm{~K}\right)$. Divide the $\left[T_{0}, T_{\text {melt }}\right]$ temperature interval into $n$ equal parts. Then

$$
\Delta T=\frac{T_{\mathrm{melt}}-T_{0}}{n}
$$

The time $\Delta t_{i}$ corresponds to the $i$-th temperature interval. Thus, the time elapsed until melting is:

$$
t=\sum_{i=1}^{n} \Delta t_{i}
$$

The values of $\Delta t_{i}$ can be calculated based on (5):

$$
\Delta t_{i}=\frac{1}{(I / A)^{2}\left(\varrho_{0} / c \varrho_{r}\right)} \cdot \frac{\Delta T}{1+\alpha\left(T_{i}-T_{0}\right)},
$$

where $T_{i}=T_{0}+i \Delta T=T_{0}+i \frac{T_{\mathrm{melt}}-T_{0}}{n}$ is the characteristic temperature of the $i$-th interval.

Finally, based on the previous two expressions,

$$
i=\frac{1}{\left(\varrho_{0} / c\right)(I / A)^{2}}\left(T_{\mathrm{melt}}-T_{0}\right) \sum_{i=1}^{n} \frac{1}{n+i \alpha\left(T_{\mathrm{melt}}-T_{0}\right)}
$$

Based on expression (6), for $n=250$ divisions, we get the following numerical value for $t$ using a pocket calculator: $t=$ $23,01 \mathrm{~s} \approx 23 \mathrm{~s}$.

Due to the inhomogeneity of the copper wire structure, not the entire wire begins to melt, but only a certain point. The melting time is very short, approximately $1-2$ s. (Numerically, we can only give the order of magnitude of the melting time.) Therefore, the time until the wire breaks is in the range of 23 - 25 s.

II. solution. Start from expression (5). For $\Delta t \rightarrow 0$, the limit on the left side is

$$
\lim _{\Delta t \rightarrow 0} \frac{\Delta T}{\Delta t}=\frac{\mathrm{d} T}{\mathrm{~d} t}
$$

This is the time derivative of the temperature. Thus, based on (5),

$$
\frac{\mathrm{d} T}{\mathrm{~d} t}=\left(\frac{I}{A}\right)^{2} \frac{\varrho_{0}}{c \varrho_{r}}\left[1+\alpha\left(T-T_{0}\right)\right]
$$

Introduce the new variable $T^{*}=1+\alpha\left(T-T_{0}\right)$. Then the new form of (7) is

$$
\frac{\mathrm{d} T^{*}}{\mathrm{~d} t}=B \cdot T^{*}, \quad \text { where } B=\left(\frac{I}{A}\right)^{2} \frac{\varrho_{0} \alpha}{c \varrho_{r}}
$$

Functions of the form $T^{*}(t)=D e^{B t}$ (where $D$ can be any constant) satisfy this differential equation. It is also known that if a function satisfies the above differential equation on an interval, it must be of the form $T^{*}(t)=D e^{B t}$.

The coefficient $D$ can be determined based on the initial value of $T^{*}$ at $t=0$. $T^{*}(0)=D=1$. The latter equality comes from the fact that at $t=0$, $T=T_{0}$, and thus $T^{*}(0)=1$. Therefore,

$$
T(t)=T_{0}+(1 / \alpha)\left(e^{B t}-1\right)
$$

If the time elapsed until melting is $t$, then $T(t)=T_{\text {melt }}$. From this,

$$
t=(1 / B) \ln \left[1+\alpha\left(T_{\mathrm{melt}}-T_{0}\right)\right]=\frac{1}{(I / A)^{2}\left(\varrho_{0} \alpha / c \varrho_{r}\right)}\left[\ln 1+\alpha\left(T_{\mathrm{melt}}-T_{0}\right)\right]
$$

Numerically, $t=23,1 \mathrm{~s}$.