A copper wire having cross-sectional area of 0.5 mm^2 and a length of 0.1 m is initially at 25°C and

Q: Why does the length of the wire not affect the melting time?

Because both the resistance (which generates heat) and the mass (which absorbs heat) are directly proportional to the length. When you equate the two, the length cancels out. A longer wire generates more heat, but also has more mass to heat up, perfectly balancing out.

Q: Why do we assume all the electrical energy goes into heating the wire?

The problem explicitly states that the wire is 'thermally insulated from the surrounding'. This means no heat is lost to the environment via conduction, convection, or radiation. 100% of the electrical work is converted into internal thermal energy.

Q: Why did we multiply the specific heat by 4.18?

The specific heat was given in calories per kg per degree Celsius. The standard SI unit for energy is the Joule. To make our equation dimensionally consistent (since $I^2 R t$ gives Joules), we must convert calories to Joules using the mechanical equivalent of heat, $1 \text{ cal} \approx 4.18 \text{ J}$.

Q: What if the resistance changed with temperature?

In reality, the resistance of copper increases with temperature. If we accounted for this, the heat generation rate ($I^2 R$) would increase over time, causing the wire to melt slightly faster. However, the problem instructs us to neglect this change to simplify the integration into a simple algebraic equation.

Q: Why did we use $9 \times 10^{-2}$ for specific heat when it's physically unrealistic for copper?

You have a sharp eye! The specific heat of copper is actually around $0.09 \text{ cal/g}^\circ\text{C}$, which translates to $90 \text{ cal/kg}^\circ\text{C}$. The question contains a typo in its units, providing $9 \times 10^{-2} \text{ cal/kg}^\circ\text{C}$. However, in competitive exams, you must strictly use the numerical values provided in the problem statement, even if they seem physically anomalous.

Question

A copper wire having cross-sectional area of $0.5 	ext{ mm}^2$ and a length of $0.1 	ext{ m}$ is initially at $25^\circ	ext{C}$ and is thermally insulated from the surrounding. If a current of $1.0 	ext{ A}$ is set up in this wire, (a) find the time in which the wire will start melting. The change of resistance with the temperature of the wire may be neglected. (b) What will this time be, if the length of the wire is doubled ?

Melting point of copper $= 1075^\circ	ext{C}$, Specific resistance of copper $= 1.6 	imes 10^{-8} \Omega	ext{m}$, Density of copper $= 9 	imes 10^3 	ext{ kg/m}^3$, Specific heat of copper $= 9 	imes 10^{-2} 	ext{ cal/kg}^\circ	ext{C}$

Sigma Percentile · Accepted Answer

The problem of a melting wire is a classic intersection of electrodynamics and thermodynamics. It beautifully demonstrates how electrical energy transforms into thermal energy, and how the physical dimensions of a conductor influence its heating rate.

### Analyzing the Setup

Imagine a copper wire, completely wrapped in thermal insulation. A steady current of $1.0 	ext{ A}$ starts flowing through it. Because it is perfectly insulated, all the heat generated by the electrical resistance is trapped inside. This trapped heat causes the wire's temperature to rise steadily from its initial $25^\circ	ext{C}$ towards its melting point of $1075^\circ	ext{C}$.

The total change in temperature required is:
$$ \Delta 	heta = 1075^\circ	ext{C} - 25^\circ	ext{C} = 1050^\circ	ext{C} $$

### The Master Equation

To find out when the wire starts melting, we must balance the electrical heat generated with the thermal heat required to raise its temperature.

From Joule's Law of Heating, the heat produced is:
$$ H = I^2 R t $$

From the principles of calorimetry, the heat absorbed is:
$$ Q = m S \Delta 	heta $$

Since the wire is insulated, no heat is lost to the surroundings. Therefore, we can equate the two:
$$ I^2 R t = m S \Delta 	heta $$

### Breaking Down the Variables

Now, let's express resistance and mass in terms of their fundamental properties. The resistance $R$ of the wire is given by:
$$ R = ho_1 \frac{l}{A} $$
where $ho_1$ is the specific resistance (resistivity), $l$ is the length, and $A$ is the cross-sectional area.

The mass $m$ of the wire is its volume multiplied by its density $ho_2$:
$$ m = (A \cdot l) ho_2 $$

Substituting these into our heat balance equation, we get:
$$ I^2 \left( ho_1 \frac{l}{A} ight) t = (A \cdot l) ho_2 S \Delta 	heta $$

### The Elegance of Cancellation

Notice something fascinating here? The length of the wire, $l$, appears on both sides of the equation.

$$ I^2 \frac{ho_1}{A} t = A ho_2 S \Delta 	heta $$

This means the length cancels out completely! The time it takes for the wire to melt is **entirely independent of its length**. A longer wire generates more heat (because it has higher resistance), but it also has proportionally more mass to heat up. These two effects perfectly balance each other out.

Rearranging the equation to solve for time $t$:
$$ t = \frac{A^2 ho_2 S \Delta 	heta}{I^2 ho_1} $$

### Final Calculation

Now, we carefully substitute the given values. We must ensure all units are in standard SI units. The cross-sectional area must be converted to square meters:
$$ A = 0.5 	ext{ mm}^2 = 0.5 	imes 10^{-6} 	ext{ m}^2 $$

The specific heat is given as $9 	imes 10^{-2} 	ext{ cal/kg}^\circ	ext{C}$. To convert this to Joules, we multiply by the mechanical equivalent of heat ($4.18 	ext{ J/cal}$):
$$ S = 9 	imes 10^{-2} 	imes 4.18 	ext{ J/kg}^\circ	ext{C} $$

Plugging everything into our time equation:
$$ t = \frac{(0.5 	imes 10^{-6})^2 (9 	imes 10^3) (9 	imes 10^{-2} 	imes 4.18 	imes 1050)}{(1.0)^2 (1.6 	imes 10^{-8})} $$

Evaluating the numerator:
$$ t = \frac{(0.25 	imes 10^{-12}) (9 	imes 10^3) (395.01)}{1.6 	imes 10^{-8}} $$
$$ t = \frac{8.888 	imes 10^{-7}}{1.6 	imes 10^{-8}} $$
$$ t = 55.5 	ext{ s} $$

So, the wire will start melting in exactly **$55.5 	ext{ s}$**.

### What if the length is doubled?

For the second part of the question, we are asked what happens if the length of the wire is doubled. As we discovered during our derivation, the length $l$ completely cancels out of the equation. Therefore, the time remains exactly the same.

Even if the wire is twice as long, it will still take **$55.5 	ext{ s}$** to melt!

Visualized Solution

$T_{i} \to T_{f}$

$H = Q$

$R = ρ_{1} \frac{l}{A}$

$t = \frac{A ^{2} ρ _{2} S Δ θ}{I ^{2} ρ _{1}}$

$S = 9 \times 1 0^{- 2} \times 4.18$

$t = 55.5 s$

$l^{'} = 2 l$

The Sigma Insight: Heating Effect of Current

Analyzing the Setup

The Master Equation

Breaking Down the Variables

The Elegance of Cancellation

Final Calculation

What if the length is doubled?

Similar Questions

Visualized Solution

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