Question8 Power cables normally operate at temperatures up to about $10^{\circ} \mathrm{G}$ If a short circuit occurs, considerable heat may be evolved which cannot escape quickly. The temperature of the conductor must never rise above $250^{\circ} \mathrm{C}$ or the electrical insulation will be damaged.
Under short-circuit conditions, a current of 7000 A lasts for 1.0 s in a cable that has an effective area of cross-section $50 \mathrm{~mm}^{2}$, resistance per unit length $0.40 \Omega \mathrm{~km}^{-1}$ and density $8.9 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. a Calculate the heat evolved in 1.0 m of cable during this short circuit. b Calculate the rise in temperature of the cable.

Studdy Solution
Calculate the rise in temperature of the cable using the formula for heat energy and specific heat capacity:
The formula for temperature rise is $\Delta T = \frac{Q}{m c}$, where $Q = 19600 \, \text{J}$, $m$ is the mass of 1.0 m of cable, and $c$ is the specific heat capacity.
First, calculate the mass $m$ of 1.0 m of cable:
The volume of 1.0 m of cable is $V = \text{cross-sectional area} \times \text{length} = 50 \times 10^{-6} \, \text{m}^2 \times 1.0 \, \text{m} = 50 \times 10^{-6} \, \text{m}^3$.
The mass is $m = \text{density} \times \text{volume} = 8.9 \times 10^3 \, \text{kg/m}^3 \times 50 \times 10^{-6} \, \text{m}^3 = 0.445 \, \text{kg}$.
Assume the specific heat capacity $c$ of the cable material is approximately $385 \, \text{J/kg}^\circ\text{C}$ (typical for copper).
$$\Delta T = \frac{19600}{0.445 \times 385}$$ $$\Delta T \approx \frac{19600}{171.325}$$ $$\Delta T \approx 114.4^\circ \text{C}$$
The rise in temperature of the cable is approximately:
$$\boxed{114.4^\circ \text{C}}$$