Solution
Let us consider a cylindrical conductor of length $L$ and radius $r$, connected across a constant voltage source $V$. The resistance of the conductor is given by:
$$ R = \frac{\rho L}{A} = \frac{\rho L}{\pi r^2} $$where $\rho$ is the resistivity of the material.
1. Power Generation
The rate of heat generation (Power) in the conductor is:
$$ P = \frac{V^2}{R} = \frac{V^2 \pi r^2}{\rho L} $$2. Heat Dissipation in Steady State
In the steady state, the rate of heat generation equals the rate of heat loss to the surroundings. Assuming Newton’s Law of Cooling, the rate of heat loss $H$ is proportional to the surface area $S$ and the temperature excess $\Delta \theta$:
$$ H = h S \Delta \theta $$Where $h$ is the heat transfer coefficient. Neglecting the end caps for a long wire, the surface area is $S = 2\pi r L$. Thus:
$$ P = H \implies \frac{V^2 \pi r^2}{\rho L} = h (2\pi r L) \Delta \theta $$Rearranging to find the dependence on length $L$ (keeping $V$, $r$, $\rho$, $h$ constant, as the radius is unchanged):
$$ \frac{1}{L} \propto L \Delta \theta $$ $$ \Delta \theta \propto \frac{1}{L^2} $$This implies that $L^2 \Delta \theta = \text{constant}$.
3. Analysis of the Change
The length of the conductor is reduced to $\eta$ times its initial value by cutting off the extra length. Therefore, the new length is:
$$ L’ = \eta L $$The radius $r$ remains unchanged.
Using the relationship $L^2 \Delta \theta = \text{constant}$:
$$ L^2 \Delta \theta = (L’)^2 \Delta \theta’ $$ $$ L^2 \Delta \theta = (\eta L)^2 \Delta \theta’ $$ $$ \Delta \theta’ = \frac{\Delta \theta}{\eta^2} $$4. Calculation
We are given:
- Initial temperature excess $\Delta \theta = 16^\circ \text{C}$
- Scaling factor $\eta = 0.8$
The change in temperature is asked. First, we find the new temperature excess $\Delta \theta’$:
$$ \Delta \theta’ = \frac{16}{(0.8)^2} = \frac{16}{0.64} = \frac{1600}{64} = 25^\circ \text{C} $$The increase in temperature is:
$$ \text{Change} = \Delta \theta’ – \Delta \theta = 25^\circ \text{C} – 16^\circ \text{C} = 9^\circ \text{C} $$Alternatively, using the formula derived:
$$ \Delta \theta_{\text{change}} = \left( \frac{1}{\eta^2} – 1 \right) \Delta \theta = \left( \frac{1 – \eta^2}{\eta^2} \right) \Delta \theta $$ $$ = \left( \frac{1 – 0.64}{0.64} \right) \times 16 = \frac{0.36}{0.64} \times 16 = \frac{9}{16} \times 16 = 9.0^\circ \text{C} $$