## anonymous one year ago Aluminum cylinder: length 1 m, diameter 1 cm Current: 1000 A Conductivity: σ=35 Sm/〖mm〗^2 Linear temperature coefficient: α=3.8∙〖10〗^(-3) 1/K Quadratic temperature coefficient: β=1.3∙〖10〗^(-6) 1/K^2 Heat transfer coefficient to the ambient air: λ=1 K/(Wm^2 ) 1)• Temperature of the aluminum rod with linear temperature dependency? 2).• Temperature of the aluminum rod with quadratic temperature dependency?

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1. Michele_Laino

is the current inside the aluminum rod a cause or is it an effect?

2. Michele_Laino

if it is a cause, then inside the aluminum rod we have this electric field $${\mathbf{E}}$$: ${\mathbf{j}} = \sigma {\mathbf{E}}$ where $${\mathbf{j}}$$ is the current density vector, and $$\sigma$$ is the conductivity of aluminum. Furthermore, the power over volume, dissipated in such rod of aluminum is given by the subsequent formula: $W = \frac{{{{\mathbf{j}}^2}}}{\sigma }$ provided that we are in static condition of electricity

3. anonymous

Hint: The electrical losses of the aluminum rod P_A transferred via heat to the ambiance can be computed with the temperature at the surface ϑ_S, the ambient temperature ϑ_A, and the surface A with the following equation: P_A=(ϑ_S-ϑ_A)/λA

4. Michele_Laino

so, thermal energy is: $dQ = Wdt = \frac{{{{\mathbf{j}}^2}}}{\sigma }dt$ where the magnitude of $${\mathbf{j}}$$, is: $\frac{I}{S} = \frac{I}{{\pi {r^2}}}$ and: $$2r=1\;cm$$

5. Michele_Laino

I meant $$dQ$$ is thermal energy over volume, of course