A long homogeneous resistance wire of radius \(r_{o}=\) \(5 \mathrm{~mm}\) is being used to heat the air in a room by the passage of electric current. Heat is generated in the wire uniformly at a rate of \(5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) as a result of resistance heating. If the temperature of the outer surface of the wire remains at \(180^{\circ} \mathrm{C}\), determine the temperature at \(r=3.5 \mathrm{~mm}\) after steady operation conditions are reached. Take the thermal conductivity of the wire to be \(k=6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The rate at which heat is generated in the wire is given as \(q_g = 5\times 10^7~\text{W}/\text{m}^3\). This heat must be conducted outward through the wire radially.

For radial heat conduction, the equation is given by: \(\frac{d}{dr}(r k \frac{dT}{dr}) + q_g = 0.\) Here, \(r\) is the distance from the center of the wire, \(T\) is the temperature, and \(k\) is the thermal conductivity of the wire.

Substitute the value of \(q_g\) and \(k\) into the equation and solve for \(\frac{dT}{dr}\): \(\frac{d}{dr}(r k \frac{dT}{dr}) + 5\times 10^7 = 0\) Integration yields: \(r \frac{dT}{dr} = -\frac{5\times 10^7}{6}{r^2} + C_1\) Next, integrate with respect to \(r\): \(T(r) = -\frac{5\times 10^7}{12}{r^3} + C_1 \ln r +C_2\) Now, we need to apply boundary conditions to get the values of \(C_1\) and \(C_2\).

At the outer radius (\(r=r_o = 5 \text{ mm} = 5\times 10^{-3}~\text{m}\)), the temperature is \(180^\circ\text{C}\): \(T(r_o) = T(5\times 10^{-3})= 180^\circ\text{C} = 453.15\text{K}\) (converting Celsius to Kelvin) Substitute values into the equation: \(453.15 = -\frac{5\times 10^7}{12}(5\times 10^{-3})^3 + C_1 \ln (5\times 10^{-3}) +C_2\) There is no source of heat at the center of the wire (since it is homogenous). Hence, as \(r \rightarrow 0,\) \(\frac{dT}{dr} \rightarrow 0\): \(0 = -\frac{5\times 10^7}{6}(0)^2 + C_1\) This means \(C_1 = 0\). So we can rewrite temperature equation as: \(T(r) = -\frac{5\times 10^7}{12}{r^3} + C_2\) Now substitute the values again to get \(C_2\): \(453.15 = -\frac{5\times 10^7}{12}(5\times 10^{-3})^3 + C_2\) Solving for \(C_2\), we get \(C_2 = 453.15 + \frac{5\times 10^7}{12}(5\times 10^{-3})^3 = 453.9247\) The temperature equation can now be written as: \(T(r) = -\frac{5\times 10^7}{12}{r^3} + 453.9247\)

We are asked to find the temperature at \(r = 3.5\text{~mm} = 3.5\times10^{-3}~\text{m}\). Substitute this value into the temperature equation: \(T(3.5\times10^{-3}) = -\frac{5\times 10^7}{12}(3.5\times 10^{-3})^3 + 453.9247\) Evaluating the expression, we get the temperature in Kelvin: \(T(3.5\times10^{-3}) = 452.009\text{K}\) Now convert the temperature back to Celsius: \(T = 452.009\text{K} - 273.15 = 178.86^\circ\text{C}\) So, the temperature at a radius of \(3.5\text{~mm}\) from the center of the wire is \(178.86^\circ\text{C}\).