The thermal conductivity of stainless steel has been characterized experimentally to vary with temperature as \(k(T)=9.14+0.021 T\) for \(273

First, we need to integrate the given function over the temperature range from 300 to 1200 K. The integral will give us the cumulative thermal conductivity, and dividing by the temperature range will give us the average thermal conductivity. Integrate \(k(T)\) with respect to temperature \(T\): $$\int_{300}^{1200} (9.14 + 0.021T) dT$$

Evaluate the integral: $$\int_{300}^{1200} (9.14 + 0.021T) dT = \left[9.14T + 0.0105T^2\right]_{300}^{1200}$$ Plug in the limits of the integration: $$(9.14(1200) + 0.0105(1200)^2) - (9.14(300) + 0.0105(300)^2)$$

Calculate the value: $$ (10968 + 151200) - (2742 + 9450) = 160168 - 12192 = 147976$$

Divide the result by the temperature range (1200 - 300) to find the average thermal conductivity: $$\frac{147976}{(1200-300)} = \frac{147976}{900} = 164.418$$ So the average thermal conductivity between 300 and 1200 K is \(164.418 \;\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\)

Now we need to rewrite the expression for the thermal conductivity in the form \(k(T) = k_{0}(1 + \beta T)\). Since \(k(273) = k_0\), we can find \(k_0\) by plugging T = 273 K into the given expression: $$k_0 = 9.14 + 0.021(273) = 9.14 + 5.733 = 14.873$$ Now we have \(k(T) = 14.873(1 + \beta T)\). To find \(\beta\), we can equate this expression with the original one and solve for \(\beta\): $$9.14 + 0.021T = 14.873(1 + \beta T)$$ From this, we can determine that \(\beta = \frac{0.021 - \frac{9.14}{14.873}}{14.873} = 6.12 \times 10^{-4}\) Now we have the desired expression: $$k(T) = 14.873(1 + 6.12 \times 10^{-4} T)$$ where \(k_0 = 14.873 \;\mathrm{W}/\mathrm{m}\cdot\mathrm{K}\) and \(\beta = 6.12 \times 10^{-4} \;\mathrm{K}^{-1}\).