1. **State the problem:** Calculate the heat loss through a stainless steel pipe wall at steady state given the inside diameter, length, wall thickness, inside and outside temperatures, and thermal conductivity. 2. **Formula used:** Heat loss by conduction through a cylindrical wall is given by $$Q = \frac{2 \pi k L (T_1 - T_2)}{\ln\left(\frac{r_2}{r_1}\right)}$$ where $k$ is thermal conductivity, $L$ is length, $T_1$ and $T_2$ are inside and outside temperatures, $r_1$ and $r_2$ are inside and outside radii. 3. **Given data:** - Inside diameter $d_1 = 0.04$ m, so inside radius $r_1 = \frac{0.04}{2} = 0.02$ m - Wall thickness $t = 0.002$ m, so outside radius $r_2 = r_1 + t = 0.02 + 0.002 = 0.022$ m - Length $L = 5$ m - Inside temperature $T_1 = 90$ °C - Outside temperature $T_2 = 88$ °C - Thermal conductivity $k = 16$ W/(m°C) 4. **Calculate the logarithmic term:** $$\ln\left(\frac{r_2}{r_1}\right) = \ln\left(\frac{0.022}{0.02}\right) = \ln(1.1) \approx 0.09531$$ 5. **Calculate heat loss $Q$:** $$Q = \frac{2 \pi \times 16 \times 5 \times (90 - 88)}{0.09531} = \frac{2 \pi \times 16 \times 5 \times 2}{0.09531}$$ 6. **Simplify numerator:** $$2 \pi \times 16 \times 5 \times 2 = 2 \times 3.1416 \times 16 \times 5 \times 2 = 1005.31$$ 7. **Final calculation:** $$Q = \frac{1005.31}{0.09531} \approx 10548.5 \text{ W}$$ **Answer:** The heat loss through the pipe wall is approximately $10548.5$ watts.