1. **Problem statement:** Calculate the total heat lost per unit length by a horizontal steel pipe of radius 2.5 cm at 50°C in a room at 20°C, with surface emissivity 0.8 and convection heat-transfer coefficient $h=6.5$ W/m²·°C. 2. **Formulas used:** The total heat loss per unit length is the sum of convective and radiative heat losses: $$Q_{total} = Q_{conv} + Q_{rad}$$ Convective heat loss per unit length: $$Q_{conv} = h A_s (T_s - T_\\infty)$$ where $A_s$ is the surface area per unit length, $T_s$ is surface temperature, and $T_\infty$ is ambient temperature. Radiative heat loss per unit length: $$Q_{rad} = \varepsilon \sigma A_s (T_s^4 - T_\text{sur}^4)$$ where $\varepsilon$ is emissivity, $\sigma = 5.67 \times 10^{-8}$ W/m²·K⁴ is Stefan-Boltzmann constant, $T_s$ and $T_\text{sur}$ are absolute temperatures in Kelvin. 3. **Convert units and calculate surface area:** Radius $r = 2.5$ cm = 0.025 m Surface area per unit length (circumference): $$A_s = 2 \pi r = 2 \pi \times 0.025 = 0.1571 \text{ m}^2/\text{m}$$ 4. **Convert temperatures to Kelvin:** $$T_s = 50 + 273.15 = 323.15 \text{ K}$$ $$T_\infty = T_\text{sur} = 20 + 273.15 = 293.15 \text{ K}$$ 5. **Calculate convective heat loss:** $$Q_{conv} = 6.5 \times 0.1571 \times (323.15 - 293.15) = 6.5 \times 0.1571 \times 30 = 30.6 \text{ W/m}$$ 6. **Calculate radiative heat loss:** Calculate $T_s^4 - T_\text{sur}^4$: $$323.15^4 - 293.15^4 = 1.089 \times 10^{10} - 7.374 \times 10^{9} = 3.516 \times 10^{9}$$ Then, $$Q_{rad} = 0.8 \times 5.67 \times 10^{-8} \times 0.1571 \times 3.516 \times 10^{9}$$ $$= 0.8 \times 5.67 \times 10^{-8} \times 0.1571 \times 3.516 \times 10^{9} = 25.1 \text{ W/m}$$ 7. **Calculate total heat loss:** $$Q_{total} = 30.6 + 25.1 = 55.7 \text{ W/m}$$ **Final answer:** The total heat lost by the pipe per unit length is approximately **55.7 watts per meter**.