Question

A \(3-\mathrm{m}^{2}\) black surface at \(140^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(35^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(16 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat loss from the surface is (a) \(5105 \mathrm{~W}\) (b) \(2940 \mathrm{~W}\) (c) \(3779 \mathrm{~W}\) (d) \(8819 \mathrm{~W}\) (e) \(5040 \mathrm{~W}\)

Short answer

Answer: The total rate of heat loss from the black surface due to convection and radiation is approximately 5940 W.

Step-by-step solution

Convert temperatures to Kelvin

To work with the given temperatures, we need to convert them from Celsius to Kelvin. We do this by adding 273.15 to each temperature. T_s = 140 + 273.15 = 413.15 K T_inf = 35 + 273.15 = 308.15 K T_sr = 15 + 273.15 = 288.15 K

Calculate heat loss due to convection

Now, we will calculate the heat loss due to convection using the formula q_conv = h*A*(T_s - T_inf). h = 16 W/m^2K A = 3 m^2 q_conv = h*A*(T_s - T_inf) q_conv = 16 * 3 * (413.15 - 308.15) q_conv = 5040 W

Calculate heat loss due to radiation

Next, we will calculate the heat loss due to radiation using the formula q_rad = σ*A*ε*(T_s^4 - T_sr^4). σ = 5.67 * 10^{-8} W/m^2K^4 (Stefan-Boltzmann constant) ε = 1 (for a black surface) q_rad = σ*A*ε*(T_s^4 - T_sr^4) q_rad = 5.67 * 10^{-8} * 3 * 1 * (413.15^4 - 288.15^4) q_rad = 881.89 W

Calculate total heat loss

Finally, we will find the total rate of heat loss by adding the heat loss due to convection and radiation. q_total = q_conv + q_rad q_total = 5040 + 881.89 q_total = 5921.89 W ≈ 5940 W Thus, the correct answer is closest to (b) \(2940 \mathrm{~W}\), with a more precise value of 5940 W.