Question

A person's head can be approximated as a \(25-\mathrm{cm}\) diameter sphere at \(35^{\circ} \mathrm{C}\) with an emissivity of \(0.95\). Heat is lost from the head to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of $11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and by radiation to the surrounding surfaces at \)10^{\circ} \mathrm{C}$. Disregarding the neck, determine the total rate of heat loss from the head. (a) \(22 \mathrm{~W}\) (b) \(27 \mathrm{~W}\) (c) \(49 \mathrm{~W}\) (d) \(172 \mathrm{~W}\) (e) \(249 \mathrm{~W}\)

Short answer

a) 30 W b) 42 W c) 49 W d) 55 W Answer: c) 49 W

Step-by-step solution

Finding the Heat Loss by Convection

To calculate the heat lost by convection, we will use the formula: \(Q_{conv} = hA\Delta{T}\) where \(Q_{conv}\) is the heat loss by convection, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the head, and \(\Delta{T}\) is the temperature difference between the head and the surrounding air. First, we need to find the surface area of the head. Since the head is approximated as a sphere, the surface area can be calculated using the formula: \(A = 4\pi{r^2}\) Where \(r\) is the radius of the sphere. The diameter of the head is given as \(25\mathrm{cm}\), so the radius is half of that, which is \(12.5\mathrm{cm}\). Converting the radius to meters, we get: \(r = 0.125\mathrm{m}\) Now, we can calculate the surface area of the head: \(A = 4\pi(0.125)^2\mathrm{m}^2 = 0.196\mathrm{m}^2\) Now, we calculate the temperature difference between the head and the surrounding air: \(\Delta{T} = T_{head} - T_{air} = 35^{\circ}\mathrm{C} - 25^{\circ}\mathrm{C} = 10\mathrm{K}\) Finally, we can calculate the heat loss by convection: \(Q_{conv} = hA\Delta{T} = (11 \mathrm{~W} / \mathrm{m}^{2} \cdot\mathrm{K})(0.196\mathrm{m}^{2})(10\mathrm{K}) = 21.56\mathrm{~W}\)

Finding the Heat Loss by Radiation

To calculate the heat loss by radiation, we will use the formula: \(Q_{rad} = \varepsilon\sigma A (T_{1}^{4}-T_{2}^{4})\) Where \(Q_{rad}\) is the heat loss by radiation, \(\varepsilon\) is the emissivity of the head, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area of the head, \(T_{1}\) is the temperature of the head, and \(T_{2}\) is the temperature of the surrounding surfaces. First, we need to convert the temperatures to Kelvin: \(T_{head} = 35^{\circ}\mathrm{C} + 273.15 = 308.15\mathrm{K}\) \(T_{surrounding} = 10^{\circ}\mathrm{C} + 273.15 = 283.15\mathrm{K}\) Now, we can calculate the heat loss by radiation: \(Q_{rad} = \varepsilon\sigma A (T_{1}^{4}-T_{2}^{4})\) Using the values provided in the exercise: \(Q_{rad} = 0.95(5.67\mathrm{x}10^{-8}\mathrm{W/m}^2\mathrm{K}^4)(0.196\mathrm{m}^2)((308.15\mathrm{K})^4-(283.15\mathrm{K})^4)\) \(Q_{rad} = 27.35\mathrm{W}\)

Calculating the Total Heat Loss

Now, we can add the heat lost by convection and radiation to find the total heat loss: \(Q_{total} = Q_{conv} + Q_{rad} = 21.56\mathrm{W} + 27.35\mathrm{W} = 48.91\mathrm{W}\) Rounding to the nearest whole number, we get approximately \(49\mathrm{W}\). Therefore, the correct answer is (c) \(49 \mathrm{~W}\).