A 10 -cm-high and 20-cm-wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of \(0.12 \mathrm{~W}\) and transferring it by convection and radiation to the surrounding medium at \(40^{\circ} \mathrm{C}\). Heat transfer from the back surface of the board is negligible. If the combined convection and radiation heat transfer coefficient on the surface of the board is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the average surface temperature of the chips is (a) \(41^{\circ} \mathrm{C}\) (b) \(54^{\circ} \mathrm{C}\) (c) \(67^{\circ} \mathrm{C}\) (d) \(76^{\circ} \mathrm{C}\) (e) \(82^{\circ} \mathrm{C}\)

We are given that each of the 100 chips generates heat at a rate of \(0.12\mathrm{~W}\). Therefore, the total heat generation (\(q\)) for all the chips can be calculated as follows: \(q = N \times Q_{chip}\) where \(N\) is the total number of chips and \(Q_{chip}\) is the heat generation per chip. \(q = 100 \times 0.12\mathrm{~W} = 12\mathrm{~W}\) The total heat generation for all the chips is 12 W.

The circuit board is 10 cm high and 20 cm wide. To calculate the area for heat transfer, we need to convert these measurements to meters and then multiply them together: \(A = H \times W\) where \(A\) is the heat transfer area, \(H\) is the height, and \(W\) is the width. \(A = 0.1\mathrm{~m} \times 0.2\mathrm{~m} = 0.02\mathrm{~m}^2\) The heat transfer area is 0.02 \(m^2\).

The heat transfer equation for combined convection and radiation is: \(q = hA(T_s - T_{\infty})\) where \(q\) is the total heat generation, \(h\) is the heat transfer coefficient, \(A\) is the heat transfer area, \(T_s\) is the average surface temperature of the chips, and \(T_{\infty}\) is the surrounding medium temperature. We are given that the combined convection and radiation heat transfer coefficient is \(22\mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K}\). We can rearrange the heat transfer equation to find the temperature difference (\(\Delta{T}\)) between the chips and the surrounding medium: \(\Delta{T} = T_s - T_{\infty} = \frac{q}{hA}\) Substituting the given values: \(\Delta{T} = \frac{12\mathrm{~W}}{22\mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K} \times 0.02\mathrm{~m}^2} = 27.27\mathrm{~K}\) The temperature difference between the chips and the surrounding medium is 27.27 K.

We can now add the temperature difference to the surrounding medium temperature to find the average surface temperature of the chips: \(T_s = T_{\infty} + \Delta{T}\) We are given that the surrounding medium temperature is \(40^{\circ} \mathrm{C}\). \(T_s = 40^{\circ} \mathrm{C} + 27.27\mathrm{~K} \approx 67.27^{\circ}\mathrm{C}\) The average surface temperature of the chips is approximately 67.27 °C. Comparing this to the given options, we can see that the closest answer is: (c) \(67^{\circ} \mathrm{C}\)