Determine the thermal conductivity of the carbon nanotube of Example 3.4 when the heating island temperature is measured to be \(T_{h}=332.6 \mathrm{~K}\), without evaluating the thermal resistances of the supports. The conditions are the same as in the example.

Data From Example 3.4:-

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The thermal conductivity of a D = 14-nm-diameter carbon nanotube is measured with an instrument that is fabricated of a wafer of silicon nitride at a temperature of T = 300 K. The 20-m-long nanotube rests on two 0.5-um-thick, 10 m x 10 m square islands that are separated by a distance s = 5 m. A thin layer of platinum is used as an electrical resis- tor on the heated island (at temperature T,) to dissipate q = 11.3 W of electrical power. On the sensing island, a similar layer of platinum is used to determine its temperature, T. The platinum's electrical resistance, R(T) = E/I, is found by measuring the voltage drop and electrical current across the platinum layer. The temperature of the sensing island, T., is then determined from the relationship of the platinum electrical resistance to its tem- perature. Each island is suspended by two L = 250-m-long silicon nitride beams that are Wsn = 3 m wide and t = 0.5 m thick. A platinum line of width w = 1 m and thickness fpt = 0.2 m is deposited within each silicon nitride beam to power the heated island or to detect the voltage drop associated with the determination of T. The entire experiment is performed in a vacuum with Tsur = 300 K and at steady state, T, = 308.4 K. Estimate the thermal conductivity of the carbon nanotube.