Correct Answer - Option 2 : 0.3 W/mK
Concept:
According to Fourier’s Law:
\(Q = - kA\frac{{dT}}{{dx}}\)
where k = Thermal conductivity, A = cross-section area, \(\frac {dT}{dx}\) = Temperature gradient
Calculation:
Given:
dx = 10 cm = 0.1 m, Q = 30 W/m2, dT = 10°C
\(Q = -kA\frac{{dT}}{{dx}}\)
\( \frac{Q}{A} =- k\frac{{dT}}{{dx}} \)
\( \Rightarrow 30 = k \times \frac{{10}}{{0.1}}\)
∴ k = 0.3 W/mK