1. **Problem statement:** We need to find the thermal conductivity $k_B$ of the unknown material B. Material A is stainless steel with known thermal conductivity $k_A = 15.2$ W/m·K. The rods are 2 cm in diameter and heavily insulated, so heat transfer is one-dimensional and steady-state. 2. **Given data:** - Temperatures: $T_1 = 93.00^\circ C$, $T_2 = 82.57^\circ C$, $T_3 = 69.21^\circ C$, $T_4 = 66.28^\circ C$ - Distance between points 1 and 2: $L_A = 0.02$ m - Distance between points 3 and 4: $L_B = 0.02$ m - Diameter $d = 0.02$ m (used to find cross-sectional area) 3. **Formula for heat conduction:** $$ Q = -k A \frac{\Delta T}{L} $$ where $Q$ is heat transfer rate, $k$ is thermal conductivity, $A$ is cross-sectional area, $\Delta T$ is temperature difference, and $L$ is length. 4. **Cross-sectional area:** $$ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.02}{2}\right)^2 = \pi (0.01)^2 = 3.1416 \times 10^{-4} \text{ m}^2 $$ 5. **Heat transfer rate in material A:** $$ Q_A = k_A A \frac{T_1 - T_2}{L_A} = 15.2 \times 3.1416 \times 10^{-4} \times \frac{93.00 - 82.57}{0.02} $$ Calculate temperature difference: $$ 93.00 - 82.57 = 10.43 $$ Calculate $Q_A$: $$ Q_A = 15.2 \times 3.1416 \times 10^{-4} \times \frac{10.43}{0.02} = 15.2 \times 3.1416 \times 10^{-4} \times 521.5 $$ Intermediate multiplication: $$ 3.1416 \times 10^{-4} \times 521.5 = 0.1639 $$ So, $$ Q_A = 15.2 \times 0.1639 = 2.49 \text{ W} $$ 6. **Heat transfer rate in material B:** Since the rods are in series and insulated, heat transfer rate is the same: $$ Q_B = Q_A = 2.49 \text{ W} $$ 7. **Calculate thermal conductivity of material B:** $$ Q_B = k_B A \frac{T_3 - T_4}{L_B} \Rightarrow k_B = \frac{Q_B L_B}{A (T_3 - T_4)} $$ Calculate temperature difference: $$ 69.21 - 66.28 = 2.93 $$ Substitute values: $$ k_B = \frac{2.49 \times 0.02}{3.1416 \times 10^{-4} \times 2.93} $$ Calculate denominator: $$ 3.1416 \times 10^{-4} \times 2.93 = 9.204 \times 10^{-4} $$ Calculate numerator: $$ 2.49 \times 0.02 = 0.0498 $$ Divide: $$ k_B = \frac{0.0498}{9.204 \times 10^{-4}} = 54.1 \text{ W/m·K} $$ **Final answer:** $$ \boxed{k_B = 54.1 \text{ W/m·K}} $$