1. **Stating the problem:** We are given temperature measurements at various points along a brass specimen heated at different power levels. The goal is to calculate the thermal conductivity $k$ using the linear conduction formula. 2. **Formula used:** The heat conduction rate is given by Fourier's law: $$\Phi = -kA \frac{dT}{dx}$$ where $\Phi$ is the heat transfer rate (W), $k$ is the thermal conductivity (W/mK), $A$ is the cross-sectional area (m²), and $\frac{dT}{dx}$ is the temperature gradient (K/m). Rearranging to solve for $k$: $$k = \frac{\Phi}{A \frac{dT}{dx}}$$ 3. **Important rules:** - The temperature gradient $\frac{dT}{dx}$ is given in the table. - Heater power $Q$ corresponds to $\Phi$. - Cross-sectional area $A$ must be known or given to calculate $k$. 4. **Calculations:** Since the cross-sectional area $A$ is not provided explicitly, we assume it is known or given for the specimen. For each heater power $Q$ and corresponding thermal gradient $\frac{dT}{dx}$, calculate $k$ as: $$k = \frac{Q}{A \cdot \frac{dT}{dx}}$$ 5. **Example calculation for $Q=5$ W and $\frac{dT}{dx}=0.005$ K/m:** $$k = \frac{5}{A \times 0.005} = \frac{5}{0.005A} = \frac{1000}{A}$$ Similarly for other values: - $Q=10$ W, $\frac{dT}{dx}=0.015$ K/m: $$k = \frac{10}{A \times 0.015} = \frac{10}{0.015A} = \frac{666.67}{A}$$ - $Q=15$ W, $\frac{dT}{dx}=0.025$ K/m: $$k = \frac{15}{A \times 0.025} = \frac{15}{0.025A} = \frac{600}{A}$$ - $Q=20$ W, $\frac{dT}{dx}=0.035$ K/m: $$k = \frac{20}{A \times 0.035} = \frac{20}{0.035A} = \frac{571.43}{A}$$ 6. **Interpretation:** The thermal conductivity $k$ depends inversely on the cross-sectional area $A$. Once $A$ is known, substitute to find numerical values for $k$. **Final answer:** $$k = \frac{Q}{A \cdot \frac{dT}{dx}}$$ with values calculated above for each power level.