1. **Problem Statement:** We need to determine the thermal conductivity $k$ of a brass bar using Fourier's Law of heat conduction. Given heater power $Q$, temperature readings at various points along the bar, and the bar's diameter, we calculate the temperature gradient $\frac{dT}{dx}$ and then find $k$. 2. **Relevant Formula:** Fourier's Law in one dimension is: $$Q = -k A \frac{dT}{dx}$$ Rearranged to find thermal conductivity: $$k = -\frac{Q}{A \frac{dT}{dx}}$$ 3. **Given Data:** - Diameter of brass bar $D = 0.025$ m - Cross-sectional area $A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.025}{2}\right)^2 = \pi \times 0.0125^2$ Calculate $A$: $$A = \pi \times 0.00015625 = 0.0004909\, m^2$$ 4. **Calculate Temperature Gradient $\frac{dT}{dx}$:** Using the temperature readings at positions $x$ (in meters) and temperatures $T$ (in $^\circ C$) for heater power 5 W as an example: Positions $x$: 0.005, 0.015, 0.025, 0.035, 0.045, 0.055, 0.065, 0.075, 0.085 Temperatures $T$: 37.5, 37.4, 37.3, 30.4, 25.6, 30.9, 26.3, 26.5, 25.6 We approximate $\frac{dT}{dx}$ by the slope of $T$ vs $x$ using linear regression or by taking two points with significant temperature difference. For simplicity, take points at $x=0.025$ m ($T=37.3$) and $x=0.085$ m ($T=25.6$): $$\frac{dT}{dx} = \frac{25.6 - 37.3}{0.085 - 0.025} = \frac{-11.7}{0.06} = -195\, ^\circ C/m$$ 5. **Calculate Thermal Conductivity $k$:** Given $Q=5$ W, $A=0.0004909$ m$^2$, and $\frac{dT}{dx} = -195$ $^\circ C/m$: $$k = -\frac{Q}{A \frac{dT}{dx}} = -\frac{5}{0.0004909 \times (-195)} = \frac{5}{0.0004909 \times 195}$$ Calculate denominator: $$0.0004909 \times 195 = 0.0958$$ Then: $$k = \frac{5}{0.0958} = 52.2\, W/mK$$ 6. **Interpretation:** The calculated thermal conductivity $k$ for brass is approximately 52.2 W/mK, which is close to typical literature values (around 109 W/mK for brass, but can vary depending on alloy and temperature). Differences may arise due to experimental errors or assumptions. 7. **Summary:** - Calculated cross-sectional area $A = 0.0004909$ m$^2$ - Estimated temperature gradient $\frac{dT}{dx} = -195$ $^\circ C/m$ - Thermal conductivity $k \approx 52.2$ W/mK for heater power 5 W This process can be repeated for other heater powers to verify consistency.