The Density Of Lead Is 1.13 X 104 Kg/M3 At 20.0°C. Find Its Density (In Kg/M3) At 111°C. (Use Α = 29 X 10-6 (OC)-1 For The Coefficient Of Linear Expansion. Give Your Answer To At Least Four Significant Figures.)
The correct answer and explanation is :
To find the density of lead at 111°C, we can use the concept of thermal expansion and the relationship between temperature and volume change. The density of a material is inversely proportional to its volume. Therefore, when the temperature increases, the volume expands, and the density decreases.
Formula:
The formula for the change in volume with temperature is based on the coefficient of volumetric thermal expansion, and it can be given as: ΔV=V0βΔT\Delta V = V_0 \beta \Delta T
Where:
- ΔV\Delta V is the change in volume,
- V0V_0 is the initial volume,
- β\beta is the coefficient of volumetric expansion, and
- ΔT\Delta T is the change in temperature.
For most materials, the coefficient of volumetric expansion (β\beta) is approximately three times the coefficient of linear expansion (α\alpha): β=3α\beta = 3 \alpha
Step 1: Calculate the Coefficient of Volumetric Expansion
We are given the coefficient of linear expansion (α\alpha) for lead: α=29×10−6 (°C)−1\alpha = 29 \times 10^{-6} \, (\text{°C})^{-1}
Therefore, the coefficient of volumetric expansion (β\beta) is: β=3×29×10−6=87×10−6 (°C)−1\beta = 3 \times 29 \times 10^{-6} = 87 \times 10^{-6} \, (\text{°C})^{-1}
Step 2: Use the Relationship Between Density and Volume
Density (ρ\rho) is inversely proportional to volume. Thus, we can relate the initial and final densities using the formula: ρ2=ρ1×(11+βΔT)\rho_2 = \rho_1 \times \left( \frac{1}{1 + \beta \Delta T} \right)
Where:
- ρ1\rho_1 is the initial density at 20.0°C20.0°C,
- ρ2\rho_2 is the final density at 111°C111°C,
- β\beta is the coefficient of volumetric expansion,
- ΔT\Delta T is the temperature change.
Step 3: Calculate the Change in Temperature
The temperature change is: ΔT=111.0°C−20.0°C=91.0°C\Delta T = 111.0°C – 20.0°C = 91.0°C
Step 4: Calculate the Final Density
We are given the initial density of lead at 20.0°C20.0°C: ρ1=1.13×104 kg/m3\rho_1 = 1.13 \times 10^4 \, \text{kg/m}^3
Now, we can plug the values into the equation: ρ2=1.13×104×11+87×10−6×91\rho_2 = 1.13 \times 10^4 \times \frac{1}{1 + 87 \times 10^{-6} \times 91} ρ2=1.13×104×11+0.007917\rho_2 = 1.13 \times 10^4 \times \frac{1}{1 + 0.007917} ρ2=1.13×104×11.007917\rho_2 = 1.13 \times 10^4 \times \frac{1}{1.007917} ρ2=1.13×104×0.9921\rho_2 = 1.13 \times 10^4 \times 0.9921 ρ2≈1.121×104 kg/m3\rho_2 \approx 1.121 \times 10^4 \, \text{kg/m}^3
Final Answer:
The density of lead at 111°C111°C is approximately: ρ2≈1.121×104 kg/m3\rho_2 \approx 1.121 \times 10^4 \, \text{kg/m}^3
This is the final density of lead at the higher temperature, with four significant figures.
Explanation:
The density of lead decreases with an increase in temperature because the material expands. The expansion of the material’s volume causes a reduction in density since density is the mass per unit volume. We used the coefficient of linear expansion to calculate the change in volume and then applied the inverse relationship between volume and density to find the new density at the higher temperature.
