Answer
The volume of water in the graduated cylinder will increase due to thermal expansion of both the water and the cylinder. The change in volume of water can be calculated using the formula:
ΔV = V₀β(ΔT)
where ΔV is the change in volume of water, V₀ is the initial volume of water, β is the volumetric coefficient of thermal expansion of water, and ΔT is the change in temperature.
The change in volume of the cylinder can be calculated using the formula:
ΔV = V₀α(ΔT)
where α is the linear coefficient of thermal expansion of Pyrex glass.
We can assume that the initial volume of the cylinder is also 25.0000 mL since it is much larger than the volume of the water in the cylinder.
Using the values from the problem:
V₀ = 25.0000 mL
β = 2.07×10⁻⁴ K⁻¹ (at 20°C)
α = 3.3×10⁻⁶ K⁻¹
ΔT = 5°C
The change in volume of water is:
ΔV = 25.0000 mL × 2.07×10⁻⁴ K⁻¹ × 5°C = 0.025875 mL
The change in volume of the cylinder is:
ΔV = 25.0000 mL × 3.3×10⁻⁶ K⁻¹ × 5°C = 0.0004125 mL
The total change in volume is the sum of the changes in volume of the water and the cylinder:
ΔV_total = ΔV_water + ΔV_cylinder = 0.025875 mL + 0.0004125 mL = 0.0262875 mL
Therefore, the new volume of water in the graduated cylinder is:
V_new = V₀ + ΔV_total = 25.0000 mL + 0.0262875 mL = 25.0263 mL
So the cylinder will measure 25.0263 mL for the new volume of water.
Work Step by Step
The volume of water in the graduated cylinder will increase due to thermal expansion of both the water and the cylinder. The change in volume of water can be calculated using the formula:
ΔV = V₀β(ΔT)
where ΔV is the change in volume of water, V₀ is the initial volume of water, β is the volumetric coefficient of thermal expansion of water, and ΔT is the change in temperature.
The change in volume of the cylinder can be calculated using the formula:
ΔV = V₀α(ΔT)
where α is the linear coefficient of thermal expansion of Pyrex glass.
We can assume that the initial volume of the cylinder is also 25.0000 mL since it is much larger than the volume of the water in the cylinder.
Using the values from the problem:
V₀ = 25.0000 mL
β = 2.07×10⁻⁴ K⁻¹ (at 20°C)
α = 3.3×10⁻⁶ K⁻¹
ΔT = 5°C
The change in volume of water is:
ΔV = 25.0000 mL × 2.07×10⁻⁴ K⁻¹ × 5°C = 0.025875 mL
The change in volume of the cylinder is:
ΔV = 25.0000 mL × 3.3×10⁻⁶ K⁻¹ × 5°C = 0.0004125 mL
The total change in volume is the sum of the changes in volume of the water and the cylinder:
ΔV_total = ΔV_water + ΔV_cylinder = 0.025875 mL + 0.0004125 mL = 0.0262875 mL
Therefore, the new volume of water in the graduated cylinder is:
V_new = V₀ + ΔV_total = 25.0000 mL + 0.0262875 mL = 25.0263 mL
So the cylinder will measure 25.0263 mL for the new volume of water.
