Initially when 1000.00 mL of water at 1O° C are poured into a glass cylinder, the height of the water column is 1000.0 mm. The water and its container are heated to 8O°C. Assuming no evaporation, what then will be the depth of the water column if the coefficient of thermal expansion for the glass is 3.6 x 10^-6 mm/mm per °C?

To solve this problem, we need to consider the thermal expansion of both the water and the glass cylinder.

1. First, we calculate the expansion of the glass cylinder. The formula for linear thermal expansion is ΔL = α * L0 * ΔT, where ΔL is the change in length, α is the coefficient of thermal expansion, L0 is the initial length, and ΔT is the change in temperature.

   For the glass cylinder, we have: α = 3.6 x 10^-6 mm/mm per °C L0 = 1000.0 mm ΔT = 80°C - 10°C = 70°C

   Substituting these values into the formula, we get: ΔL (glass) = 3.6 x 10^-6 mm/mm per °C * 1000.0 mm * 70°C = 0.252 mm

2. Next, we calculate the expansion of the water. The coefficient of volume expansion for water is approximately 0.000214 per °C. The formula for volume expansion is ΔV = β * V0 * ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V0 is the initial volume, and ΔT is the change in temperature.

   For the water, we have: β = 0.000214 per °C V0 = 1000.0 mL (since 1 mL of water has a height of 1 mm in the cylinder) ΔT = 70°C

   Substituting these values into the formula, we get: ΔV (water) = 0.000214 per °C * 1000.0 mL * 70°C = 15.0 mL

3. Finally, we calculate the new height of the water column. The change in height is equal to the change in volume (since 1 mL of water corresponds to a 1 mm increase in height), plus the change in height of the glass cylinder.

   ΔH = ΔV (water) + ΔL (glass) = 15.0 mm + 0.252 mm = 15.252 mm

   The new height of the water column is therefore 1000.0 mm + 15.252 mm = 1015.252 mm.

So, the depth of the water column after heating to 80°C will be approximately 1015.252 mm.