If a 500-mL glass beaker is filled to the brim with methanol at a temperature of 5.00ºC, how much will overflow when its temperature reaches 22.0ºC? (Suppose the coefficient of volume expansion for methanol is 1200x10^(-6) 1/ºCRequired to answer. Single choice.
(1 Point)

10.2 mL.

Explanation:

From the question given above, the following data were obtained:

Original Volume (V₁) at temperature θ₁ = 500 mL

Initial temperature (θ₁) = 5 °C

Final temperature (θ₂) = 22 °C

coefficient of volume expansion (γ) = 1200×10¯⁶ / °C

Rise in volume (ΔV) =?

Next, we shall determine the volume (V₂) of the methanol at 22 °C. This can be obtained as follow:

Original Volume (V₁) at temperature θ₁ = 500 mL

Initial temperature (θ₁) = 5 °C

Final temperature (θ₂) = 22 °C

coefficient of volume expansion (γ) = 1200×10¯⁶ / °C

Volume 2 (V₂) at temperature θ₂ =?

γ = V₂ – V₁ / V₁ (θ₂ – θ₁)

1200×10¯⁶ = V₂ – 500 / 500 (22 – 5)

1200×10¯⁶ = V₂ – 500 / 500 × 17

1200×10¯⁶ = V₂ – 500 / 8500

Cross multiply

V₂ – 500 = 1200×10¯⁶ × 8500

V₂ – 500 = 10.2

Collect like terms

V₂ = 10.2 + 500

V₂ = 510.2 mL

Thus, the volume of the methanol at 22 °C is 510.2 mL

Finally, we shall determine the rise in volume of methanol as illustrated below:

Original Volume (V₁) at temperature θ₁ = 500 mL

Volume (V₂) at temperature θ₂ = 510.2 mL

Rise in volume (ΔV) =?

ΔV = V₂ – V₁

ΔV = 510.2 – 500

ΔV = 10.2 mL

Thus, 10.2 mL of the methanol will overflow when the temperature reaches 22 °C.

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