Volumetric Expansion and Overflow

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The discussion focuses on calculating the overflow of turpentine from a glass beaker due to thermal expansion. Given a beaker volume of 1.90 L at an initial temperature of 11.0 °C, the volumetric expansion coefficient for turpentine is 9.72 × 10–4 K–1. The correct calculation for the overflow when the temperature rises to 30.0 °C results in an overflow volume of 3.41e-05 m3. Participants noted that the initial calculation of 3.51e-05 m3 was incorrect because it did not account for the glass beaker's expansion.

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A glass beaker of volume V = 1.90 L is filled with turpentine when the temperature is T1 = 11.0 °C. The linear expansion coefficient for glass is αglass = 9.13 × 10–6 K–1 and the volumetric expansion coefficient for turpentine is βturpentine = 9.72 × 10–4 K–1. How much turpentine will overflow the beaker when the temperature will reach T2 = 30.0 °C?

The answer is 3.41e-05 m3.

I used deltaV = β x deltaT x V
deltaV = 9.72e-4 x (30-11) x (1.9e-3) = 3.51e-05 m3. It's not the same as the answer but its close. Is that just a coincidence? What have I done wrong?
 
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To determine the overflow of turpentine you have to consider the volume expansion of the beaker along with that of turpentine.
 

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