Question

You are making pesto for your pasta and have a cylindrical measuring cup 10.0 cm high made of ordinary glass [β=2.7×10−5(C∘)−1] that is filled with olive oil [β=6.8×10−4(C∘)−1] to a height of 3.00 mm below the top of the cup. Initially, the cup and oil are at room temperature (22.0∘C). You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

Answer 1

Answer:

The temperature will the olive oil start to spill out of the cup is 51.4 C

Explanation:

Knowing

βglass = $(2.7 * 10^{-5} C )^{-1}$ .... βoil = $(6.8 * 10^{-4} C )^{-1}$ .... T1 = 20 $C^{-1}$

Knowing the coefficient volume expansion of oil is greater than glass

δVoil =  δVglass + $0.3 * 10^{-2} * A$

Using the first equation and

δVoil = δVoil * βoil * δT

δVglass = δVglass * βglass * δT

δT(δVoil * βoil - δVglass * βglass) = $0.3 * 10^{-2} * A$

Solving

δT = $\frac{0.3 * 10^{-2} * A }{9.7 * 10^{-2}m * A * 6.8 * 10^{-4}C^{-1} * 10 * 10^{-2}m * A * 25 * 10^{-5} C^{-1} } = 31.4 C$

T2 = T1 + δT

T2 = 20 + 31.4 = 51.4 C