You are making pesto for your pasta and have a cylindrical measuring cup 12.0 cmcm high made of ordinary glass [β=2.7×10−5(C∘)−1][β=2.7×10−5(C∘)−1 ] that is filled with olive oil [β=6.8×10−4(C∘)−1][β=6.8×10−4(C∘)−1 ] to a height of 4.50 below the top of the cup. Initially, the cup and oil are at room temperature (22.0∘C∘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?

Express your answer in degrees Celsius to three significant figures

81.8°C

Explanation:

βglass = 2.7×10^(−5) (C∘)−1

βoil = 6.8 × 10^(−4) (C∘)−1

T1 = 22°C

Total cup height = 12cm = 0.12m

Height below top of cup= 4.5mm or 0.0045m

From the question, we can see that the coefficient volume of expansion of oil is greater than that of glass and therefore, the heat absorbed by oil will be greater.

Thus;

δVoil =δVglass + 0.0045A

So, 0.0045A = δVoil - δVglass

0.0045A = δV(oil) - δVglass

Now,

δV(oil) is expressed fully as;

δT x V(oil) xβ(oil)

Likewise, δV(glass) is expressed fully as;

δT x V(glass) xβ(glass). Thus;

0.0045A = [δT x V(oil) xβ(oil)] - [δT x V(glass) xβ(glass)]

So,

0.0045 = δT[V(oil) xβ(oil)] - [V(glass) xβ(glass)]

So;

δT = 0.0045/[V(oil) xβ(oil)] - [V(glass) xβ(glass)]

Plugging in the relevant values to obtain ;

δT = 0.0045/[V(oil) xβ(oil)] - [V(glass) xβ(glass)]

V(oil) = Area of cup x height = A(0.12 - 0.0045) = 0.1155A

V(glass) = 0.12A

So,δT = 0.0045/[(0.1155A) x 6.8 × 10^(−4) ] - [0.12A x 2.7×10^(−5))]

= 0.0045/[0.7854 -x 10^(−4) - 0.0324 x 10^(−4)] = 0.0045/(0.753 x 10^(-4) = 59.76°C

Now, δT is the change in temperature and it's ;

δT = T2 - T1

Thus, T2 = δT + T1

T2 = 59.76 + 22 = 81.76°C

Approximating to 3 significant figures, T2 = 81.8°C

because if the tool that you are using is not balanced it there would be false mass.