You are making pesto for your pasta and have a cylindrical measuring cup 10.9 cm

Question

You are making pesto for your pasta and have a cylindrical measuring cup 10.9 cm high made of ordinary glass (=2.7×105(C)1) that is filled with olive oil (=6.8×104(C)1) to a height of 1.10 mm below the top of the cup. Initially, the cup and oil are at a kitchen temperature of 22.3 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?

Explanation / Answer

given that ::

height of the cylindrical cup, h1 = 10.9 cm

Expansion of ordinary glass, B1(beta) = 2.7 x 10-5 C-1

height below the top of the cup, h2 = 1.1 mm = 0.11 cm

Expansion of ordinary glass, B2(beta) = 6.8 x 10-4 C-1

initial temperature for cup & oil which at a kitchen, Tinitial = 22.3 degree celsius

the difference in increase in heights of oil and glass for a rise of temperature Ttotal is given as ::

[ (h1 - h2) x B2(beta) - h1 x B1(beta) ] Ttotal = h2                                     { eq. 1 }

inserting the values in above eq,

[ (10.9 cm - 0.11 cm) (6.8 x 10-4 C-1) - (10.9 cm) (2.7 x 10-5 C-1) ] Ttotal = (0.11 cm)

(73.37 x 10-4) - (29.43 x 10-5) Ttotal = 0.11

Ttotal = 0.11/(0.0070427)

Ttotal = 15.61 Degree celsius

the olive oil will start to spill out of the cup at temperature which is given as ::

Ttotal = Tfinal - Tinitial

or Tfinal = Ttotal + Tinitial                                { eq. 2 }

inserting the values in eq.2,

Tfinal = (15.61 + 22.3) degree celsius

Tfinal = 37.91 Degree celsius

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