Question

You are making pesto for your pasta and have a cylindrical measuring cup 10.1 cm high made of ordinary glass (B(beta)=2.7x10^-5(C)^-1) and that is filled with olive oil (B(beta)=6.8x10^-4(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 21.4 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?

T=_____________ degrees C

Answer 1

given that ::

height of the cylindrical cup, h₁ = 10.1 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, T_initial = 21.4 degree celsius

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

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

inserting the values in above eq,

[ (10.1 cm - 0.11 cm) (6.8 x 10^-4 C^-1) - (10.1 cm) (2.7 x 10^-5 C^-1) ] T_total = (0.11 cm)

(67.932 x 10^-4) - (27.27 x 10^-5) T_total = 0.11

(0.0067932) - (0.0002727) T_total = 0.11

(0.0065205) T_total = 0.11

T_total = 0.11 / 0.00652 = 16.87 degree celsius

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

T_total = T_final - T_initial

or T_final = T_total + T_initial                                { eq. 2 }

inserting the values in eq.2,

T_final = (16.87 + 21.4) degree celsius

T_final = 38.27 degree celsius