Answer:
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
