Question

A vertical glass tube of length L = 1.3m is half filled with a liquid at 22.5 ˚C. How much will the height of the liquid column change when the tube is heated to 32.5 ˚C? Take αglass = 1.34× 10-5/K and βliquid = 3.74× 10-5/K.

Answer 1

h = V/A = (V0 + dV)/(A0 + dA)

dA = A0*2*alpha*dT

dV = V0*beta*dT

h = (V0 + V0*beta*dT)/(A0 + 2*A0*alpha*dT)

h = V0*(1 + beta*dT)/(A0*(1 + 2*alpha*dT))

h = h0*(1 + beta*dT)/(1 + 2*alpha*dT)

dh = h - h0

dh = h0*(1 + beta*dT)/(1 + 2*alpha*dT) - h0

dh = h0*[(1 + beta*dT)/(1 + 2*alpha*dT) - 1]

Using the given values

h0 = 1.3/2 = 0.65 m

dT = 32.5 - 22.5 = 10

alpha = 1.34*10^-5

beta = 3.74*10^-5

dh = 0.65*[(1 + 3.74*10^-5*10)/(1 + 2*1.34*10^-5*10) - 1]

dh = 6.888*10^-5