Question

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.220 × 10-3 m3. Into this cavity is placed 1.100 × 10-3 m3 of benzene [β = 1240 × 10-6 (C°)-1]. Both the copper and the benzene have the same temperature. By what amount ΔT should the temperature of the sphere and the benzene within it be increased, so that the liquid just begins to spill out?

Answer 1

Δ should be 0.1009

Step-by-step explanation:

The change in the units volume when temperature change can be expressed as:

∆v = v0Δ

with v0 = the initial volume

with = the volumetric temperature expansion coefficient

with Δ = the change of temperature.

To calculate the final volume vf we'll get:

v = v0 + ∆ = v0(1 + Δ)

The liquid just begins to spill out if v(benzene) = :

v()(1 + Δ) = = v() (1 + Δ)

(v(cavity)-v(benzene))/(((benzene) -(copper)) = Δ

((1.22*10^-3)-(1.1*10^-3))/((1240*10^-6)-(51*10^-6)) = Δ

Δ = 0.1009

Δ should be 0.1009