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?

Δ should be 0.1009

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

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.