You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ c until its volume contracts by 0.507 % of its initial value. what is the final temperature of the substance? the coefficient of volume expansion for carbon disulfide is 1.15 × 10 − 3 ( ∘ c ) − 1 .

= 85.89 ° C

Explanation:

The linear thermal expansion process is given by

      ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

     ΔV = V β ΔT

Let's apply this equation to our case

     ΔV / V = -0.507% = -0.507 10-2

     ΔT = (ΔV / V)  1 /β

     ΔT = -0.507 10⁻²  1 / 1.15 10⁻³

     ΔT = -4.409

      –T₀ = 4,409

      = T₀ - 4,409

      = 90.3-4409

      = 85.89 ° C

are there options? cause you ain’t. real specific

the answer is b i think