Question

Pour into a container with a longitudinal expansion coefficient of five times ten to the negative power of five, a liquid with a longitudinal expansion coefficient of three to fifty-five hundredths multiplied by ten to the negative power of four. Increase the temperature of the set by a few degrees Kelvin to increase the height of the liquid inside the container by five percent? (Liquid does not spill out of the container)

Answer 1

Answer:

 ΔT = 200ºC to  ΔT = 11ºC

Explanation:

This is a thermal expansion exercise, for a material with volume the expression

          ΔV = β V₀ ΔT

for when the changes are small

         β = 3 aα

in the exercise they indicate the coefficient of thermal expansion of the liquid and solid

         α_solid = 5 10⁻⁵ ºC⁻¹

         α_liquid = 3 to 55 10⁻⁴ ºC⁻¹ = 30 to 550 10⁻⁵ ºC⁻¹

it is requested to increase the temperature so that the height of the liquid rises 5% = 0.05 inside the container

The expression for the volume of a body is its area for the height

        V₀ = π r² h

the final volume when heated changes as the radius and height change, suppose the radius is equal to the height

        r = h

        V_f = π (h + 0.05h)² (h + 0.05)

        V_f = pi h² 1,05³

     

        V_f = 1.157525 V₀

the volume variation is

         V_f - V₀ = 1.15 V₀ - V₀

         ΔV = 0.15 V₀

therefore, so that the total change in the volume of liquid and solid is the desired

         ΔV = ΔV_{liquid} - ΔV_{solid}

let's write the dilation equations for the two elements

         ΔV_{liquid} = 3 α_{liquid} V₀ ΔT

         ΔV_{solid} =  3 α_{solid} V₀ ΔT

we are assuming that the volumes of the solid and liquid are initial, as well as their temperatures

we substitute

          0.15 V₀ = 3 α_{liquid} V₀ ΔT - 3 α_{solid} V₀ ΔT

          0.15 = 3 ΔT ( α_{liquid} - α_{solid})

          ΔT = $\frac{0.05}{\alpha_{liquid} - \alpha_{solid} }$

let's calculate for the extreme values ​​of the liquid

α_{liquid} = 30 10⁻⁵ ºC⁻¹

          ΔT = $\frac{0.05}{( 30 -5 ) 10^{-5}}$

          ΔT = 5000/25

          ΔT = 200ºC

α_{liquid} = 5.5 10-5

           ΔT = \frac{0.05}{( 550 -5 ) 10^{-5}}

           ΔT = 5000 / 450

           ΔT = 11ºC