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

ΔT = 200ºC to ΔT = 11ºC

Step-by-step 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 =
`(0.05)/(\alpha_(liquid) - \alpha_(solid) )`

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

α_{liquid} = 30 10⁻⁵ ºC⁻¹

ΔT =
`(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