Answer:
the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm
Explanation:
The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:
[tex]V = \frac{m}{ \rho}[/tex]
Given that ;
the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g
density of the glycerine ([tex]\rho[/tex]) = 1.260 kg/L = 1.260 g/cm³
Then:
V = [tex]\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}[/tex]
V = [tex]0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})[/tex]
V = 0.0873 L
Now;Initially the volume of the glycerin before the ice cube starts to melt is:
[tex]V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L[/tex]
However; the volume of the water produced by the 0.11 kg ice cube = [tex]0.11*10^3 \ cm^3[/tex]
The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:
[tex]V_2 = V_i + V"[/tex]
replacing [tex]V"[/tex] with [tex]0.11*10^3 \ cm^3[/tex] ; we have:
[tex]V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})[/tex]
[tex]V_2 = V_i + 0.11 \ L[/tex]
The overall total change in the volume of the glycerin is illustrated as:
[tex]V_f = V_2 - V_1[/tex]
Now; from the foregoing ; lets replace the respective value of [tex]V_2[/tex] and [tex]V_1[/tex] in the above equation ; we have;
[tex]V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L[/tex]
The formula usually known to be the volume of a cylinder is :
[tex]V = \pi r ^2 h[/tex]
For the question ; we will have:
[tex]V_f = \pi r ^2 h[/tex]
making h the subject of the formula ; we have:
[tex]h = \frac{V_f}{\pi r^2}[/tex]
replacing 0.0227 L for [tex]V_f[/tex]and the given value of radius which is = 3.70 cm; we have:
[tex]h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}[/tex]
[tex]h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm[/tex]
Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm
