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4 years ago

How is that speed affected by medium density and medium temperature?

1 answer:

5 0

Speed decreases more the denser the medium.
Since density of a material normally decreases with temperature, it is not surprising that the speed of light in a liquid will normally increase as the temperature increases. (Thus, the index of refraction normally decreases)

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A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The w

miss Akunina [59]

The question is incomplete. The complete question is :

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).

Solution :

From flow over sphere, the mass transfer equation can be written as :

$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$

where, Sherood number, $Sh = \frac{K_L d}{D_{eff}}$

Reynolds number, $Re=\frac{Vd\rho}{\mu}$

Schmid number, $Sc= \frac{\mu}{\rho D_{eff}}$

So,

$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$

Diameter, d = 1 cm = $1 \times 10^{-2}$ m

V = 1 m/s

$\rho = 1000 \ kg/m^3$

$\mu = 10^{-3} \ kg/m/s$

$D_{eff} = 2 \times 10^{-9} \ m^2/s$

$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$

$K_L \times 5 \times 10^6=478.22$

$K_L=9.5644 \times 10^{-5}$ m/s

So the mass transfer coefficient is 9.5644 $\times 10^{-5}$ m/s. It is given solubility,

$\Delta C = 2 \ kg/m^3$

$N = Md^2 \times \Delta C \times K_L$

$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$

$N= 6 \times 10^{-8}$ kg/s (dissolution rate)

6 0

3 years ago

What is the effect of stirring on supercooling?

garik1379 [7]

I believe that the answer to the question provided above is in a supercooling the effect of stirring is that is acts as catalyst that hastens the cooling effect.

Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

5 0

3 years ago

The average distance between collisions for atoms in a real gas is known as the mean free path (see pp. 298-9 in McKay). Which o

dusya [7]

Answer:

300 nm

Explanation:

R = Gas constant = 8.314 J/molK

r = Atomic radii = $1\times 10^{-10}\ m$

d = Atomic diameter = $2r=2\times 10^{-10}\ m$

At STP

T = Temperature = 273.15 K

P = Pressure = 100 kPa

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

The mean free path is given by

$\lambda=\frac{RT}{\sqrt2d^2N_AP}\\\Rightarrow \lambda=\frac{8.314\times 273.15}{\sqrt2 \pi \times (2\times 10^{-10})^2\times 6.022\times 10^{23}\times 100000}\\\Rightarrow \lambda=2.12165\times 10^{-7}\ m=212.165\times 10^{-9}\ m=212.165\ nm$