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Monica [59]
3 years ago
5

What is the entropy of a closed system in which 25 distinguishable grains of sand are distributed among 1000 distinguishable equ

alsized compartments with only one grain per compartment?
Engineering
2 answers:
Veronika [31]3 years ago
7 0

Answer:

S = 172.69 J/K

Explanation:

For a closed system, the entropy is given as the natural logarithm of microstates.

The number of particles (grains of sand); N = 25

The number of boxes (compartments); M = 1000

Entropy is the logarithm of number of microstates; Thus,

S = In ψ

However, the number of microstates is given by the formula;

ψ = Mⁿ

Thus, S = In Mⁿ

Plugging in the relevant values to obtain;

S = In (1000)^(25)

S = 172.69 J/K

charle [14.2K]3 years ago
3 0

Answer:

The entropy of the closed system is

S = (1.58 × 10⁻²¹) J/K

Explanation:

Entropy of a system is proportional to the number of different microstates. The constant of proportionality is the Boltzmann's constant.

S = K In W

K = Boltzmann's constant = (1.381 × 10⁻²³) J/K.

Number of different microstates possible for distributing 25 distinguishable grains of sand among 1000 distinguishable equal sized compartments with only one grain per compartment = ¹⁰⁰C₂₅ = (4.7642 × 10⁴⁹)

S = (1.381 × 10⁻²³) In (4.7642 × 10⁴⁹)

S = 1.381 × 10⁻²³ × 114.4

S = (1.58 × 10⁻²¹) J/K

Hope this Helps!!!

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Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

  = 0.0214 * 3.738

  = 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

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3 years ago
What is the most important part of a successful Election Day?
wlad13 [49]

Answer: voting of course

Explanation:

8 0
2 years ago
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A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti
Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = \frac{0.03194}{0.315}

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m  

here h is heat required and m is mass flow rate

H = 2680.36  × 0.10139

H =  271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

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net boiler heat = \frac{271.75}{0.90}

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5 0
3 years ago
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical
dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6  N

For Brass spacer

Pressure = 42201.6  N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6  N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 =  291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 =  370.758 + 16^2

d^2 =   626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

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What can be the main disadvantage of pulse amplitude modulation?​
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Answer:

transmission bandwidth required is very large.

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