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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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Answer:

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3 years ago
A reciprocating engine of 750mm stroke runs at 240 rpm. If the length of the connecting rod is 1500mm find the piston speed and
Sedbober [7]

Answer:

speed = 16.44 m/s

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Explanation:

Given data

Speed ( N) = 240 rpm

angle  = 45°

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To find out

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Solution

we find speed by this formula

speed = r ω (sin(θ) + (sin2(θ)/ 2n))  ...................1

here we have find  r and ω

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so ω = 2\pi × 240 / 60

ω =  25.132 rad/s

n = l/r =  1500/750 = 2

we know  L = 2r

so r = L/2 = 750/2 = 375 mm

put these value in equation 1

speed = 375 × 25.132 (sin(45) + (sin2(45)/ 2×2))  

speed = 16444.811823 mm/s = 16.44 m/s

Acceleration = r ω² (cos(θ) + (cos2(θ)/ n))  ...................2

put the value  r, ω and n in equation 2

Acceleration = r ω² (cos(θ) + (cos2(θ)/ n))

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3 0
3 years ago
A food department is kept at -12 °C by a refrigerator in an environment at 30 °C. The total heat gain to the food department is
Nataly [62]

Answer: P = 0.416 kW

Explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

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Шnet = 4800 - 3300 = 1500 kJ/h

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COPr = 2.2

The Power as required gives;

P = Qn - Ql  = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1

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3 years ago
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Decolagem e pouso


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