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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?

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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Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

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4 years ago

Triss [41]

Answer:

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3 years ago

Water is being heated in a closed pan on top of a range whilebeing stirred by a paddle wheel. During the process 30kJ of heat is

My name is Ann [436]

Answer:

38 kJ

Explanation:

The solution is obtained using the energy balance:

ΔE=E_in-E_out

U_2-U_1=Q_in+W_in-Q_out

U_2=U_1+Q_in+W_in-Q_out

=38 kJ

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3 years ago

Consider the rigid pressure vessel containing air in the diagram at the

vredina [299]

Answer:

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Explanation:

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3 years ago

Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

nikitadnepr [17]

Answer:

a) $m_e= 3.05 Kg$

b) $\rho=1072.3kg/m^3$

c) $m_e= 3.05 Kg$

Explanation:

From the question we are told that:

Beaker Mass $m_b=1.20$

Liquid Mass $m_l=1.85$

Balance D:

Mass $m_d=3.10$

Balance E:

Mass $m_e=7.50$

Volume $v=4.15*10^{-3}m^3$

a)

Generally the equation for Liquid's density is mathematically given by

$m_e=m_b+m_l+(\rho*v)$

$\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}$

$\rho=1072.3kg/m^3$

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

$m_d=m- (\rho *v )$

$m=3.10+ (1072.30 *4.15*10^{-3}m^3 )$

$m=18.10kg$

c)

Generally the equation for E's Reading at A pulled is mathematically given by

$m_e=m_b+m_l$

$m_e = 1.20 + 1.85$

$m_e= 3.05 Kg$

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3 years ago