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2 years ago

Can someone help me answer this??

Physics

1 answer:

Bingel [31]2 years ago

7 0

Answer:

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What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

<u>For process 1:</u>

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

<u>For process 2:</u>

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

<u>For process 3:</u>

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

<u>For process 4:</u>

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

<u>For process 5:</u>

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

Please answer this question for me and explain why.

horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass $m_2$ and the 4kg mass $m_1$ . If the tension in the string is $T$ then for the mass $m_2$

(1). $T-m_2g=-m_2a$ <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass $m_1$

(2). $T-m_1g =m_1a$ <em> (the acceleration is upwards, hence the positive sign)</em>

Solving for $T$ in the 2nd equation we get:

$T =m_1a+m_1g$ ,

and putting this into the 1st equation we get:

$m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g$

$\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}$

Back to the question:

Using the formula for the acceleration we find

$a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g$

$a = \dfrac{g}{9},$

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0

3 years ago

Suppose you analyze standardized test results for a country and discover almost identical distributions of physics scores for fe

zaharov [31]

Answer:

go ask a teacher

Explanation:

go to school

go find the teacher

ask him/her for help

finish question done :)

7 0

3 years ago

When frequency is decreased what happens to the amplitude?

guajiro [1.7K]

Answer:

it increases the amplitude of the wave as it propagates.

Explanation:

7 0

2 years ago

Read 2 more answers

un astronauta cuya masa es 80kg permanece inmovil en el espacio exterior, al activar la unidad propulsora que lleva en la espald

beks73 [17]

Answer:

Explanation:

80(0) = 2(15) + (80-2)v

v = - 0.38 m/s

5 0

2 years ago