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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]

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

Explanation:

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:

For process 1:

We are given:

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

Putting values in equation 1, we get:

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

For process 2:

We are given:

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

Putting values in equation 2, we get:

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

For process 3:

We are given:

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

Putting values in equation 1, we get:

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

For process 4:

We are given:

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

Putting values in equation 2, we get:

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

For process 5:

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

A microfarad capacitor is charged via a 12v battery. there is 1 kohms resistor in series with the capacitor. how much current is

olga2289 [7]

The answer completely depends on the number that belongs in the space before the word "microfarad".

7 0

3 years ago

An 80.0 kg hiker walks a distance of 400.0 m along a road that slopes 5.0 degrees upward, and then stops. What is the hiker's fi

lana66690 [7]

The height difference is found by
$\delta H=400sin(5 \°)=34.86m$
Then the change in potential energy is
$E=mgh=(80.0kg)(9.8 \frac{kgm}{s^2})(34.86)= 27332J$

4 0

3 years ago

Now imagine a person dragging a 50 kg box along the ground with a rope, as

ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

= 50 x 9.8

= 490 N

The coefficient of friction,

μ = f/ η

= 30 / 490

= 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0

3 years ago

A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top

Advocard [28]

Answer:

a) Em₀ = 42.96 104 J , b) $W_{fr}$ = -2.49 105 J , c) vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

Em = K + U

a) Let's look for the initial mechanical energy

Em₀ = K + U

Em₀ = ½ m v2 + mg and

Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

Em₀ = 36 104 + 6.96 104

Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

$W_{fr}$ = Em₂ -Em₀

Em₂ = K + U

Em₂ = ½ m v₂² + m g y₂

Em₂ = ½ 50 85 2 + 50 9.8 427

Em₂ = 180.625 + 2.09 105

Em₂ = 1,806 105 J

$W_{fr}$ = Em₂ -Em₀

$W_{fr}$ = 1,806 105 - 4,296 105

$W_{fr}$ = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

$W_{fr}$ = ΔEm

$W_{fr}$ = Emf - Emo

-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

½ 50.0 vf² = 0.561

vf = √ 0.561 25

vf = 3.75 m / s

6 0

3 years ago

Plzzz help will mark brainliest ​

Plzzz help will mark brainliest