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3 years ago

I’ll give brainliest and 10 points

Physics

1 answer:

leonid [27]3 years ago

7 0

Answer:

what's these its D okkkk

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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 car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0

3 years ago

Give 1 real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is c

OverLord2011 [107]

Answer:

On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket

How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period

When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned he uses more force and the ball also spends less time on the racket to produce the same momentum

Explanation:

The impulse of a force, ΔP is given by the following formula;

ΔP = F × Δt

Where ΔP is constant, we have;

F ∝ 1/Δt

Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.

7 0

3 years ago

Can lamp that works on a 2.5 v work on a 1.12 v ?

12345 [234]

Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.

5 0

2 years ago

If a object is at rest, does it mean there are no forces acting on it

Ganezh [65]

Yes that is correct :)

4 0

2 years ago

Read 2 more answers