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2 years ago

Which of the following is an example of a chemical change?

Physics

1 answer:

tankabanditka [31]2 years ago

3 0

b. burning paper

hope it helps

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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 force of 185 n is needed to keep a small boat moving at 2.39 m/s. what is the power required to keep the boat moving at the st

mash [69]

We know, P = F . ΔV

P = 185 * 2.39 = 442.15 watt

5 0

3 years ago

if thanos lost in End Game" what would happend to the gountlet and who would take over the world with it?????

AleksandrR [38]

Have you seen Endgame?

3 0

3 years ago

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The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th

MaRussiya [10]

Answer:

a). $P=78.4W$

b). $P=392kPa$

c.) It must be at the bottom

Explanation:

Given:

Volume flow $V_f=2.0x10^{-4}m^3/s$

Well depp $h=40.m$

The power output of the pum

$W=F*d$

$F=m*g$

$m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg$

$W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2$

$W=78.4J$

$P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W$

The pressure of difference the pum

Δ $P=p*g*y'$

Δ $P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa$

$P=392kPa$

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way

4 0

3 years ago

How is the scientific use of the term digital different from the common use

Bess [88]

We commonly know refer to something 'digital' has to something electronic that can be visibly seen such as a watch, clock, camera, screen, etc. It really refers to stored energy or electricity that's not natural. But the word 'digital' in science refers to the depiction of data or information in figures (such as in a table) in contrast to as a chart, graph, drawing, or other pictorial form.

7 0

3 years ago

Read 2 more answers