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Alex_Xolod [135]
3 years ago
8

a glass beaker has a mass of 50g. a liquid of density 1.8g/cm3 is poured into the beaker until it reaches the 200cm3 mark. calcu

late the total mass of the beaker and its contents​
Physics
1 answer:
xz_007 [3.2K]3 years ago
5 0

<u>Answer:</u>

total mass = 410 g

<u>Explanation:</u>

density = 1.8 g/cm³

volume = 200 cm³

                              density = mass / volume

                              mass (of liquid) = density   x    volume

                                        = 1.8 x 200

                                        = 360 g

          total mass (beaker + liquid) = 50 + 360 = 410 g     [Ans]

Hope this helps!

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

<u>Explanation:</u>

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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)

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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)

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\Delta S = Entropy change

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

T_2 = final temperature

T_1 = initial temperature

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

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\Delta S = Entropy change

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

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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

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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

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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

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We are given:

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Putting values in equation 2, we get:

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

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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

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