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3 years ago

11

At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel

ors measure the two "dry" legs of a right triangle. What is the length in meters of the swimming course in the figure below?

1 answer:

Assoli18 [71]3 years ago

3 0

Answer:

47.17 m

Explanation:

From the diagram of the question attached, The length of the legs are 25 m and 40 m . This legs form a right angle triangle with the length of the swimming course (L).

Pythagoras theorem states that for a right angle triangle with hypotenuse a and legs b and c, then:

a² = b² + c²

In the triangle, the length of the swimming course (L) is the hypotenuse and the two legs are 25 m and 40 m. Using Pythagoras:

L² = 25² + 40²

L² = 625 + 1600

L² = 2225

L = √2225

L = 47.17 m

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

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

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$\Delta S$ = Entropy change

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

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

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We are given:

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

Putting values in equation 2, we get:

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

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

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

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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

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