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08 Jul 2024, 13:02
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Humidity percentage is the water vapor mass contained within the total mass of air inside a specified volume of air. An observatory has three large containers each containing 100 tons of compressed air. The first two containers have the same humidity percentage. In an experiment, the air from all three containers was mixed into a container that had practically no air before the experiment. The humidity percentage of the combined air was found to be 62%.
In the table, select a humidity percentage for the First two containers and a humidity percentage for the Third container that are jointly consistent with the given information. Make only two selections, one in each column.
Since equal amounts of air (100 tons each) from the three containers have been compressed into the new container
Let x be the humidity percentage of the first 2 containers and y be the humidity percentage of the third container.
Therefore, x + x + y = 3(0.62)
2x + y = 1.86
Considering the options, if y = 0.58, then x = 0.64 as solved below:
2x + 0.58 = 1.86
2x = 1.86 - 0.58
2x = 1.28
x = 0.64
Therefore, the humidity percentage of the second container is 64% and that of the third container is 58%.
In the table, select a humidity percentage for the First two containers and a humidity percentage for the Third container that are jointly consistent with the given information. Make only two selections, one in each column.
Since equal amounts of air (100 tons each) from the three containers have been compressed into the new container
Let x be the humidity percentage of the first 2 containers and y be the humidity percentage of the third container.
Therefore, x + x + y = 3(0.62)
2x + y = 1.86
Considering the options, if y = 0.58, then x = 0.64 as solved below:
2x + 0.58 = 1.86
2x = 1.86 - 0.58
2x = 1.28
x = 0.64
Therefore, the humidity percentage of the second container is 64% and that of the third container is 58%.
08 Jul 2024, 13:06
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Let the first two containers' humidity percentage be x, the third container's humidity percentage y and each container's total mass M. After combining the three containers we obtain a mix of a total air mass 3M and the amount of water mass coming from:
\(z =\frac{2*x*M + y*M}{3M} = \frac{2}{3}*x+\frac{1}{3}*y = 62\%\)
By substituting the list of options in the eqution above, it can be induced that the values \(x=64\%\) and \( y=58\%\) are the only solution to the equation and therefore correspond to the right answer.
- The first two containers = \(2*x*M\)
- The third container =\( y*M\)
\(z =\frac{2*x*M + y*M}{3M} = \frac{2}{3}*x+\frac{1}{3}*y = 62\%\)
By substituting the list of options in the eqution above, it can be induced that the values \(x=64\%\) and \( y=58\%\) are the only solution to the equation and therefore correspond to the right answer.
08 Jul 2024, 13:09
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In order to get the percentage of humidity, we need to take the vapour mass \(V \)and divide it by the total air volume \(T\).
We know that \( T = 100 \) for each of the containers, and we also know that \( V_1 = V_2\) for the two containers and \( V_3\) is unknown.
To get the combined humidity, we divide the total vapour mass by the total air volume:
\(\frac{V_1 + V_2 + V_3 }{ 100 + 100 + 100} =\frac{ V_1 + V_2 + V_3 }{ 300} = 0.62\)
=> \(V_1 + V_2 + V_3 = 300 * 0.62 = 186\)
From here, let's say \(V_1 = V_2 = x \) (as stipulated in the task) and\( V_3 = y\)
Therefore, we need to select such values from the anwer options that \(2x + y = 186\)
Let's start with the highest possible \(x \)of 65%:
\(65*2 = 130 = 186 - 56\) [56 is not a suitable option]
Then we try \(x\) of 64%:
\(64*2 = 128 = 186 - 58\) [and 58 fits, well done!]
Great! this lets us answer that first two containers have 64% of humidity and the third one has 58%.
We know that \( T = 100 \) for each of the containers, and we also know that \( V_1 = V_2\) for the two containers and \( V_3\) is unknown.
To get the combined humidity, we divide the total vapour mass by the total air volume:
\(\frac{V_1 + V_2 + V_3 }{ 100 + 100 + 100} =\frac{ V_1 + V_2 + V_3 }{ 300} = 0.62\)
=> \(V_1 + V_2 + V_3 = 300 * 0.62 = 186\)
From here, let's say \(V_1 = V_2 = x \) (as stipulated in the task) and\( V_3 = y\)
Therefore, we need to select such values from the anwer options that \(2x + y = 186\)
Let's start with the highest possible \(x \)of 65%:
\(65*2 = 130 = 186 - 56\) [56 is not a suitable option]
Then we try \(x\) of 64%:
\(64*2 = 128 = 186 - 58\) [and 58 fits, well done!]
Great! this lets us answer that first two containers have 64% of humidity and the third one has 58%.
