G ters to volume Goog x aln coe ttu.edu/andersor X c aln coettu.edu/anderson T-M

Question

G ters to volume Goog x aln coe ttu.edu/andersor X c aln coettu.edu/anderson T-Model 10-1-Encripted/tmodel10-1.swf 3.20: The 10 liter rigid container shown here contains 9.8 liters of water PO vapor. The temperature of the container contents is 1750C. What is the mass of the water in this container? Variables pane appears to be complete. Closed Multi-Phase None Required. Answer Unis LabelEqu Label/Equ Value Units Unknown variables ALSO ON GRAPHIC Known Variables kg 0.01 0.0098 lm 3 Equations 0.001121 mA3/kg 0.0002 m 3 0.21659 m 3 kg Post-Calculations: (Sequental equations) Pre-Calculations: 175 equations) (Sequental 0.1784 kg

Explanation / Answer

Volume of Water in the container = Total Volume of container - Volume of Water Vapour

===> Volume of Water in the container = 10 - 9.8 litres = 0.2 litres = 0.0002 m^3

At 175°C from the property table of water, the specific volume of water is v = 0.0012 m^3/kg

Now, Mass of water in the container is M = Volume of Water in the container / Specific Volume of water

===> M = 0.0002/0.0012 kg

===> M = 0.1786 kg or 178.6 grams

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