2. At a pressure of 1 atm = 1.01 x 105 Pa, water boils at a temperature of 100° C. At what pressure will it boil at 115° C? The heat of vaporization of water in 539 kcal/kg = 22.6 x 105 J/kg, and we will assume this is constant. Two possibly useful numbers are R = 8.315 × 10³ J/(kmole K) and 1 kmole of water = 18 kg. =
2. At a pressure of 1 atm = 1.01 x 105 Pa, water boils at a temperature of 100° C. At what pressure will it boil at 115° C? The heat of vaporization of water in 539 kcal/kg = 22.6 x 105 J/kg, and we will assume this is constant. Two possibly useful numbers are R = 8.315 × 10³ J/(kmole K) and 1 kmole of water = 18 kg. =
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Transcribed Image Text:2. At a pressure of 1 atm = 1.01 x 105 Pa, water boils at a temperature of 100°
C. At what pressure will it boil at 115° C? The heat of vaporization of water
in 539 kcal/kg = 22.6 x 105 J/kg, and we will assume this is constant. Two
possibly useful numbers are R = 8.315 x 103 J/(kmole K) and 1 kmole of water
18 kg.
=
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