ThermodynamicsMedium
Question
When water is boiled at 2 atm pressure, the letant heat of vaprization is 2. × 106 J/ kg and the boiling point is 120oC. At 2 atm pressure, 1 atm pressure, 1 kg of water has volume of 10-3 m3 and 1 kg of steam has a volume of 0.824 m3. The increase in internal energy of 1 kg of water when it is converted into steam at 2atm pressure and 120oC is [1 atm pressure = 1.013 × 105N/m2]
Options
A.2.033 J
B.2.033 × 106 J
C.0.167 × 106 J
D.2.267 × 106 J
Solution
Total heat given to convert water into steam at 120oC is Q = mL 1 × 2.2 × 106 = 2.2 × 106 J
The work done by the system against the surrounding is
PᐃV = 2 × 1.013 × 105(0.824 × 0.001) = 0.167 × 106 J ∴ ᐃU = Q - W = 2.033 × 106 J
The work done by the system against the surrounding is
PᐃV = 2 × 1.013 × 105(0.824 × 0.001) = 0.167 × 106 J ∴ ᐃU = Q - W = 2.033 × 106 J
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