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Calculate the increase in the internal energy of 10 g of water when it is heated from `0^(0)C to 100^(0)C` and converted into steam at 100 kPa. The density of steam `=0.6 kg m^(-3)` specific heat capacity of water `=4200 J kg^(-1 ^(0)C^(-3)` latent heat of vaporization of water `=2.25xx10 6 J kg^(-1)`

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The correct Answer is:
A, B, D
Mass` = 10kg = 0.01kg`
`P = 10^5kpa`
`Delta Q = QH_2 O 0^@-100^@+QH_(2) O-steam`
`0.01xx4200xx100+0.01xx2.5xx10^3`
` = 4200+2500 = 29200`
`Delta W = paV`
`Delta V = ((0.01)/(06))-((0.01)/(1000)) = 0.01699`
`DeltaW = pDeltaV = 0.01699xx10^5 = 1699J`
` = 27501 = 2.75xx10^4J
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