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Discussion Response Post to: Thermal Physics

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ramyani chakrabarty (3105)

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volm of 10 gm of water = 0.01/1000 = 10 ^{- 5} m³

volm of 10 gm of steam = 0.01/ 0.6 = 1/ 60 m³

Increase in volm = ( 1/ 60 - 10 ^{- 5} ) m³ = 9944 / 600000 m³

work done = p dv = 1665.67 J

Now, heat reqd to raise the temp of water to 100 degree = ms dt
                                                                                 = 0.01 * 4200 * 100 J
                                                                                 = 4200 J

heat reqd to convert 10 gm of water to steam = mL = 0.01* 2.5* 10⁶ J
                                                                          = 2.5* 10⁴ J

Total heat reqd = 4200 J + 2.5* 10⁴ J = 29200 J

Therefore, increase in internal energy = ( 29200 J - 1665.67 J )
                                                      = 2.75 X 10⁴ J

P.S. The data supplied in the book for L = 2.25 X 10⁶for which the result is
2.5 X 10⁴ .

NIT silchar electrical engineering

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