The volume of steam produced by 1g of water at 100°c is 1650 cm3.Calculate the change in internal energy during the change of state givenJ = 4.2 × 107 erg cal–1 g = 98 J cm/s2latent heat of steam = 540 cal/g

Question

Arun · Answer

Dear Vaibhav Mass of water = 1 gm = .001 kg Volume of water = mass/density = .001/1000=.000001 m&sup3; = 1 cm&sup3; Change in volume = 1650 -1 = 1649 cm&sup3; dQ = m*L = 540 * 4.2 * 10^7 erg. P = 1 atm = 76* 13.6 *9.8 dU = dQ - P* dV = 22.68 * 10^9 - 1.67 * 10^9 = 21.01 * 10^9 erg Regards Arun(askIITians forum expert)

ankit singh · Answer

From first law of thermodynamics, δ Q = d U + δ W 5 4 × 4 . 1 8 4 = d U + P d V 5 4 × 4 . 1 8 4 = d U + 1 . 0 1 3 × 1 0 5 × 1 6 7 . 1 × 1 0 − 6 0 On solving, we will get d U = 2 0 8 . 7 J

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The volume of steam produced by 1g of water at 100°c is 1650 cm3.Calculate the change in internal energy during the change of state givenJ = 4.2 × 107 erg cal–1 g = 98 J cm/s2latent heat of steam = 540 cal/g

The volume of steam produced by 1g of water at 100°c is 1650 cm3.Calculate the change in internal energy during the change of state givenJ = 4.2 × 107 erg cal–1 g = 98 J cm/s2latent heat of steam = 540 cal/g

2 Answers

From first law of thermodynamics,

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