Question

What is the entropy change of 450 g of water when it changes from liquid to steam at its usual boiling point? For water, $L_F$ = 0.334 MJ/kg, $L_V$ = 2.26 MJ/kg.

Answer 1

Answer:

2726.54 J/K

Explanation:

Entropy change = $\frac{m\times L_{v}}{T}$

where

m = mass of water = 450g = 0.45kg

$L_{v}$ = 2.26 MJ/kg = 2260000 J/kg

T = temperature at which water change from liquid to steam = 100 degree Celsius = 272 + 100 = 373 degree kelvin.

Entropy change = $\frac{0.4\times 2260000}{373}$ = 2726.54 J/K