Question

What is the entropy change (in J/K) of 71 g of steam at 100°C when it condenses to water at the same temperature?

Answer 1

Answer : The change in entropy is $4.29\times 10^2J/K$

Explanation :

Formula used :

$\Delta S=\frac{m\times L_v}{T}$

where,

$\Delta S$ = change in entropy = ?

m = mass of water = 71 g

$L_v$ = heat of vaporization of water = $2256J/g$

T = temperature = $100^oC=273+100=373K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{(71g)\times (2256J/g)}{373K}$

$\Delta S=429.4J/K=4.29\times 10^2J/K$

Therefore, the change in entropy is $4.29\times 10^2J/K$