Question

Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the heat of vaporization of water is 2256×103J/kg.

Answer 1

Answer : The change in entropy is $6.05\times 10^3J/K$

Explanation :

Formula used :

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

where,

$\Delta S$ = change in entropy = ?

m = mass of water = 1.00 kg

$L_v$ = heat of vaporization of water = $2256\times 10^3J/kg$

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

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

$\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}$

$\Delta S=6048.25J/K=6.05\times 10^3J/K$

Therefore, the change in entropy is $6.05\times 10^3J/K$