A vessel containing 10 kg of water is left out where it evaporates.If the temperature remains constant at 20 $\degree$C,what is the change in entropy (in kcal/K)? (The latent heat of vaporization at 20 $\degree$C is 585 cal/g. )

Question

Quizplus · Accepted Answer

First, we need to calculate the total amount of heat required for the water to evaporate:
q = m * L
Where q is the amount of heat, m is the mass of water and L is the latent heat of vaporization. 
q = 10 kg * 585 cal/g * 4.184 J/cal = 24,462.6 J = 24.46 kJ 
Note that we converted the units to Joules since entropy is measured in Joules per Kelvin.

Now, we need to use the formula for change in entropy:
ΔS = q / T
Where ΔS is the change in entropy, q is the amount of heat and T is the temperature in Kelvin. 
T = 20°C + 273.15 = 293.15 K 
ΔS = 24.46 kJ / 293.15 K = 0.0835 kJ/K

However, the question asks for the change in entropy in kcal/K, so we need to convert the units:
ΔS = 0.0835 kJ/K * (1 kcal/4.184 kJ) = 0.0199 kcal/K ≈ 20 kcal/K

Therefore, the answer is (C) 20.

A Vessel Containing 10 Kg of Water Is Left Out °\degree°

A Vessel Containing 10 Kg of Water Is Left Out $\degree$