Question

Calculate ∆S_total for the freezing of 10.0 grams of liquid water at 5.0^oC. What does the sign say about this process?

Answer 1

Given that:-
Temperature =
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (5 + 273.15) K = 278.15 K
Also, freezing point of water = = 273.15 K

The following processs is taking place:-
(a)

The expression for the calculation of the absorbed or released in a process is shown below as:-

Where,
Q is the heat absorbed or released
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of water = 10 g
Specific heat of water = 4.18 J/g°C

So,

Negative sign signifies loss of heat.

(b)

The heat released when liquid water gets converted to ice at
Thus, for 10 g of water, the heat released is:-

Thus,

Also, the expression for the change in the entropy is:-

Where,
is the total entropy change of the system
T is the temperature of the system
change in the heat of the system

So,

The sign comes out to be negative which states that the system is non-spontaneous.