Question

A mixture of 1773g of water and 227g of ice is in an initial equilibrium state at ${{0.000}^{0}}C$. The mixture is then, in a reversible process, brought to a second equilibrium state where the water-ice ratio, by mass, is 1.00:1.00 at. (a) Calculate the entropy change of the system in this process (The heat of fusion for water is 333kJ/kg.) (b) The system is then returned to the initial equilibrium state in an irreversible process (say, by using Bunsen burner). Calculate the entropy change of the system during this process. (c) Are your answers consistent with the second law of thermodynamics?

Answer 1

Hint:. The answer to this question is based on the calculation of entropy change according to second law of thermodynamics and this is given by the formula $\Delta S=\dfrac{Q}{T}$ and final mass of water and ice are equal wherein their sum can be taken as same.

Complete step by step answer:
In our previous chapters we have come across the concepts of latent heat of fusion, vaporisation etc., thus, from the data given above we can write as,

(a) Initial mass and temperature of water ${{m}_{{{i}_{1}}}} = 1.773kg$ and ${{T}_{{{i}_{1}}}} = 273K$
Similarly for ice, ${{m}_{{{i}_{2}}}} = 0.227kg$ and ${{T}_{{{i}_{2}}}} = 273K$
Since, the phase changing process is at constant temperature at final temperature, we have ${{m}_{{{f}_{1}}}}={{m}_{{{f}_{2}}}}$ [Since, ratio is 1.00:1.00] …..(1)

Now, ${{m}_{i}}_{_{1}}+{{m}_{{{i}_{2}}}}={{m}_{{{f}_{1}}}}+{{m}_{{{f}_{2}}}}$ and based on condition number (1) we can write this formula as,${{m}_{{{f}_{1}}}}=\dfrac{{{m}_{{{i}_{1}}}}+{{m}_{{{i}_{2}}}}}{2}$
${{m}_{{{f}_{1}}}}=\dfrac{1.773 = 0.227}{2} = 1kg$
Also, total mass in process is constant, thus, $m={{m}_{{{i}_{1}}}}-{{m}_{{{f}_{1}}}}=1.773-1=0.773kg$ ……(2)
Thus, the final mass of ice is $\dfrac{1773+227}{2} = 1000g$ .
This means that 773g of water froze.
let the energy left in the form of heat from system be$m{{L}_{f}}$
By the second law of thermodynamics, we have $\Delta S = \dfrac{Q}{T}$$\Rightarrow \dfrac{-m{{L}_{f}}}{T} = \dfrac{-\left( 0.773 \right)\times \left( 333\times {{10}^{3}} \right)}{273} = -943J/K$
[Since By substituting the values of equation (2) ]

(b) Now, since the ice is being melted, change in the entropy is given as
$\Delta S=\dfrac{Q}{T}=\dfrac{m{{L}_{f}}}{T} = +943J/K$

(c) The answer is Yes. The values obtained are consistent with the second law of thermodynamics as the overall change in the water-ice system is zero even if part of this cycle is irreversible.

Note: Latent heat of fusion is applicable only if any amount of substance melts when it undergoes change in the enthalpy and for unit mass this is called specific heat of fusion. This point may be helpful for solving such types of questions.