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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A mixture of 1773g of water and 227g of ice is in an initial equilibrium state at ${\mathbf{0}}{\mathbf{.}}{\mathbf{000}}{\mathbf{°}}{\mathbf{C}}$. 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 ${\mathbf{0}}{\mathbf{.}}{\mathbf{000}}{\mathbf{°}}{\mathbf{C}}$ . (a)Calculate the entropy change of the system during this process. (The heat of fusion for water is 333 kJ/kg.)(b) The system is then returned to the initial equilibrium state in an irreversible process (say, by using a Bunsen burner). Calculate the entropy change of the system during this process. (c)Are your answers consistent with the second law of thermodynamics?

1. Entropy change during reversible process is $-943\mathrm{J}/\mathrm{K}$.
2. Entropy change during irreversible process is $943\mathrm{J}/\mathrm{K}$ .
3. Yes, they are consistent with the second law of thermodynamics.
See the step by step solution

## Step 1: The given data

1. Mass of water ${m}_{1}=1773\mathrm{g}$
2. Mass of ice ${m}_{2}=227\mathrm{g}$
3. Initial temperature of mixture of water and ice $T=0°\mathrm{C}$ or $273\mathrm{K}$
4. The heat of fusion of water is ${L}_{F}=333\mathrm{kJ}/\mathrm{kg}$

## Step 2: Understanding the concept of entropy

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We use the second law of thermodynamics to calculate the entropy.

Formulae:

The entropy change of the gas using second law of thermodynamics, $∆S=\frac{Q}{T}$ …(i)

The heat released by the body due to latent heat, $Q=m{L}_{F}$ …(ii)

## Step 3: (a) Calculation of the entropy change during reversible process

We have a mixture of 1773g of water and 227g of ice.

The final mass of ice is given as:

$m=\frac{{m}_{1}+{m}_{2}}{2}\phantom{\rule{0ex}{0ex}}=\frac{1773\mathrm{g}+227\mathrm{g}}{2}\phantom{\rule{0ex}{0ex}}=1000\mathrm{g}$

This shows that 773g of water is frozen. The energy in the form of heat left by the system is by an amount $m{L}_{F}$ where, is the mass of water that froze and ${L}_{F}$ is the heat of fusion of water.

Substituting the value of equation (ii) in equation (i), we get the entropy change during the isothermal reversible process is given as:

$∆S=\frac{-m{L}_{F}}{T}$

Minus sign shows heat is left by the system.

Hence, the entropy change during reversible process is $-943\mathrm{J}/\mathrm{K}$

## Step 4: (b) Calculation of entropy change during an irreversible process

If an irreversible process occurs in a closed system, the entropy of the system always increases. So, substituting the value of equation (ii) in equation (i) gives the entropy change during irreversible process is given as:

$∆S=\frac{m{L}_{F}}{T}\phantom{\rule{0ex}{0ex}}=943\mathrm{J}/\mathrm{K}\left(\mathrm{from}\mathrm{part}\left(\mathrm{a}\right)\mathrm{calculations}\right)$

Hence, the entropy change during that irreversible process is $943\mathrm{J}/\mathrm{K}$

## Step 5: (c) Checking whether the above answers is consistent or not.

Yes, they are consistent with the second law of thermodynamics. Over the entire cycle, the change in entropy of the water–ice system is zero even though part of the cycle is irreversible. However, the system is not closed. To consider a closed system, we must include the energy exchanged with ice and water. Suppose it is a constant temperature heat reservoir during the freezing portion of the cycle and a Bunsen burner during the melting portion. During freezing, the entropy of the reservoir increases by 943 J/K. As far as the reservoir–water–ice system is concerned, the process is adiabatic and reversible, so its total entropy does not change. The melting process is irreversible so the total entropy of the burner–water–ice system increases. The entropy of the burner either increases or else decreases by less than 943 J/K.