The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K

Q: Why can't we just use one formula for the whole process?

Entropy change depends on whether the heat added is changing the temperature or changing the phase. The formulas for these two processes are fundamentally different, so we must calculate them separately for each distinct physical stage and then add them up.

Q: Why do we multiply by 2.303 in the temperature change formula?

The thermodynamic formula $\Delta S = C_p \ln(T_2/T_1)$ uses the natural logarithm ($\ln$ or $\log_e$). However, the values given in the question are for base-10 logarithms ($\log_{10}$). The factor $2.303$ is the mathematical conversion factor between them: $\ln(x) \approx 2.303 \log_{10}(x)$.

Q: Why is the entropy change for vaporization ($\Delta S_3$) so much larger than the others?

Entropy is a measure of randomness or disorder. When a liquid turns into a gas, the molecules spread out significantly and move chaotically, leading to a massive increase in disorder compared to just heating a liquid or melting a solid.

Q: What if the specific heat was given in Joules instead of kilo Joules?

You must ensure all units are consistent across all steps. If specific heat is in Joules and enthalpy of fusion is in kilo Joules, you must convert one to match the other before adding the entropy changes together. In this problem, everything is conveniently provided in kilo Joules.

Q: Is entropy a state function or a path function?

Entropy is a state function. This means the total entropy change only depends on the initial state (ice at $273\text{ K}$) and the final state (vapour at $383\text{ K}$). We can calculate it by breaking it down into any reversible path of our choice, which is exactly what we did by splitting it into four logical steps!

Question

The entropy change associated with the conversion of $1	ext{ kg}$ of ice at $273	ext{ K}$ to water vapours at $383	ext{ K}$ is (Specific heat of water liquid and water vapour are $4.2	ext{ kJ K}^{-1}	ext{kg}^{-1}$ and $2.0	ext{ kJ K}^{-1}	ext{kg}^{-1}$; heat of liquid fusion and vapourisation of water are $334	ext{ kJ kg}^{-1}$ and $2491	ext{ kJ kg}^{-1}$ respectively).
($\log 273 = 2.436, \log 373 = 2.572, \log 383 = 2.583$)

Options:

(A) $9.26	ext{ kJ kg}^{-1}	ext{K}^{-1}$

(B) $8.49	ext{ kJ kg}^{-1}	ext{K}^{-1}$

(C) $7.90	ext{ kJ kg}^{-1}	ext{K}^{-1}$

(D) $2.64	ext{ kJ kg}^{-1}	ext{K}^{-1}$

Sigma Percentile · Accepted Answer

### The Thermodynamic Journey: Calculating Total Entropy Change

Imagine you are holding a block of ice at $273	ext{ K}$ ($0^\circ	ext{C}$). Your goal is to transform this solid block into superheated steam at $383	ext{ K}$ ($110^\circ	ext{C}$). This transformation doesn't happen in a single magical leap. It is a fascinating thermodynamic journey consisting of four distinct steps.

Because entropy ($S$) is a **state function**, the total change in entropy for this entire process is simply the sum of the entropy changes of each individual step. Let's break down this journey and calculate the entropy change at each milestone.

### Step 1: Melting the Ice (Phase Change)

First, the ice must melt into liquid water. During this phase change, the temperature remains constant at $273	ext{ K}$. The heat added goes entirely into breaking the solid crystal lattice.

The formula for entropy change during a reversible phase transition at constant temperature is:

$$\Delta S_1 = \frac{\Delta H_{	ext{fusion}}}{T_{	ext{fusion}}}$$

Substituting the given values:

$$\Delta S_1 = \frac{334	ext{ kJ kg}^{-1}}{273	ext{ K}} = 1.22	ext{ kJ kg}^{-1}	ext{K}^{-1}$$

### Step 2: Heating the Water (Temperature Change)

Now we have liquid water at $273	ext{ K}$. We need to heat it up to its boiling point, $373	ext{ K}$. When the temperature of a substance changes without a phase change, the entropy change is calculated using its specific heat capacity ($C_p$):

$$\Delta S_2 = C_p(l) \ln\left(\frac{T_2}{T_1}ight)$$

Since we are given base-10 logarithm values, we must convert the natural logarithm ($\ln$) to base-10 ($\log$) by multiplying by $2.303$:

$$\Delta S_2 = 4.2 	imes 2.303 \log\left(\frac{373}{273}ight)$$

$$\Delta S_2 = 4.2 	imes 2.303 (\log 373 - \log 273)$$

$$\Delta S_2 = 4.2 	imes 2.303 (2.572 - 2.436) = 1.31	ext{ kJ kg}^{-1}	ext{K}^{-1}$$

### Step 3: Boiling the Water (Phase Change)

At $373	ext{ K}$, the water begins to boil and turn into vapour. This is another phase change at a constant temperature. Notice how large the enthalpy of vaporization is compared to fusion; it takes a lot of energy to completely separate liquid molecules into a gas!

$$\Delta S_3 = \frac{\Delta H_{	ext{vap}}}{T_{	ext{vap}}}$$

$$\Delta S_3 = \frac{2491	ext{ kJ kg}^{-1}}{373	ext{ K}} = 6.67	ext{ kJ kg}^{-1}	ext{K}^{-1}$$

**Observe:** This step produces the largest jump in entropy ($6.67$). This makes perfect physical sense because converting a relatively ordered liquid into a highly chaotic gas drastically increases the randomness (entropy) of the system.

### Step 4: Heating the Vapour (Temperature Change)

Finally, we take our water vapour at $373	ext{ K}$ and heat it further to $383	ext{ K}$. We use the temperature change formula again, but this time we must use the specific heat capacity of the *vapour* ($2.0	ext{ kJ K}^{-1}	ext{kg}^{-1}$).

$$\Delta S_4 = C_p(g) \ln\left(\frac{T_2}{T_1}ight)$$

$$\Delta S_4 = 2.0 	imes 2.303 \log\left(\frac{383}{373}ight)$$

$$\Delta S_4 = 2.0 	imes 2.303 (2.583 - 2.572) = 0.05	ext{ kJ kg}^{-1}	ext{K}^{-1}$$

### The Final Calculation

To find the total entropy change for the entire process, we simply sum up the entropy changes from all four steps:

$$\Delta S_{	ext{Total}} = \Delta S_1 + \Delta S_2 + \Delta S_3 + \Delta S_4$$

$$\Delta S_{	ext{Total}} = 1.22 + 1.31 + 6.67 + 0.05 = 9.25 \approx 9.26	ext{ kJ kg}^{-1}	ext{K}^{-1}$$

By systematically breaking down a complex thermodynamic process into its fundamental atomic steps, we can easily navigate through the calculations and arrive at the correct answer!

Select Answer:

Visualized Solution

\text{The Thermodynamic Journey}

\Delta S_1: \text{Melting of Ice}

\Delta S_2: \text{Heating Water}

\Delta S_3: \text{Vaporisation}

\Delta S_4: \text{Heating Vapour}

\Delta S_{\text{Total}}

The Sigma Insight: Entropy and Free Energy

The Thermodynamic Journey

Step 1

Step 2

Step 3

Step 4

The Final Calculation

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