Find the change in the entropy in the following process 100 g of ice at `0^@C` melts when dropped in a bucket of water at `50^@C` (Assume temperature of water does not change)

Options

A. -4.5 cal/K
B. + 4.5 cal/K
C. +5.4 cal/K
D. -5.4 cal/K

Explanation

To find the change in entropy when 100 g of ice at 0°C melts in a bucket of water at 50°C, we can follow these steps:

### Step 1: Calculate the heat absorbed by the ice during melting

The heat absorbed by the ice (Q) when it melts can be calculated using the formula:
\[ Q = m \cdot L_f \]
where:
- \( m \) = mass of the ice = 100 g = 0.1 kg (since we need to convert grams to kilograms for SI units)
- \( L_f \) = latent heat of fusion of ice = 80 cal/g = 80,000 cal/kg

Calculating \( Q \):
\[ Q = 0.1 \, \text{kg} \cdot 80,000 \, \text{cal/kg} = 8000 \, \text{cal} \]

### Step 2: Calculate the change in entropy for the melting ice (S1)

The change in entropy for the melting process can be calculated using the formula:
\[ S_1 = \frac{Q}{T_1} \]
where:
- \( T_1 \) = temperature of the ice during melting = 0°C = 273 K

Calculating \( S_1 \):
\[ S_1 = \frac{8000 \, \text{cal}}{273 \, \text{K}} \]

### Step 3: Calculate the change in entropy for the water (S2)

The water loses heat when the ice melts. The change in entropy for the water can be calculated using the formula:
\[ S_2 = -\frac{Q}{T_2} \]
where:
- \( T_2 \) = temperature of the water = 50°C = 323 K

Calculating \( S_2 \):
\[ S_2 = -\frac{8000 \, \text{cal}}{323 \, \text{K}} \]

### Step 4: Calculate the total change in entropy (ΔS)

The total change in entropy is given by:
\[ \Delta S = S_1 + S_2 \]

Substituting the values we calculated:
\[ \Delta S = \frac{8000}{273} - \frac{8000}{323} \]

### Step 5: Simplify the expression

Factoring out the common term:
\[ \Delta S = 8000 \left( \frac{1}{273} - \frac{1}{323} \right) \]

### Step 6: Calculate the numerical value

Calculating the fractions:
\[ \Delta S = 8000 \left( \frac{323 - 273}{273 \cdot 323} \right) \]
\[ \Delta S = 8000 \left( \frac{50}{273 \cdot 323} \right) \]

Now, calculating the values:
1. \( 273 \cdot 323 = 88059 \)
2. \( \Delta S = 8000 \cdot \frac{50}{88059} \approx 4.5 \, \text{cal/K} \)

### Final Answer

Thus, the change in entropy \( \Delta S \) is approximately:
\[ \Delta S \approx 4.5 \, \text{cal/K} \]