Question

Calculate ∆s for the melting of 5g of ice (heat of fusion = 79.7 cal/g) at 0 degrees Celsius: a) ∆s = 398.5 cal b) ∆s = 15.94 cal c) ∆s = 1594 cal d) ∆s = 0.0797 cal

Answer 1

The change in entropy
`(\( \Delta S \))` for the melting of 5 grams of ice at 0 degrees Celsius is approximately 1.46 cal/K.

To calculate the change in entropy
`(\( \Delta S \))` for the melting of ice, we use the formula:

`\[ \Delta S = (q)/(T) \]`

where
`\( q \)` is the heat absorbed during the process (at constant temperature), and
`\( T \)` is the temperature in Kelvin.

Given:

- Heat of fusion of ice
`(\( \Delta H_{\text{fusion}} \))` = 79.7 cal/g

- Mass of ice = 5 g

- Temperature
`(\( T \))` = 0°C, which is 273.15 K (since Kelvin = Celsius + 273.15)

Step 1: Calculate the heat absorbed
`(\( q \))` during the melting of ice.

`\[ q = \text{Mass} * \Delta H_{\text{fusion}} \]`

`\[ q = 5 \text{ g} * 79.7 \text{ cal/g} \]`

Step 2: Convert
`\( q \)` into calories (which it already is).

Step 3: Calculate
`\( \Delta S \)` using the formula.

`\[ \Delta S = (q)/(T) \]`

Let's perform these calculations.

The change in entropy
`(\( \Delta S \))` for the melting of 5 grams of ice at 0 degrees Celsius is approximately 1.46 cal/K. However, this value does not match any of the provided options (a) 398.5 cal, (b) 15.94 cal, (c) 1594 cal, or (d) 0.0797 cal.

The calculated value is correct based on the standard formula for entropy change, and it's expressed in cal/K, which is the standard unit for entropy change.