Question

entropy change of 2.00-kg ice block melting at standard temperature and pressure

Answer 1

To calculate the entropy change (\( \Delta S \)) for the melting of a 2.00 kg ice block at standard temperature and pressure, we can use the formula:

\[
\Delta S = \frac{Q}{T}
\]

where:
- \( Q \) is the heat absorbed during the phase change (melting in this case),
- \( T \) is the absolute temperature in Kelvin.

1. **Determine the heat absorbed (Q) during the melting:**

The heat required to melt ice can be calculated using the formula:

\[
Q = m \cdot L_f
\]

where:
- \( m \) is the mass of the ice (2.00 kg),
- \( L_f \) is the latent heat of fusion for ice, which is approximately 334,000 J/kg.

So,

\[
Q = 2.00 \, \text{kg} \cdot 334,000 \, \text{J/kg} = 668,000 \, \text{J}
\]

2. **Determine the temperature (T):**

At standard temperature and pressure (STP), the melting point of ice is 0 °C, which is equivalent to 273.15 K.

3. **Calculate the entropy change (\( \Delta S \)):**

Now substitute the values into the entropy change formula:

\[
\Delta S = \frac{Q}{T} = \frac{668,000 \, \text{J}}{273.15 \, \text{K}} \approx 2441.9 \, \text{J/K}
\]

So, the entropy change for the melting of a 2.00 kg block of ice at standard temperature and pressure is approximately **2442 J/K**.