A thermally isolated vessel contains 100 g of water at 0° C. When air above the water is pumped out, some of the water freezes and some evaporate at 0° C itself. Calculate the mass of the ice formed if no water is left in the vessel. (Latent heat of vaporization of water = 2.10 × 10⁶ J/kg and latent heat of fusion of ice = 3.36 × 10⁶ J/kg).

To solve this problem, we need to understand the energy exchanges that occur when water freezes and evaporates in a thermally isolated system. In this scenario, we have 100 g of water at 0°C, and when the air is pumped out, some of this water will freeze into ice, while some will evaporate. The key is to balance the energy lost during freezing with the energy gained during evaporation.

Understanding the Energy Changes

First, let's convert the mass of water into kilograms since the latent heat values are given in J/kg. We have:

• Mass of water, m = 100 g = 0.1 kg

Next, we need to define the latent heats involved:

• Latent heat of vaporization of water, L_v = 2.10 × 10⁶ J/kg
• Latent heat of fusion of ice, L_f = 3.36 × 10⁶ J/kg

Energy Balance Equation

When some water freezes, it releases energy equal to the latent heat of fusion. Conversely, when some water evaporates, it absorbs energy equal to the latent heat of vaporization. If we let x be the mass of water that freezes into ice, then the mass of water that evaporates will be (0.1 - x) kg. The energy released by freezing must equal the energy absorbed by evaporation:

Energy released by freezing = Energy absorbed by evaporation

Mathematically, this can be expressed as:

x × L_f = (0.1 - x) × L_v

Substituting Values

Now, substituting the known values into the equation:

x × (3.36 × 10⁶) = (0.1 - x) × (2.10 × 10⁶)

Expanding this gives:

3.36 × 10⁶ x = 2.10 × 10⁶ (0.1 - x)

Distributing the right side:

3.36 × 10⁶ x = 2.10 × 10⁶ × 0.1 - 2.10 × 10⁶ x

Which simplifies to:

3.36 × 10⁶ x + 2.10 × 10⁶ x = 2.10 × 10⁵

Combining like terms:

(3.36 + 2.10) × 10⁶ x = 2.10 × 10⁵

5.46 × 10⁶ x = 2.10 × 10⁵

Calculating the Mass of Ice Formed

Now, we can solve for x:

x = (2.10 × 10⁵) / (5.46 × 10⁶)

Calculating this gives:

x ≈ 0.0385 kg

To express this in grams:

x ≈ 38.5 g

Final Result

Thus, the mass of ice formed when the air is pumped out of the vessel is approximately 38.5 grams. This result shows how energy conservation principles apply in phase changes, demonstrating the balance between freezing and evaporation in a closed system.