A thermally isolated vessel contains 100 gm water at 0 deg Celsius. Wen air above d water is pumpd out, sum o’ d water freezes n sum evaporates at 0 deg Celsius, calculate d mass o’ d ice formed if no water at 0 deg Celsius= 2.10*10^6 j/kg. latnt heat of fusion of ice= 3.36*10^5 j/kg.

To solve this problem, we need to analyze the energy exchanges occurring in the thermally isolated vessel containing water at 0 degrees Celsius. When the air above the water is pumped out, some of the water will freeze, while some will evaporate. The key here is to balance the energy lost by the water that freezes with the energy gained by the water that evaporates.

Understanding the Energy Changes

First, let's define the relevant parameters:

• Mass of water (m): 100 grams = 0.1 kg
• Specific latent heat of vaporization (L_v): 2.10 × 10⁶ J/kg
• Latent heat of fusion (L_f): 3.36 × 10⁵ J/kg

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. We can set up an equation based on the principle of conservation of energy.

Setting Up the Equation

Let’s denote:

• m_f: mass of water that freezes (in kg)
• m_e: mass of water that evaporates (in kg)

The energy released by the freezing water can be expressed as:

E_f = m_f × L_f

The energy absorbed by the evaporating water can be expressed as:

E_e = m_e × L_v

Since the system is thermally isolated, the energy lost by the freezing water must equal the energy gained by the evaporating water:

m_f × L_f = m_e × L_v

Relating Masses of Ice and Vapor

We also know that the total mass of water remains constant, so:

m_f + m_e = 0.1 kg

From this equation, we can express m_e in terms of m_f:

m_e = 0.1 kg - m_f

Substituting and Solving

Now, we can substitute m_e into our energy equation:

m_f × L_f = (0.1 kg - m_f) × L_v

Plugging in the values for L_f and L_v:

m_f × 3.36 × 10⁵ = (0.1 - m_f) × 2.10 × 10⁶

Expanding this gives:

3.36 × 10⁵ m_f = 2.10 × 10⁶ × 0.1 - 2.10 × 10⁶ m_f

3.36 × 10⁵ m_f + 2.10 × 10⁶ m_f = 2.10 × 10⁶ × 0.1

(3.36 × 10⁵ + 2.10 × 10⁶) m_f = 2.10 × 10⁵

Now, calculating the left side:

2.10 × 10⁶ + 3.36 × 10⁵ = 2.4336 × 10⁶

Now, we can solve for m_f:

m_f = (2.10 × 10⁵) / (2.4336 × 10⁶)

m_f ≈ 0.0863 kg

Final Calculation

To find the mass of ice formed, we convert this to grams:

m_f ≈ 86.3 grams

Thus, the mass of ice formed when the air above the water is pumped out is approximately 86.3 grams. This demonstrates how energy conservation principles can be applied to phase changes in a thermally isolated system.