A thermally insulated vessel contains 150 g of water at 0°C. Then, the air from the vessel is pumped

Q: Why does the water freeze when we pump the air out?

Pumping air out lowers the pressure, causing water to evaporate. Evaporation requires heat (latent heat of vaporization). Since the vessel is insulated, this heat is taken from the remaining water, lowering its energy and causing it to freeze.

Q: Why don't we consider the specific heat capacity ($mc\Delta T$) in this problem?

Because the water is already at $0^\circ\text{C}$, and both freezing and evaporation in this scenario occur at $0^\circ\text{C}$. There is no change in temperature, only phase changes. Hence, we only use latent heats.

Q: Can we use the masses in grams even though the latent heats are given per kg?

Yes! As long as you use the same mass unit on both sides of the equation, the conversion factor (like $10^{-3}$ for kg) will cancel out perfectly.

Q: Why is the mass of ice $(150 - x)$?

The total mass of water is $150\text{ g}$. If $x$ grams evaporate into vapor, the remaining mass that stays in the liquid/solid phase is exactly $150 - x$ grams.

Q: What does 'adiabatically' mean in this context?

Adiabatically means there is no heat exchange with the surroundings. This guarantees that the heat required for evaporation comes entirely and exclusively from the freezing water.

Question

A thermally insulated vessel contains $150	ext{ g}$ of water at $0^\circ	ext{C}$. Then, the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at $0^\circ	ext{C}$ itself. The mass of evaporated water will be closest to (Latent heat of vaporisation of water $= 2.10 	imes 10^6	ext{ J kg}^{-1}$ and latent heat of fusion of water $= 3.36 	imes 10^5	ext{ J kg}^{-1}$)

Options:

(A) $150	ext{ g}$

(B) $20	ext{ g}$

(C) $130	ext{ g}$

(D) $35	ext{ g}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing in front of a perfectly thermally insulated vessel. Inside this vessel, there is exactly $150	ext{ g}$ of water, chilling at precisely $0^\circ	ext{C}$. Now, we introduce a twist: we start pumping the air out of the vessel adiabatically. What happens next is a beautiful dance of thermodynamics!

As the pressure inside the vessel drops due to the vacuum pump, the water begins to evaporate. But here is the catch—evaporation is an endothermic process; it requires heat! Where does this heat come from? Since the vessel is thermally insulated, no heat can enter from the outside world. The water has no choice but to sacrifice its own internal energy.

### The Master Equation
As the evaporating water steals heat, the remaining water loses heat. Because the water is already at $0^\circ	ext{C}$, any loss of heat will force it to undergo a phase change and freeze into ice. This is the principle of calorimetry in action: the heat lost by the freezing water must perfectly balance the heat gained by the evaporating water.

Let's translate this physical reality into mathematics. Suppose $x$ grams of water evaporates. This means the remaining $(150 - x)$ grams of water must freeze.

The heat gained by the evaporating water is given by $m_{	ext{vapor}} L_v$, where $L_v$ is the latent heat of vaporization. The heat lost by the freezing water is $m_{	ext{ice}} L_f$, where $L_f$ is the latent heat of fusion. Equating the two, we get our master equation:
$$m_{	ext{ice}} L_f = m_{	ext{vapor}} L_v$$

### Final Calculation
Now, let's substitute the given values into our equation. We know $L_f = 3.36 	imes 10^5	ext{ J kg}^{-1}$ and $L_v = 2.10 	imes 10^6	ext{ J kg}^{-1}$.
$$(150 - x) 	imes 3.36 	imes 10^5 = x 	imes 2.10 	imes 10^6$$

Don't let the large numbers intimidate you. Notice the powers of ten! We can easily cancel $10^5$ from both sides, leaving a factor of $10$ on the right side.
$$(150 - x) 	imes 3.36 = x 	imes 21$$

Let's divide both sides by $3.36$ to isolate the terms.
$$150 - x = \frac{21}{3.36} x$$

Calculating the fraction, we find that $21 / 3.36$ is exactly $6.25$.
$$150 - x = 6.25 x$$

Moving $x$ to the right side, we get:
$$7.25 x = 150$$

Finally, solving for $x$:
$$x = \frac{150}{7.25} \approx 20.69	ext{ g}$$

Looking at our options, the closest value to $20.69	ext{ g}$ is $20	ext{ g}$. And just like that, by trusting the conservation of energy, we've cracked the problem!

Select Answer:

Visualized Solution

$m_{total} = 150 g$

\text{Adiabatic Expansion}

\text{Principle of Calorimetry}

\text{Master Equation}

\text{Substitution}

\text{Simplification}

\text{Algebraic Manipulation}

\text{Final Calculation}

The Sigma Insight: Calorimetry

Analyzing the Setup Imagine you are standing in front of a perfectly thermally insulated vessel

The Master Equation As the evaporating water steals heat, the remaining water loses heat

Final Calculation Now, let's substitute the given values into our equation

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