2 kg of ice at -20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible

Q: Why do we first calculate heat up to $0^\circ\text{C}$?

Because $0^\circ\text{C}$ is the phase change temperature. We need to check if the mixture reaches this point before any melting or freezing occurs.

Q: What if the heat released by water was less than the heat required to warm the ice?

Then the water would start freezing, and the final temperature would be below $0^\circ\text{C}$.

Q: Why didn't all the ice melt?

The remaining heat after warming the ice was $80\text{ kcal}$, but melting all $2\text{ kg}$ of ice requires $160\text{ kcal}$. Thus, only a fraction of it could melt.

Q: Is the final temperature exactly $0^\circ\text{C}$?

Yes, because both ice and water coexist in the final mixture, which only happens at the melting point, $0^\circ\text{C}$.

Q: Why is the specific heat of water $1\text{ kcal/kg}^\circ\text{C}$?

It's a standard property of water. $1\text{ kcal}$ is defined as the heat required to raise the temperature of $1\text{ kg}$ of water by $1^\circ\text{C}$.

Question

2 kg of ice at –20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg /°C and 0.5 kcal/kg /°C while the latent heat of fusion of ice is 80 kcal/kg

Options:

(A) 7 kg

(B) 6 kg

(C) 4 kg

(D) 2 kg

Sigma Percentile · Accepted Answer

Welcome to one of the most classic and fascinating scenarios in thermodynamics: the ice-water mixture. Imagine a perfectly insulated container—a battleground where hot water and freezing ice are forced to interact, exchanging heat until they reach a peaceful thermal equilibrium.

I know these calorimetry problems can sometimes feel like a maze of formulas, but let's take a breath and walk through it logically. We don't just plug numbers into equations; we follow the journey of the heat.

### Analyzing the Setup

Let's look at our contenders. On one side, we have $5	ext{ kg}$ of water sitting comfortably at $20^\circ	ext{C}$. On the other side, we have $2	ext{ kg}$ of ice, shivering at $-20^\circ	ext{C}$.

Because the vessel is perfectly insulating, no heat can escape to the surroundings, and no heat can enter. It's a closed system. The fundamental rule here is simple: **Heat lost by the hot body equals heat gained by the cold body**. The water has thermal energy to spare, and the ice is hungry for it.

### The Race to Zero

Before any ice can melt, or any water can freeze, both substances must reach the critical battleground temperature: $0^\circ	ext{C}$, the melting point of ice.

Let's first ask the water: "How much heat can you give up if you cool all the way down to $0^\circ	ext{C}$?"
We use the specific heat formula:
$$Q_1 = m_w c_w \Delta T_w$$

Substituting our values (remembering that the specific heat of water $c_w$ is $1	ext{ kcal/kg}^\circ	ext{C}$):
$$Q_1 = 5 	imes 1 	imes (20 - 0) = 100	ext{ kcal}$$

So, the water can provide a maximum of $100	ext{ kcal}$ of heat before it starts turning into ice itself.

Now, let's ask the ice: "How much heat do you need just to warm up to $0^\circ	ext{C}$?"
Using the same formula, but with the specific heat of ice ($c_i = 0.5	ext{ kcal/kg}^\circ	ext{C}$):
$$Q_2 = m_i c_i \Delta T_i$$
$$Q_2 = 2 	imes 0.5 	imes (0 - (-20)) = 20	ext{ kcal}$$

The ice only needs $20	ext{ kcal}$ to reach $0^\circ	ext{C}$.

### The Melting Phase

This is where it gets interesting. The water has $100	ext{ kcal}$ to give, but the ice only needed $20	ext{ kcal}$ to warm up. We have a surplus of heat!

Let's calculate the remaining heat available:
$$Q_{	ext{rem}} = Q_1 - Q_2 = 100 - 20 = 80	ext{ kcal}$$

This $80	ext{ kcal}$ of heat isn't just going to disappear. It will start attacking the solid structure of the ice, breaking the hydrogen bonds and melting it into water.

But is it enough to melt *all* the ice? Let's check. To melt all $2	ext{ kg}$ of ice, we would need:
$$Q_{	ext{melt\_all}} = m_i L = 2 	imes 80 = 160	ext{ kcal}$$

We only have $80	ext{ kcal}$ available, which is less than $160	ext{ kcal}$. This tells us a crucial fact: **the ice will only partially melt**. Because there is still ice left over, the final temperature of the mixture must be exactly $0^\circ	ext{C}$.

### Final Calculation

Now, let's find out exactly how much ice melts. We use the latent heat formula rearranged for mass:
$$m_{	ext{melt}} = \frac{Q_{	ext{rem}}}{L}$$
$$m_{	ext{melt}} = \frac{80}{80} = 1	ext{ kg}$$

Exactly $1	ext{ kg}$ of ice melts and turns into liquid water at $0^\circ	ext{C}$.

The question asks for the final mass of water remaining in the container. We started with $5	ext{ kg}$ of water, and we just gained another $1	ext{ kg}$ from the melted ice.
$$m_{w,	ext{final}} = 5 + 1 = 6	ext{ kg}$$

And there we have it! The final mixture consists of $6	ext{ kg}$ of water and $1	ext{ kg}$ of ice, all resting peacefully at $0^\circ	ext{C}$. By breaking the problem down into logical phases—warming/cooling, checking the balance, and then phase change—we navigated the thermodynamics flawlessly.

Select Answer:

Visualized Solution

Visualizing the Mixture

Maximum Heat Released by Water

Heat Required to Warm Ice

Remaining Heat for Melting

Mass of Ice Melted

Final Composition

The Way Forward

The Sigma Insight: Calorimetry

Analyzing the Setup

The Race to Zero

The Melting Phase

Final Calculation

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