In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final

Q: Why do we check the heat at $273\text{ K}$?

Because $273\text{ K}$ is the phase transition temperature for water and ice. It acts as a natural checkpoint to balance the heat equations before assuming the final state.

Q: What if the heat available was less than $Q_3$?

If the heat available was less than the heat required to warm the ice to $273\text{ K}$, the final temperature would be below $273\text{ K}$, and the steam would completely freeze into ice.

Q: How much ice actually melts?

The heat available to melt the ice is $32\text{ kcal} - 4.5\text{ kcal} = 27.5\text{ kcal}$. Since the latent heat of fusion is $80\text{ cal/g}$, the mass of ice melted is $27500 / 80 = 343.75\text{ g}$.

Q: Why don't we use Joules instead of calories?

You can use either! However, the latent heats and specific heats are given in both units. Using calories often simplifies the arithmetic since the specific heat of water is exactly $1\text{ cal/g-K}$.

Q: Does the vessel absorb any heat?

The problem states it is an "insulated vessel", which implies no heat is exchanged with the surroundings or the vessel itself.

Question

In an insulated vessel, $0.05	ext{ kg}$ steam at $373	ext{ K}$ and $0.45	ext{ kg}$ of ice at $253	ext{ K}$ are mixed. Find the final temperature of the mixture (in kelvin).
Given, $L_{	ext{fusion}} = 80	ext{ cal/g} = 336	ext{ J/g}$,
$L_{	ext{vaporisation}} = 540	ext{ cal/g} = 2268	ext{ J/g}$,
$S_{	ext{ice}} = 2100	ext{ J/kg K} = 0.5	ext{ cal/g-K}$
and $S_{	ext{water}} = 4200	ext{ J/kg K} = 1	ext{ cal/g-K}$.

Sigma Percentile · Accepted Answer

The problem of mixing steam and ice is a classic battle of heat exchange. On one side, we have scorching hot steam ready to release a massive amount of energy. On the other side, we have freezing ice, hungry to absorb that energy to warm up and melt. Our goal is to find the final equilibrium state of this mixture.

### The Calorimetry Battlefield

According to the principle of calorimetry, in an insulated vessel, the heat lost by the hot body must exactly equal the heat gained by the cold body.

$$ 	ext{Heat Lost} = 	ext{Heat Gained} $$

Because water undergoes phase changes at $273	ext{ K}$ (melting/freezing) and $373	ext{ K}$ (boiling/condensation), we can't just plug numbers into a single equation. Instead, we must evaluate the heat exchange step-by-step, using $273	ext{ K}$ as our primary checkpoint.

### Analyzing the Heat Source: Steam

Let's determine the maximum heat the steam can provide if it cools all the way down to $273	ext{ K}$.

**Step 1: Condensation of Steam**
First, the $50	ext{ g}$ of steam at $373	ext{ K}$ condenses into water at $373	ext{ K}$. The heat released ($Q_1$) depends on the latent heat of vaporization ($L_{	ext{vap}} = 540	ext{ cal/g}$).

$$ Q_1 = m_{	ext{steam}} 	imes L_{	ext{vap}} $$
$$ Q_1 = 50	ext{ g} 	imes 540	ext{ cal/g} = 27000	ext{ cal} = 27	ext{ kcal} $$

**Step 2: Cooling the Hot Water**
Next, this newly formed $50	ext{ g}$ of hot water cools from $373	ext{ K}$ to $273	ext{ K}$. The heat released ($Q_2$) depends on the specific heat of water ($S_{	ext{water}} = 1	ext{ cal/g-K}$).

$$ Q_2 = m_{	ext{water}} 	imes S_{	ext{water}} 	imes \Delta T $$
$$ Q_2 = 50	ext{ g} 	imes 1	ext{ cal/g-K} 	imes (373	ext{ K} - 273	ext{ K}) $$
$$ Q_2 = 50 	imes 100 = 5000	ext{ cal} = 5	ext{ kcal} $$

**Total Heat Available:**
The total heat the steam can provide to reach $273	ext{ K}$ is:
$$ Q_{	ext{available}} = Q_1 + Q_2 = 27	ext{ kcal} + 5	ext{ kcal} = 32	ext{ kcal} $$

### Analyzing the Heat Sink: Ice

Now, let's see how much heat the ice needs to reach $273	ext{ K}$ and melt completely.

**Step 3: Warming the Ice**
The $450	ext{ g}$ of ice must first warm from $253	ext{ K}$ to $273	ext{ K}$. The heat required ($Q_3$) depends on the specific heat of ice ($S_{	ext{ice}} = 0.5	ext{ cal/g-K}$).

$$ Q_3 = m_{	ext{ice}} 	imes S_{	ext{ice}} 	imes \Delta T $$
$$ Q_3 = 450	ext{ g} 	imes 0.5	ext{ cal/g-K} 	imes (273	ext{ K} - 253	ext{ K}) $$
$$ Q_3 = 450 	imes 0.5 	imes 20 = 4500	ext{ cal} = 4.5	ext{ kcal} $$

**Step 4: Melting the Ice**
To completely melt the $450	ext{ g}$ of ice at $273	ext{ K}$ into water, the heat required ($Q_4$) depends on the latent heat of fusion ($L_{	ext{fusion}} = 80	ext{ cal/g}$).

$$ Q_4 = m_{	ext{ice}} 	imes L_{	ext{fusion}} $$
$$ Q_4 = 450	ext{ g} 	imes 80	ext{ cal/g} = 36000	ext{ cal} = 36	ext{ kcal} $$

**Total Heat Required:**
The total heat required to completely melt the ice is:
$$ Q_{	ext{required}} = Q_3 + Q_4 = 4.5	ext{ kcal} + 36	ext{ kcal} = 40.5	ext{ kcal} $$

### The Grand Heat Balance

Now we compare the heat available with the heat required:
- Heat available from steam: **$32	ext{ kcal}$**
- Heat required to warm ice: **$4.5	ext{ kcal}$**
- Heat required to melt all ice: **$40.5	ext{ kcal}$**

Since $32	ext{ kcal} > 4.5	ext{ kcal}$, the steam provides more than enough heat to warm the ice to $273	ext{ K}$. 
However, since $32	ext{ kcal} < 40.5	ext{ kcal}$, the steam does not provide enough heat to melt *all* the ice.

### The Final Verdict

Because the available heat is exhausted while the ice is melting, the phase change is incomplete. The final mixture will consist of both water and unmelted ice coexisting in thermal equilibrium.

Whenever ice and water coexist in equilibrium, the temperature of the mixture is exactly the melting point. Therefore, the final temperature of the mixture is **$273	ext{ K}$**.

Enter Numerical Value:

Visualized Solution

$Calorimetry Setup$

$Principle of Calorimetry$

$Heat Released by Steam$

$Heat Released by Steam$

$Heat Released by Hot Water$

$Heat Released by Hot Water$

$Heat Absorbed by Ice$

$Heat Absorbed by Ice$

$Heat Required to Melt Ice$

$Heat Required to Melt Ice$

$Final State Analysis$

$Food for Thought$

The Sigma Insight: Calorimetry

The Calorimetry Battlefield

Analyzing the Heat Source

Analyzing the Heat Sink

The Grand Heat Balance

The Final Verdict

Similar Questions

$M$ grams of steam at $10 0^{\circ} C$ is mixed with $200 g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $4 0^{\circ} C$ [heat of vaporisation of water is $540 cal/g$ and heat of fusion of ice is $80 cal/g$ ], the value of $M$ is ... .

300 g of water at $2 5^{\circ} C$ is added to 100 g of ice at $0^{\circ} C$ . The final temperature of the mixture is ......... $^{\circ} C$ .

When $M_{1}$ gram of ice at $- 1 0^{\circ} C$ (specific heat $= 0.5 \text{ cal g}^{-1} ^\circ\text{C}^{-1}$ ) is added to $M_{2}$ gram of water at $5 0^{\circ} C$ , finally no ice is left and the water is at $0^{\circ} C$ . The value of latent heat of ice, in $cal g^{- 1}$ is

Steam at $10 0^{\circ} C$ is passed into $1.1 kg$ of water contained in a calorimeter of water equivalent $0.02 kg$ at $1 5^{\circ} C$ till the temperature of the calorimeter and its contents rises to $8 0^{\circ} C$ . The mass of the steam condensed in kg is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calorimetry Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Calorimetry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Released by Steam

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Released by Steam

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Released by Hot Water

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Released by Hot Water

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Absorbed by Ice

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Absorbed by Ice

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Required to Melt Ice

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Required to Melt Ice

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final State Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought

The Sigma Insight: Calorimetry

The Calorimetry Battlefield

Analyzing the Heat Source

Analyzing the Heat Sink

The Grand Heat Balance

The Final Verdict

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 300 g of water at 25∘C is added to 100 g of ice at 0∘C. The final temperature of the mixture is .........∘C.

$Calorimetry Setup$

$Principle of Calorimetry$

$Heat Released by Steam$

$Heat Released by Steam$

$Heat Released by Hot Water$

$Heat Released by Hot Water$

$Heat Absorbed by Ice$

$Heat Absorbed by Ice$

$Heat Required to Melt Ice$

$Heat Required to Melt Ice$

$Final State Analysis$

$Food for Thought$

300 g of water at $2 5^{\circ} C$ is added to 100 g of ice at $0^{\circ} C$ . The final temperature of the mixture is ......... $^{\circ} C$ .