Steam at 100°C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg

Q: Why do we add the water equivalent to the mass of the water?

The water equivalent represents the mass of water that would absorb the same amount of heat as the calorimeter for the same temperature change. Adding it to the actual water mass simplifies the calculation into a single term.

Q: Why does the steam have two terms for heat lost?

Steam undergoes a phase change. It first releases latent heat to turn from gas to liquid at $100^\circ\text{C}$. Then, the resulting liquid water cools from $100^\circ\text{C}$ to the final temperature of $80^\circ\text{C}$, releasing sensible heat.

Q: What is the value of latent heat of vaporization used here?

We use $L_v = 540 \text{ cal/g}$ for steam. It's a standard constant you should memorize for JEE.

Q: Can I solve this using SI units (Joules and kg)?

Yes, absolutely! You would use $c_w = 4200 \text{ J/kg}^\circ\text{C}$ and $L_v = 2.26 \times 10^6 \text{ J/kg}$. However, using CGS units (calories and grams) often makes the arithmetic much simpler in calorimetry problems.

Q: What if the final temperature was $100^\circ\text{C}$?

If the final temperature was $100^\circ\text{C}$, the steam would only condense and not cool down further. The heat lost would just be $m \cdot L_v$.

Question

Steam at $100^{\circ}	ext{C}$ is passed into $1.1	ext{ kg}$ of water contained in a calorimeter of water equivalent $0.02	ext{ kg}$ at $15^{\circ}	ext{C}$ till the temperature of the calorimeter and its contents rises to $80^{\circ}	ext{C}$. The mass of the steam condensed in kg is

Options:

(A) 0.130

(B) 0.065

(C) 0.260

(D) 0.135

Sigma Percentile · Accepted Answer

The problem of mixing steam and water is a classic application of the **Principle of Calorimetry**. It beautifully illustrates how energy is conserved as it flows from a hot body to a cold body, causing temperature changes and phase transitions.

### Analyzing the Setup

Imagine a well-insulated calorimeter containing some water. The water and the calorimeter are initially at a cool $15^\circ	ext{C}$. This is our **cold system**.

We then introduce steam at $100^\circ	ext{C}$ into this setup. The steam is our **hot system**. As the steam enters, it will lose heat, and the water-calorimeter system will gain that exact same amount of heat.

The final equilibrium temperature of the entire mixture is given as $80^\circ	ext{C}$.

### Calculating Heat Gained

Let's first determine how much heat the cold system absorbs. The cold system consists of the water and the calorimeter itself.

To make things simpler, the problem provides the **water equivalent** of the calorimeter, $W = 0.02 	ext{ kg}$. This means the calorimeter absorbs heat exactly like $0.02 	ext{ kg}$ of water would.

We can combine the mass of the water and the water equivalent to get the total effective mass being heated:
$m_{	ext{eff}} = 1.1 	ext{ kg} + 0.02 	ext{ kg} = 1.12 	ext{ kg}$

Converting this to grams (since we will use CGS units for simplicity), we have $1120 	ext{ g}$.

The heat gained by this effective mass of water as it heats from $15^\circ	ext{C}$ to $80^\circ	ext{C}$ is:
$Q_{	ext{gained}} = m_{	ext{eff}} \cdot c_w \cdot \Delta T$

$Q_{	ext{gained}} = 1120 \cdot 1 \cdot (80 - 15)$

$Q_{	ext{gained}} = 1120 \cdot 65 = 72800 	ext{ cal}$

### Calculating Heat Lost

Now, let's trace the journey of the steam. The steam doesn't just cool down; it undergoes a **phase change**.

First, the steam at $100^\circ	ext{C}$ condenses into liquid water at $100^\circ	ext{C}$. This process releases **latent heat**. If the mass of the steam is $m$ (in kg), its mass in grams is $1000m$. The heat released during condensation is:
$Q_1 = 1000m \cdot L_v = 1000m \cdot 540 = 540000m 	ext{ cal}$

Next, this newly formed boiling water at $100^\circ	ext{C}$ must cool down to the final mixture temperature of $80^\circ	ext{C}$. This releases **sensible heat**:
$Q_2 = 1000m \cdot c_w \cdot (100 - 80) = 1000m \cdot 1 \cdot 20 = 20000m 	ext{ cal}$

The total heat lost by the steam is the sum of these two quantities:
$Q_{	ext{lost}} = Q_1 + Q_2 = 540000m + 20000m = 560000m 	ext{ cal}$

### The Final Calculation

According to the **Principle of Calorimetry**, the total heat lost must equal the total heat gained:
$Q_{	ext{lost}} = Q_{	ext{gained}}$

Substituting the values we calculated:
$560000m = 72800$

Solving for $m$:
$m = \frac{72800}{560000} = \frac{728}{5600} = 0.130 	ext{ kg}$

Thus, exactly **$0.130 	ext{ kg}$** of steam must condense to raise the temperature of the system to $80^\circ	ext{C}$.

Select Answer:

Visualized Solution

$Q_{lost} = Q_{gained}$

$Q_{gained}$

$Q_{gained}$

$Q_{gained}$

$Q_{lost}$

$Q_{lost}$

$Q_{lost}$

$Q_{lost} = Q_{gained}$

$m$

$T_{final}$

The Sigma Insight: Calorimetry

Analyzing the Setup

Calculating Heat Gained

Calculating Heat Lost

The Final Calculation

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 ... .

The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. Calculate the mass of the steam required for this purpose.

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

Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qlost​=Qgained​

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

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The Sigma Insight: Calorimetry

Analyzing the Setup

Calculating Heat Gained

Calculating Heat Lost

The Final Calculation

Similar Questions

The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. Calculate the mass of the steam required for this purpose.

$Q_{lost} = Q_{gained}$

$Q_{gained}$

$Q_{gained}$

$Q_{gained}$

$Q_{lost}$

$Q_{lost}$

$Q_{lost}$

$Q_{lost} = Q_{gained}$

$m$

$T_{final}$