M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated c

Q: Why do we use $s_w = 1$ for the specific heat of water?

In the CGS system of units, the specific heat capacity of liquid water is exactly $1 \text{ cal/g}^{\circ}\text{C}$. Since the latent heats are given in $\text{cal/g}$, we must use CGS units throughout the calculation to maintain consistency.

Q: Why is the temperature difference for steam written as $(100 - 40)$ instead of $(40 - 100)$?

When setting up the equation $\text{Heat Lost} = \text{Heat Gained}$, we want both sides to be positive quantities. Therefore, we always subtract the lower temperature from the higher temperature: $\Delta T = T_{\text{hot}} - T_{\text{cold}}$.

Q: How do we know that all the steam condenses and all the ice melts?

The problem explicitly states that the final mixture is 'liquid water at $40^{\circ}\text{C}$'. Since $40^{\circ}\text{C}$ is strictly between the melting point ($0^{\circ}\text{C}$) and the boiling point ($100^{\circ}\text{C}$), we can be absolutely certain that no ice or steam remains in the final state.

Q: What if the final temperature wasn't given in the problem?

If the final temperature is unknown, the problem becomes a bit more complex. You would have to make an assumption (e.g., assume all ice melts) and calculate the final temperature. If your calculated temperature falls outside the $0^{\circ}\text{C}$ to $100^{\circ}\text{C}$ range, your assumption was wrong, and you'd need to recalculate assuming a phase mixture (like ice and water at $0^{\circ}\text{C}$).

Q: Why don't we include the heat capacity of the container in our equation?

The problem specifies that the container is 'thermally insulated', which means no heat is lost to the surroundings. Furthermore, since the water equivalent or heat capacity of the container itself is not provided, we assume it absorbs a negligible amount of heat.

Question

$M$ grams of steam at $100^{\circ}	ext{C}$ is mixed with $200	ext{ g}$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ}	ext{C}$ [heat of vaporisation of water is $540	ext{ cal/g}$ and heat of fusion of ice is $80	ext{ cal/g}$], the value of $M$ is ... .

Sigma Percentile · Accepted Answer

Welcome to one of the most fascinating topics in thermal physics: **Calorimetry**! This is the science of measuring heat transfer, and it governs everything from how your morning coffee cools down to how massive industrial steam engines operate.

Imagine you are standing in a laboratory. In front of you is a perfectly insulated container. Inside this container, a dramatic thermal battle is about to take place. On one side, we have $M$ grams of furious, boiling steam at $100^{\circ}	ext{C}$. On the other side, we have $200	ext{ g}$ of freezing ice at exactly $0^{\circ}	ext{C}$.

When these two extremes meet, they exchange energy until they reach a peaceful equilibrium: liquid water at a comfortable $40^{\circ}	ext{C}$. Our mission is to find out exactly how much steam ($M$) was required to achieve this perfect balance.

### The Principle of Calorimetry

The fundamental law governing this process is the conservation of energy. Because our container is thermally insulated, no heat can escape into the room, and no heat can enter from the outside.

Therefore, the total heat energy lost by the hot substance (the steam) must be exactly equal to the total heat energy gained by the cold substance (the ice).

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

Let's break down this thermal journey into two distinct paths.

### Tracking the Heat Lost (The Steam's Journey)

The steam doesn't just magically become $40^{\circ}	ext{C}$ water. It has to go through two distinct phases of heat loss.

**Phase 1: Condensation**
First, the steam at $100^{\circ}	ext{C}$ must condense into liquid water at $100^{\circ}	ext{C}$. This phase change releases a massive amount of energy known as the latent heat of vaporization ($L_v$).
$Q_1 = M \cdot L_v$

**Phase 2: Cooling**
Now that we have boiling water at $100^{\circ}	ext{C}$, it must cool down to the final temperature of $40^{\circ}	ext{C}$. This temperature change releases sensible heat, governed by the specific heat capacity of water ($s_w$).
$Q_2 = M \cdot s_w \cdot (100 - 40)$

The total heat lost by the steam is the sum of these two quantities:
$Q_{	ext{lost}} = M \cdot L_v + M \cdot s_w \cdot 60$

### Tracking the Heat Gained (The Ice's Journey)

Similarly, the ice undergoes its own two-step transformation.

**Phase 1: Melting**
The solid ice at $0^{\circ}	ext{C}$ must first melt into liquid water at $0^{\circ}	ext{C}$. This absorbs the latent heat of fusion ($L_f$).
$Q_3 = m_{	ext{ice}} \cdot L_f$

**Phase 2: Warming**
This freshly melted, ice-cold water at $0^{\circ}	ext{C}$ must now warm up to the final temperature of $40^{\circ}	ext{C}$. This absorbs sensible heat.
$Q_4 = m_{	ext{ice}} \cdot s_w \cdot (40 - 0)$

The total heat gained by the ice is:
$Q_{	ext{gained}} = m_{	ext{ice}} \cdot L_f + m_{	ext{ice}} \cdot s_w \cdot 40$

### The Master Equation

Now, we bring it all together by equating the heat lost to the heat gained:

$$M \cdot L_v + M \cdot s_w \cdot 60 = m_{	ext{ice}} \cdot L_f + m_{	ext{ice}} \cdot s_w \cdot 40$$

We are given the following constants in CGS units:
- Latent heat of vaporization, $L_v = 540 	ext{ cal/g}$
- Latent heat of fusion, $L_f = 80 	ext{ cal/g}$
- Specific heat of water, $s_w = 1 	ext{ cal/g}^{\circ}	ext{C}$
- Mass of ice, $m_{	ext{ice}} = 200 	ext{ g}$

Let's substitute these values into our master equation:

$$M(540) + M(1)(60) = 200(80) + 200(1)(40)$$

### The Final Calculation

Now, it's just a matter of simple algebra. Let's simplify both sides of the equation.

On the left side, we factor out $M$:
$$540M + 60M = 600M$$

On the right side, we calculate the total heat absorbed by the ice:
$$16000 + 8000 = 24000$$

Equating them gives us a beautifully clean linear equation:
$$600M = 24000$$

Dividing both sides by 600:
$$M = \frac{24000}{600} = 40$$

And there we have it! Exactly **$40	ext{ g}$** of steam was required to melt the ice and bring the entire mixture to a perfectly balanced $40^{\circ}	ext{C}$.

Calorimetry problems might look intimidating at first glance, but by systematically breaking down the journey of each substance into phase changes and temperature changes, the underlying math reveals itself to be incredibly elegant and logical.

Enter Numerical Value:

Visualized Solution

$Initial and Final States$

$Principle of Calorimetry$

$Heat Lost (Q_{lost})$

$Heat Gained (Q_{gained})$

$Equating Heat$

$Substituting Values$

$Simplifying the Equation$

$Calculating M$

$What if?$

The Sigma Insight: Calorimetry

The Principle of Calorimetry

Tracking the Heat Lost (The Steam's Journey)

Tracking the Heat Gained (The Ice's Journey)

The Master Equation

The Final Calculation

Similar Questions

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

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 ......... $^{\circ} C$ .

Enter Numerical Value:

Visualized Solution

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

.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 Lost (Qlost​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Gained (Qgained​)

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

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

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

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

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