A block of ice at -10°C is slowly heated and converted to steam at 100°C. Which of the following cur

Q: Why does the temperature not increase during melting or boiling?

During a phase change, all the supplied heat energy is used to break the intermolecular bonds holding the substance together, rather than increasing the kinetic energy of the molecules. Hence, the temperature remains constant.

Q: Why are the slopes of the heating ice and heating water regions different?

The slope of the Temperature vs Heat graph is inversely proportional to the specific heat capacity ($1/ms$). Since water has a higher specific heat capacity than ice, its slope will be less steep.

Q: Does the heat supplied ever decrease in this process?

No, the problem states that the block is 'slowly heated', meaning heat is continuously and steadily supplied to the system. The x-axis value always increases.

Q: What if the heating rate wasn't constant?

If we were plotting Temperature vs Time, a varying heating rate would curve the sloped lines. However, since we are plotting Temperature vs Heat Supplied ($Q$), the relationship $Q = ms\Delta T$ remains linear regardless of the rate.

Q: Why does the graph start below the origin?

The initial temperature of the ice is $-10^\circ\text{C}$. Since the y-axis represents temperature in Celsius, the starting point must be on the negative y-axis.

Question

A block of ice at $-10^\circ	ext{C}$ is slowly heated and converted to steam at $100^\circ	ext{C}$. Which of the following curves represents the phenomena qualitatively?

Sigma Percentile · Accepted Answer

The journey of a block of ice transforming into steam is one of the most beautiful and fundamental processes in thermodynamics. It is a story of energy, temperature, and the invisible bonds that hold molecules together. Let's dive deep into this phenomenon and understand exactly what happens when we continuously supply heat to ice at $-10^\circ	ext{C}$.

### Phase 1: Waking Up the Ice

Imagine a solid block of ice sitting at $-10^\circ	ext{C}$. In this state, the water molecules are locked in a rigid, crystalline lattice. They are vibrating, but they don't have enough energy to break free from their fixed positions.

When we start supplying heat, this thermal energy is absorbed by the ice molecules, causing them to vibrate more vigorously. This increase in average kinetic energy manifests as a rise in temperature.

The relationship between the heat supplied $Q$ and the change in temperature $\Delta T$ is given by the specific heat equation:
$$Q_1 = ms_i \Delta T$$
where $m$ is the mass of the ice and $s_i$ is the specific heat capacity of ice. Because the temperature $T$ increases linearly with the heat supplied $Q$, the graph of Temperature versus Heat starts as a straight line with a positive slope, rising from $-10^\circ	ext{C}$ towards $0^\circ	ext{C}$.

### Phase 2: The Melting Point

As the temperature reaches $0^\circ	ext{C}$, something fascinating happens. The ice doesn't just keep getting hotter. Instead, the temperature stops rising, even though we are still pumping heat into the system!

Why does this happen? At $0^\circ	ext{C}$, the thermal energy is no longer used to increase the kinetic energy of the molecules. Instead, it is used to break the rigid hydrogen bonds holding the crystalline lattice together. This energy is known as the **Latent Heat of Fusion**.

The heat supplied during this phase change is given by:
$$Q_2 = mL_f$$
where $L_f$ is the latent heat of fusion. Because the temperature remains perfectly constant at $0^\circ	ext{C}$ while heat is continuously added, the graph becomes a perfectly horizontal line. This flat plateau represents the coexistence of solid ice and liquid water.

### Phase 3: Heating the Liquid

Once every last crystal of ice has melted, we are left with pure liquid water at $0^\circ	ext{C}$. Now, the heat we supply once again goes into increasing the kinetic energy of the molecules. The water molecules move faster and faster, and the temperature begins to rise again.

The heat equation for this phase is:
$$Q_3 = ms_w \Delta T$$
where $s_w$ is the specific heat capacity of liquid water. Interestingly, the specific heat of water is about twice that of ice. This means it takes more heat to raise the temperature of water by one degree than it does for ice. Consequently, the slope of this line will be slightly less steep than the initial ice-heating line, but it is still a straight line rising steadily from $0^\circ	ext{C}$ to $100^\circ	ext{C}$.

### Phase 4: The Boiling Point

Finally, the water reaches $100^\circ	ext{C}$, its boiling point. Just like during melting, the temperature plateaus once again. The heat energy is now being used to completely overcome the intermolecular forces, tearing the liquid molecules apart and sending them flying into the air as a gas (steam).

This energy is the **Latent Heat of Vaporization**, and the heat supplied is:
$$Q_4 = mL_v$$
Because the latent heat of vaporization for water is massive (much larger than the latent heat of fusion), this horizontal plateau on the graph will be significantly longer than the melting plateau. The temperature remains locked at $100^\circ	ext{C}$ until every drop of water has turned into steam.

### The Final Picture

When we piece this all together, the Temperature versus Heat graph tells a clear story:
1. A sloped line from $-10^\circ	ext{C}$ to $0^\circ	ext{C}$.
2. A flat horizontal line at $0^\circ	ext{C}$.
3. A sloped line from $0^\circ	ext{C}$ to $100^\circ	ext{C}$.
4. A flat horizontal line at $100^\circ	ext{C}$.

This sequence of sloped and flat regions perfectly matches the curve shown in the first option. It is a classic representation of sensible heat (temperature change) alternating with latent heat (phase change).

Animated Solution for Physics - Properties of Solids and Liquids: A block of ice at $- 1 0^{\circ} C$ is slowly heated and converted to steam at $10 0^{\circ} C$ . Which of the following curves represents the phenomena qualitatively?

Select Answer:

Visualized Solution

$T vs Q Graph$

$Heating Ice$

$Melting of Ice$

$Heating Water$

$Boiling of Water$

$Conclusion$

The Sigma Insight: Calorimetry

Phase 1

Phase 2

Phase 3

Phase 4

The Final Picture

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