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Lesson 3

Calorimetry with Phase Changes

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Hello! Welcome to the third lesson in our calorimetry module.

In our first two lessons, we explored how to calculate the heat required to change a substance's temperature, using for constant specific heat and for variable specific heat. But what happens when adding heat doesn't raise the temperature at all? This brings us to the crucial concept of phase changes.

Today's lesson directly addresses the learning outcome: Analyze calorimetry problems with phase changes including latent heat of fusion and vaporization. We will introduce the concept of latent heat and integrate it with the principle of calorimetry to solve multi-step problems, a common feature in JEE Advanced questions.

1. Latent Heat: The Energy of Phase Transition

When a substance reaches its melting or boiling point, any additional heat energy it absorbs doesn't increase the kinetic energy of its molecules (and thus its temperature). Instead, this energy is used to change the potential energy of the molecules, breaking the bonds that hold them in a solid or liquid state. This energy is called latent heat.

There are two main types of latent heat:

  • Latent Heat of Fusion (): The heat per unit mass required to change a substance from solid to liquid at its melting point.
  • Latent Heat of Vaporization (): The heat per unit mass required to change a substance from liquid to gas at its boiling point.

The heat involved in a phase change for a mass is given by a simple formula:

To get a clear introduction to these concepts and see how to decide which formula to use ( vs. ), let's watch the first few parts of a comprehensive video.

Latent Heat of Fusion and Vaporization, Specific Heat Capacity & Calorimetry - Physics

This video from The Organic Chemistry Tutor clearly defines latent heat and specific heat, explains when to use each corresponding formula, and works through a basic example.

Please watch the video from the beginning up to the 11:28 mark. Focus on: The definitions of latent heat of fusion and vaporization (0:00 - 3:01). The conditions for using Q = mcΔT versus Q = mL (3:01 - 5:20). The worked example of converting ice at 0°C to water at 30°C, which combines both formulas (8:47 - 11:28).

As the video explains, the key is to check if the process involves a temperature change or a phase change. If temperature changes, use . If the substance is at its melting or boiling point and changing phase, use .


2. Visualizing the Process: The Heating Curve

A powerful tool for visualizing and solving problems that involve both temperature and phase changes is the heating curve. This is a graph of temperature versus the amount of heat added to a substance.

Heating Curve for Phase Changes
This heating curve shows the temperature of a substance as heat is added. The sloped sections represent temperature increases within a single phase (solid, liquid, gas), where \(Q=mc\Delta T\) applies. The flat plateaus represent phase transitions (melting, boiling) at constant temperature, where \(Q=mL\) applies.

The video you just started watching has an excellent segment that walks through this curve in detail.

Latent Heat of Fusion and Vaporization, Specific Heat Capacity & Calorimetry - Physics

Let's continue with the same video to understand the heating curve. This section provides a deep conceptual understanding of what's happening at each stage.

Please watch from 11:28 to 20:55. Pay close attention to: How the different segments of the graph correspond to either a temperature change or a phase change. Why the plateau for vaporization is much longer than for fusion (related to the values of L_v and L_f). The explanation of where the energy goes: increasing kinetic energy (temperature rise) vs. increasing potential energy (phase change).

This conceptual distinction is vital. During temperature changes, heat increases the average kinetic energy of the molecules. During phase changes, heat increases the potential energy by overcoming intermolecular forces, with no change in average kinetic energy (and thus no change in temperature).


3. A Comprehensive Multi-Step Problem

Now let's apply this to a full problem that takes a substance through all five possible stages: heating a solid, melting it, heating a liquid, boiling it, and finally heating the gas. This requires systematically calculating the heat for each step and summing them up.

Latent Heat of Fusion and Vaporization, Specific Heat Capacity & Calorimetry - Physics

The final part of the video you've been watching solves a comprehensive problem: heating ice from -30°C to steam at 200°C. This is a perfect example of a multi-step calorimetry problem.

Watch from 20:55 to the end of the video. Observe the five distinct calculations (Q1 to Q5) and how they are summed to find the total energy.

The systematic approach is:

  1. Heat ice to 0°C:
  2. Melt ice at 0°C:
  3. Heat water to 100°C:
  4. Boil water at 100°C:
  5. Heat steam above 100°C:

Total Heat:


4. Strategy for Mixing Problems with Phase Change

This is where the real challenge for JEE Advanced begins. What happens when you mix substances at different temperatures, and a phase change might occur? For example, adding ice to warm water.

The final state is not always obvious. Will all the ice melt? Will all the water freeze? Or will you end up with a mixture of ice and water at 0°C?

Here is a robust, systematic strategy to find out:

  1. Establish a reference point: The most convenient reference point is usually the phase change temperature (0°C for water/ice).
  2. Calculate Heat 'Needed': Calculate the heat required to bring the colder substance to the reference point and then completely through the phase change. For ice at 0°C mixed with warm water, this is the heat needed to melt all the ice: .
  3. Calculate Heat 'Available': Calculate the maximum heat the warmer substance can release to reach the reference point. For the warm water, this is the heat it gives off cooling to 0°C: .
  4. Compare:
    • If : There is more than enough heat to melt all the ice. All the ice melts, and the excess heat () will raise the temperature of the entire mixture (now all water) to a final temperature .
    • If : There is not enough heat to melt all the ice. Only a portion of the ice will melt. The final temperature will be exactly 0°C, with a mix of ice and water.
    • If : There is exactly enough heat to melt all the ice. The final state is all water at exactly 0°C.

Let's see this strategy in action.

Class 11 Physics | Calorimetry | #9 Solved Example-4 on Calorimetry | For JEE & NEET

This Physics Galaxy video solves a classic mixing problem using the exact strategy we just outlined. It's a foundational problem type for JEE.

Watch the entire video. Notice how the presenter first calculates the heat required to melt all the ice (q1) and then calculates the heat released by the water cooling to 0°C (q2). The comparison between q1 and q2 dictates the final state of the system.

This logical, step-by-step comparison is the key to correctly analyzing these problems and avoids confusion. It's a much more reliable approach than trying to guess the final temperature and solve a single large equation.

Test your understanding!

Let's tackle a JEE Advanced problem that involves mixing steam and ice.

Problem: M grams of steam at 100°C is mixed with 200 g of ice at its melting point (0°C) in a thermally insulated container. The final state is liquid water at 40°C. Find the value of M.
(Use: Heat of vaporization of water, cal/g; Heat of fusion of ice, cal/g; Specific heat of water, cal/g°C).

Hint: Set up the energy balance equation: Heat Lost by Steam = Heat Gained by Ice. The steam loses heat in two steps (condensing, then cooling). The ice gains heat in two steps (melting, then warming).

Show answer

Let's break down the heat exchange based on the principle of calorimetry: Heat Lost = Heat Gained.

Heat Lost by Steam (mass M):
The steam starts at 100°C and ends as water at 40°C. This involves two processes:

  1. Condensation: The steam at 100°C turns into water at 100°C.
    cal
  2. Cooling: The resulting water at 100°C cools down to 40°C.
    cal

Total Heat Lost = cal.

Heat Gained by Ice (mass 200 g):
The ice starts at 0°C and ends as water at 40°C. This also involves two processes:

  1. Melting: The ice at 0°C turns into water at 0°C.
    cal
  2. Warming: The resulting water at 0°C warms up to 40°C.
    cal

Total Heat Gained = cal.

Equating Heat Lost and Heat Gained:


So, the value of M is 40 g.

This problem is adapted from Question 3 in the BYJU's resource on JEE Advanced questions.


Conclusion

Today we've added a critical tool to your thermodynamics toolkit: the concept of latent heat. By understanding how energy is partitioned during phase changes, you can now analyze more complex thermal systems.

Key Takeaways:

  • Phase changes (melting, boiling) occur at a constant temperature.
  • The heat required for a phase change is calculated using , where is the specific latent heat.
  • The heating curve is an essential visual aid for breaking down multi-step heating problems.
  • For mixing problems involving potential phase changes, the most reliable strategy is to calculate the heat needed for a full phase change and compare it to the heat available from the other components cooling/heating to the transition temperature.

In our next lesson, we will dive deeper into the consequences of the comparison we learned today. We will focus specifically on tackling problems with partial phase changes, where you must determine not only the final temperature (which will be the melting/freezing point) but also the final masses of each phase in the mixture.