Thermal Interaction Analysis

Hello! In our last lesson, we explored how mechanical work done during a thermodynamic process can be converted into heat within a calorimeter. Today, we'll continue building on the theme of integrated problems by examining a different kind of energy transfer.

This lesson addresses the learning outcome: Analyze problems where heat from a thermodynamic cycle induces temperature or phase changes in another system. Instead of work being the input, we'll focus on the heat exchanged by a system—typically a heat engine or refrigerator—over a full cycle. This heat will then become the energy source or sink for a separate calorimetric process, like melting ice or boiling water. This is a common and practical scenario that frequently appears in JEE Advanced questions.

1. The Core Principle: Coupling Thermodynamic Systems

Imagine a heat engine running. It absorbs heat $Q_{H}$ from a hot source, performs net work $W_{net}$ , and inevitably rejects some waste heat $Q_{L}$ to a cold sink.
From the First Law of Thermodynamics applied to a cycle ( $Δ U = 0$ ), we know:
$W_{net} = Q_{net} = Q_{H} - Q_{L}$
Here, $Q_{H}$ and $Q_{L}$ are the magnitudes of heat absorbed and rejected, respectively.

The problems we're tackling today place a calorimetric system (like a container of ice and water) in the position of the cold sink. The "waste heat" from the engine is no longer wasted; it's the energy input that drives temperature or phase changes in the calorimeter.

The fundamental connection is the principle of energy conservation:
Heat rejected by the engine cycle = Heat absorbed by the calorimetric system
$Q_{L} = Q_{calorimeter}$

Similarly, for a refrigerator (a heat engine running in reverse), it uses external work $W$ to extract heat $Q_{L}$ from a cold body and rejects a larger amount of heat $Q_{H} = Q_{L} + W$ to the surroundings. In this case, the heat extracted from the cold body can be used to freeze water or cool a substance.

Heat extracted by the refrigerator = Heat removed from the calorimetric system
$Q_{L, fridge} = Q_{removed}$

2. A Quick Review of Multi-Step Calorimetry

These problems often involve multiple steps in the calorimetric system, such as heating a substance, causing a phase change, and then heating it further. A powerful tool for visualizing this is the heating curve.

If you need a refresher on the concepts of specific heat and latent heat, or how to calculate the total heat for a process involving both temperature and phase changes, the following video provides a clear and detailed explanation.

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

The Organic Chemistry Tutor provides an excellent walkthrough of calorimetry calculations. The section on the heating curve is particularly useful for visualizing the energy required at each stage.

Watch the segment from 11:28 to 17:05, which discusses the heating curve for water. Focus on: The flat regions represent phase changes (melting, boiling) where temperature is constant (Q=mL). The sloped regions represent temperature changes within a single phase (Q=mc\Delta T). Why the vaporization phase requires significantly more energy than the fusion (melting) phase.

The key takeaway is that you must sum the heat from each individual step to find the total heat required. For example, to turn ice at -10°C to steam at 110°C, you'd calculate five distinct heat quantities.

3. Worked Example 1: A Heat Engine Melting Ice

Let's consider a practical example. A heat engine uses 1 mole of a monoatomic ideal gas ( $γ = 5/3$ ) and operates on the cycle shown below. The heat rejected by the engine during each cycle is used to melt ice at 0°C. We need to find the mass of ice melted per cycle.

Heat Engine Cycle PV Diagram — A PV diagram for a cyclic process. For a heat engine, the cycle is clockwise. The net work done is the area enclosed, and the heat rejected is transferred to a cold reservoir—in our case, the ice.

Let's analyze the cycle in the diagram, which consists of three steps:

A → B: Isochoric (constant volume) heating from $P_{A} = 1$ atm to $P_{B} = 3$ atm at $V_{A} = 24.6$ L.
B → C: Isothermal expansion at temperature $T_{B}$ .
C → A: Isobaric (constant pressure) compression at $P_{A} = 1$ atm.

Our strategy is to find the heat rejected, $Q_{L}$ , and equate it to $m_{i ce} L_{f}$ . Heat is rejected during processes where $Q < 0$ .

Step 1: Find the state variables (P, V, T) at points A, B, and C.

Point A: $P_{A} = 1$ atm, $V_{A} = 24.6$ L. For 1 mole of gas, we can find $T_{A}$ using the ideal gas law $P V = n R T$ . Let's use $R \approx 0.0821 L\cdotatm/(mol\cdotK)$ .
$T_{A} = \frac{P _{A} V _{A}}{n R} = \frac{( 1 atm ) ( 24.6 L )}{( 1 mol ) ( 0.0821 L\cdotatm/(mol\cdotK) )} \approx 300 K$
Point B: $P_{B} = 3$ atm, $V_{B} = V_{A} = 24.6$ L.
$T_{B} = \frac{P _{B} V _{B}}{n R} = \frac{( 3 atm ) ( 24.6 L )}{( 1 mol ) ( 0.0821 L\cdotatm/(mol\cdotK) )} \approx 900 K$
Point C: Isothermal expansion from B, so $T_{C} = T_{B} = 900$ K. It's also on the isobar from A, so $P_{C} = P_{A} = 1$ atm.
$V_{C} = \frac{n R T _{C}}{P _{C}} = \frac{( 1 mol ) ( 0.0821 ) ( 900 K )}{1 atm} = 73.89 L$

Step 2: Calculate heat transfer for each leg.
For a monoatomic gas, $C_{V} = \frac{3}{2} R$ and $C_{P} = \frac{5}{2} R$ . Let's use $R \approx 8.314 J/(mol\cdotK)$ .

A → B (Isochoric):
$Q_{A B} = n C_{V} Δ T = (1) (\frac{3}{2} \times 8.314) (900 - 300) = (12.47) (600) \approx 7482 J (Heat absorbed)$
B → C (Isothermal): Work is done, $Δ U = 0$ , so $Q = W$ .
$Q_{B C} = n R T_{B} ln (\frac{V _{C}}{V _{B}}) = (1) (8.314) (900) ln (\frac{73.89}{24.6}) \approx (7482.6) ln (3) \approx 8220 J (Heat absorbed)$
C → A (Isobaric):
$Q_{C A} = n C_{P} Δ T = (1) (\frac{5}{2} \times 8.314) (300 - 900) = (20.785) (- 600) \approx - 12471 J (Heat rejected)$

Step 3: Identify rejected heat and perform the calorimetric calculation.
The only heat rejected is during the process C → A.
$Q_{L} = ∣ Q_{C A} ∣ = 12471 J$
This heat is used to melt ice. The latent heat of fusion for ice is $L_{f} = 334, 000 J/kg$ .
$Q_{L} = m_{i ce} L_{f}$
$m_{i ce} = \frac{Q _{L}}{L _{f}} = \frac{12471 J}{334000 J/kg} \approx 0.0373 kg = 37.3 grams$
So, about 37.3 grams of ice are melted during each cycle of this engine.

4. Worked Example 2: Refrigerator Freezing Water (JEE 2016 Problem)

Now let's analyze a problem from the perspective of a refrigerator. This problem is from a past JEE Advanced paper and is a perfect example for this topic.

JEE Advanced Thermal Physics Solved Examples

The following text from Vedantu provides a solved example from JEE Advanced 2016. It involves a Carnot freezer (a refrigerator) that uses energy to turn water into ice, directly aligning with our learning goal.

Please read problem 2 under the heading 'Previous Year Questions of Thermal Physics'. The problem starts with 'A Carnot freezer takes heat from water...'. We will walk through the solution together.

Problem Statement: A Carnot freezer takes heat from water at 0°C inside it and rejects it to the room at a temperature of 27°C. The latent heat of ice is 336×10³ J/kg. If 5 kg of water at 0°C is converted into ice at 0°C by the freezer, then the energy consumed by the freezer is close to...?

Solution Breakdown:

Identify the Goal and the Calorimetric Heat:
The goal is to freeze 5 kg of water at 0°C. This is a phase change at constant temperature. The heat that must be extracted from the water is $Q_{L}$ .
$Q_{L} = m L_{f} = (5 kg) \times (336 \times 1 0^{3} J/kg) = 1, 680, 000 J = 1.68 \times 1 0^{6} J$
This $Q_{L}$ is the heat taken from the cold reservoir (the inside of the freezer).
Identify the Thermodynamic System and Temperatures:
The system is a Carnot freezer, which is a reversible refrigerator.
- The cold reservoir temperature is $T_{L} = 0^{\circ} C = 273 K$ .
- The hot reservoir temperature (the room) is $T_{H} = 2 7^{\circ} C = 300 K$ .
Apply the Relevant Thermodynamic Relation:
The "energy consumed" by the freezer is the work, $W$ , required to run it. For a refrigerator, the key performance metric is the Coefficient of Performance (COP), $β$ . For a reversible (Carnot) refrigerator:
$β = \frac{Heat extracted}{Work done} = \frac{Q _{L}}{W}$
And for a Carnot cycle, the COP is determined solely by the reservoir temperatures:
$β_{Carnot} = \frac{T _{L}}{T _{H} - T _{L}}$
Solve for the Work Done (Energy Consumed):
We can equate the two expressions for $β$ :
$\frac{Q _{L}}{W} = \frac{T _{L}}{T _{H} - T _{L}}$
Rearranging to solve for $W$ :
$W = Q_{L} (\frac{T _{H} - T _{L}}{T _{L}})$
Plugging in the values:
$W = (1.68 \times 1 0^{6} J) (\frac{300 K - 273 K}{273 K}) = (1.68 \times 1 0^{6}) (\frac{27}{273})$
$W \approx (1.68 \times 1 0^{6}) \times (0.0989) \approx 166, 153 J \approx 1.66 \times 1 0^{5} J$
This matches the logic in the provided resource and corresponds to one of the multiple-choice options.

This problem beautifully integrates the concept of latent heat with the thermodynamics of a refrigeration cycle.

Test your understanding!

Consider the Carnot freezer problem again. How much total heat is rejected to the room at 27°C?

Show answer

According to the First Law for a refrigerator, the heat rejected to the hot reservoir ( $Q_{H}$ ) is the sum of the heat extracted from the cold reservoir ( $Q_{L}$ ) and the work done ( $W$ ).
$Q_{H} = Q_{L} + W$
$Q_{H} = (1.68 \times 1 0^{6} J) + (1.66 \times 1 0^{5} J) \approx 1.846 \times 1 0^{6} J$
So, about 1.85 million Joules of heat are exhausted into the room.

5. Advanced Problem-Solving Technique: Multi-Stage Calorimetry

JEE Advanced problems often involve multiple stages within the calorimetric part itself, requiring careful bookkeeping. The following video solves a classic problem of this type.

JEE Advanced Physics 2019 Paper 1 #17 (#7) Calorimetry

This video from Michel van Biezen tackles a JEE Advanced problem that, while not involving a cycle, is an excellent example of the complex, multi-stage calorimetry reasoning you need.

Watch the full walkthrough of the problem (0:00 - 5:53). The problem is broken into two parts: Part 1: 5 grams of liquid are heated and then fully vaporized. Heat lost by the calorimeter = Heat gained by the liquid (temperature rise + phase change). Part 2: 80 more grams of liquid are added, and the system reaches a final equilibrium temperature. Notice how the two separate energy balance equations are set up and then solved simultaneously to find the unknown ratio. This 'two-part' problem structure is very common in advanced physics.

This problem highlights that sometimes you don't need to find absolute heat values, but can use the principle of energy conservation across different stages of an experiment to find ratios of physical quantities, a common shortcut in competitive exams.

Conclusion

Today, we've connected the macroscopic world of heat engines and refrigerators to the microscopic processes of temperature and phase change. This type of problem requires you to think in terms of coupled systems, where the output of one system becomes the input for another.

Key Takeaways:

Energy Conservation is the Bridge: The heat rejected by a heat engine ( $Q_{L}$ ) or extracted by a refrigerator ( $Q_{L}$ ) is the quantity that connects the thermodynamic cycle to the calorimetry calculation.
Problem-Solving Strategy:
1. Analyze the thermodynamic cycle to find the required heat transfer ( $Q_{H}$ , $Q_{L}$ , or $Q_{n e t}$ ). This may involve calculating work, internal energy, and heat for each leg of the cycle.
2. Set up the energy balance for the calorimetric system, accounting for all temperature and phase changes ( $Q = m c Δ T$ and $Q = m L$ ).
3. Equate the heat transferred from the cycle to the heat absorbed/released by the calorimeter and solve for the unknown.
Refrigerators vs. Engines: Remember to distinguish between the two. Engines produce work and reject heat. Refrigerators consume work to extract heat.

Preview of the Next Lesson:
We will continue our exploration of integrated problems by adding another layer of complexity. In the next lesson, we will analyze problems involving piston-cylinder systems under external mechanical forces like springs or friction. This will combine thermodynamics with mechanics, forcing us to account for potential energy stored in springs and energy dissipated by friction.