Multi-Substance Phase Change Problems

Hello! Welcome to the fifth lesson in our Calorimetry module.

In the previous lesson, we established a robust, algorithmic approach for problems involving partial phase changes. The key was using the phase change temperature (like 0°C) as a checkpoint to compare the heat available versus the heat needed, allowing us to determine the final state without guesswork.

Today, we're taking on an even more classic and complex JEE Advanced scenario. Our learning goal is to solve problems involving simultaneous phase changes in multiple substances. This typically involves mixing three components, such as ice, water, and steam, where multiple energy transfers and potential phase changes happen concurrently. Mastering this is crucial for tackling the top-tier problems you're aiming for.

1. The Challenge of Multi-Component Mixtures

When you mix ice, water, and steam, the system is more complex. The steam wants to condense and cool, the ice wants to warm and melt, and the water's temperature will adjust based on the balance. The final state could be:

An ice-water mixture at 0°C.
A water-steam mixture at 100°C.
All water at a temperature between 0°C and 100°C.

Guessing the final state and solving is risky; if you guess wrong, your answer will be physically impossible (like a temperature >100°C with ice still present). We need a systematic way to find the correct final state.

2. A Deterministic Strategy: Checkpoint Analysis

Let's extend the checkpoint method from our last lesson. This approach, reminiscent of analyzing states in an algorithm, will give us a definitive answer. We'll use 0°C as our primary reference point.

The Algorithm:

Calculate Total Heat Released to reach 0°C ( $Q_{l os t}$ ):
Calculate the total heat given off by all components initially above 0°C if they were to cool down to 0°C. This includes condensation of steam.
$Q_{l os t} = \sum (Heat from condensation) + \sum (Heat from cooling to 0°C)$
Calculate Total Heat Absorbed to reach 0°C ( $Q_{g ain e d}$ ):
Calculate the total heat absorbed by all components initially below 0°C if they were to warm up to 0°C.
$Q_{g ain e d} = \sum (Heat to warm to 0°C)$
Find the Net Heat at the 0°C Checkpoint:
Let $Q_{n e t} = Q_{l os t} - Q_{g ain e d}$ . This value tells us if there is a surplus or deficit of energy once everything has reached 0°C.
Analyze $Q_{n e t}$ to Determine the Final State:
- Case A: $Q_{n e t} = 0$
  - The system is in perfect equilibrium at 0°C with the original water and ice. The final temperature is 0°C.
- Case B: $Q_{n e t} > 0$
  - There is a surplus of heat. This energy will be used to melt the ice.
  - Calculate the heat needed to melt all ice: $Q_{m e l t_a l l} = m_{i ce} \cdot L_{f}$ .
  - If $Q_{n e t} < Q_{m e l t_a l l}$ : Not enough heat to melt all the ice.
    - Final State: Partial melting. Final temperature is 0°C.
    - Mass of ice melted: $m_{m e l t e d} = Q_{n e t} / L_{f}$ .
  - If $Q_{n e t} \geq Q_{m e l t_a l l}$ : Enough heat to melt all the ice.
    - Final State: All water. The remaining heat, $Q_{r e m} = Q_{n e t} - Q_{m e l t_a l l}$ , raises the temperature of the entire mass of water above 0°C.
    - Final Temperature: Find $T_{f}$ using $Q_{r e m} = m_{t o t a l} \cdot c_{w a t er} \cdot (T_{f} - 0)$ .
- Case C: $Q_{n e t} < 0$
  - There is a heat deficit. This deficit must be covered by freezing some of the water.
  - Final State: Partial freezing. Final temperature is 0°C.
  - Mass of water frozen: $m_{f r oz e n} = ∣ Q_{n e t} ∣/ L_{f}$ .

3. Worked Example: Ice, Water, and Steam

Let's apply this algorithm to a concrete problem.

Problem: 1 kg of ice at -20°C, 2 kg of water at 50°C, and 0.5 kg of steam at 100°C are mixed in an insulated container. Find the final state.

$c_{i ce} = 2100$ J/kg°C
$c_{w a t er} = 4200$ J/kg°C
$L_{f} = 3.34 \times 1 0^{5}$ J/kg
$L_{v} = 2.26 \times 1 0^{6}$ J/kg

Step 1: Calculate $Q_{l os t}$ to reach 0°C.
Heat is lost by the steam and the warm water.

Steam condenses: $Q_{1} = m_{s} L_{v} = 0.5 kg \times 2.26 \times 1 0^{6} J/kg = 1, 130, 000 J$
Condensed water (from steam) cools from 100°C to 0°C: $Q_{2} = m_{s} c_{w} Δ T = 0.5 kg \times 4200 \times (100 - 0) = 210, 000 J$
Initial water cools from 50°C to 0°C: $Q_{3} = m_{w} c_{w} Δ T = 2 kg \times 4200 \times (50 - 0) = 420, 000 J$
Total $Q_{l os t}$ : $1, 130, 000 + 210, 000 + 420, 000 = 1, 760, 000 J$

Step 2: Calculate $Q_{g ain e d}$ to reach 0°C.
Heat is gained by the ice warming up.

Ice warms from -20°C to 0°C: $Q_{4} = m_{i} c_{i} Δ T = 1 kg \times 2100 \times (0 - (- 20)) = 42, 000 J$
Total $Q_{g ain e d}$ : $42, 000 J$

Step 3: Find Net Heat.

$Q_{n e t} = Q_{l os t} - Q_{g ain e d} = 1, 760, 000 - 42, 000 = 1, 718, 000 J$

Step 4: Analyze the Result.

We are in Case B since $Q_{n e t} > 0$ . There is a surplus of heat.
Now, we check if this surplus can melt all the ice.
Heat to melt all 1 kg of ice: $Q_{m e l t_a l l} = m_{i} L_{f} = 1 kg \times 3.34 \times 1 0^{5} J/kg = 334, 000 J$ .
Compare: $Q_{n e t} (1, 718, 000 J) > Q_{m e l t_a l l} (334, 000 J)$ .
Conclusion: Yes, all the ice melts, and there is still heat left over. The final state is all water at some $T_{f} > 0^{\circ} C$ .

Step 5: Calculate Final Temperature.

Remaining heat: $Q_{r e m} = Q_{n e t} - Q_{m e l t_a l l} = 1, 718, 000 - 334, 000 = 1, 384, 000 J$ .
Total mass of water: $m_{t o t a l} = m_{i} + m_{w} + m_{s} = 1 + 2 + 0.5 = 3.5 kg$ .
Use $Q_{r e m} = m_{t o t a l} c_{w} (T_{f} - 0)$ to find $T_{f}$ .
$1, 384, 000 = 3.5 \times 4200 \times (T_{f} - 0)$
$T_{f} = \frac{1 , 384 , 000}{3.5 \times 4200} \approx 94.1 5^{\circ} C$
(Note: If this calculation gave $T_{f} > 10 0^{\circ} C$ , it would mean our model is incomplete; the final state would actually be a water-steam mix at 100°C. Our method naturally catches these edge cases.)

4. An Alternative Strategy: Assume and Verify

The "checkpoint" method is deterministic. Another common and powerful method is to make an educated guess about the final state and then solve. If the result is physically plausible, the assumption was correct.

Physics 23 Calorimetry (5 of 5) Unknown Final State

This video from Michel van Biezen tackles a problem with three components: ice, water, and steam. The presenter uses the strategy of 'assuming' a final state (an ice-water mixture at 0°C) and then solving to verify if the assumption holds. This is a common and effective technique.

Watch from the beginning to 07:49. Pay attention to: The initial assumption about the final state (0:40). How the Q_gained = Q_lost equation is constructed based on this assumption, including a term for the fraction of ice melted (1:21). How the final calculated fraction (0.285) confirms the initial assumption because it's a value between 0 and 1 (6:40).

As you saw, the presenter assumed the final state was a mix of ice and water at 0°C. By solving for the fraction of melted ice, f, he got a value of 0.285. Since 0 < f < 1, this is a physically valid result, confirming the initial assumption. If he had calculated f < 0 or f > 1, he would know his initial assumption was wrong and would need to try a new one (e.g., all water at $T_{f} > 0^{\circ} C$ ).

Both the "checkpoint" and "assume-and-verify" methods are excellent tools for your problem-solving arsenal.

5. Practice Problems

Let's test these strategies. The following examples are drawn from typical JEE-level resources.

First, a classic two-component, simultaneous phase change problem.

Test your understanding! (Problem 1)

Problem: 5g of ice at 0°C is mixed with 5g of steam at 100°C. What is the final temperature and composition?
(Use $L_{f} = 80$ cal/g, $L_{v} = 540$ cal/g, $c_{w} = 1$ cal/g°C)

Try solving this using the "checkpoint" method, but with 100°C as the reference. What is the net heat exchange required to bring everything to 100°C?

Show answer

Let's analyze this using the 100°C checkpoint.

1. Heat Absorbed ( $Q_{g ain e d}$ ) to reach 100°C:
The ice must first melt and then the resulting water must be heated to 100°C.

Heat to melt ice: $Q_{1} = m_{i} L_{f} = 5 g \times 80 cal/g = 400 cal$
Heat to warm resulting water from 0°C to 100°C: $Q_{2} = m_{i} c_{w} Δ T = 5 g \times 1 cal/g°C \times 10 0^{\circ} C = 500 cal$
Total $Q_{g ain e d}$ : $400 + 500 = 900 cal$

2. Heat Released ( $Q_{l os t}$ ) to reach 100°C:
The only component that can release heat is the steam, by condensing.

Total $Q_{l os t}$ : $m_{s} L_{v} = 5 g \times 540 cal/g = 2700 cal$

3. Analyze:
The heat available from condensing all the steam ( $2700 cal$ ) is much greater than the heat needed to melt the ice and bring it to 100°C ( $900 cal$ ).
This means that not all the steam will condense.

Conclusion:

Final State: A mixture of water and steam.
Final Temperature: 100°C.

Final Composition:

The ice melts and warms up, absorbing 900 cal.
This heat must come from condensing some of the steam.
Mass of steam condensed: $m_{co n d e n se d} = \frac{Q _{ab sor b e d}}{L _{v}} = \frac{900 cal}{540 cal/g} \approx 1.67 g$ .
Final mass of steam: $5 g - 1.67 g = 3.33 g$ .
Final mass of water: Initial 5g from ice + 1.67g from condensed steam = 6.67 g.

The final mixture is 6.67 g of water and 3.33 g of steam, all at 100°C. This problem is also discussed in the resource "Elasticity, Thermal Expansion, Calorimetry & Heat Transfer".

Now, let's try a problem where the final state is given, which is a common way for examiners to test your ability to correctly formulate the energy balance equation.

Test your understanding! (Problem 2)

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 process produces liquid water at 40°C. Find the value of M.
(Use $L_{f} = 80$ cal/g, $L_{v} = 540$ cal/g, $c_{w} = 1$ cal/g°C)
(This is adapted from JEE Advanced Previous Year Questions on Calorimetry and Heat..., Question 3)

Show answer

Here, the final state is known (water at 40°C), so we can directly apply the principle Heat Gained = Heat Lost.

Heat Gained (by the ice):

Melting the ice at 0°C: $Q_{g ain, 1} = m_{i ce} L_{f} = 200 g \times 80 cal/g = 16000 cal$
Heating the resulting water from 0°C to 40°C: $Q_{g ain, 2} = m_{i ce} c_{w} Δ T = 200 g \times 1 cal/g°C \times (40 - 0)^{\circ} C = 8000 cal$

Total Heat Gained: $16000 + 8000 = 24000 cal$

Heat Lost (by the steam):

Condensing the steam at 100°C: $Q_{l os t, 1} = M \cdot L_{v} = M \times 540 cal/g$
Cooling the resulting water from 100°C to 40°C: $Q_{l os t, 2} = M \cdot c_{w} Δ T = M \times 1 cal/g°C \times (100 - 40)^{\circ} C = M \times 60 cal/g$

Total Heat Lost: $540 M + 60 M = 600 M cal$

Equating Heat Gained and Heat Lost:
$600 M = 24000$
$M = \frac{24000}{600} = 40 g$

So, 40 grams of steam were required.

Conclusion

In this lesson, we've developed two robust strategies to solve complex calorimetry problems involving multiple simultaneous phase changes.

Key Takeaways:

For multi-component mixtures (ice, water, steam), do not guess the final state blindly.
Use the deterministic checkpoint method: calculate the net heat exchange at a reference temperature (0°C or 100°C) to systematically determine the final state.
Alternatively, use the assume-and-verify method: propose a plausible final state, solve for an unknown (like mass melted), and check if the result is physically possible.
The principle Heat Gained = Heat Lost is the foundation for all these problems. The challenge lies in correctly identifying all the terms that contribute to each side of the equation.

In our next and final lesson of this module, we will expand our energy balance model one last time to apply energy conservation in calorimetry problems that include heat loss to the surroundings, a realistic complication often found in advanced problems.