This MCQ module is based on: Change of State
TOPIC 11 OF 33
Change of State
🎓 Class 11
Physics
CBSE
Theory
Ch 10 – Thermal Properties of Matter
⏱ ~14 min
🌐 Language: [gtranslate]
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Change of State
10.8 Change of State
Matter exists in three commonly observed states: solid, liquid and gas. A transition from one state to another is called a change of state or a phase change. The remarkable feature of a phase change is this: although the substance absorbs (or releases) heat throughout the transition, its temperature does not change.
10.8.1 Melting and Freezing
The transition solid → liquid is called melting (or fusion); liquid → solid is freezing (or solidification). The temperature at which a pure substance melts at 1 atm pressure is called its melting point (e.g. ice melts at 0 °C, lead at 327 °C, iron at 1535 °C).
For most substances the melting point increases with pressure, but ice is an exception: its melting point decreases with increasing pressure (because ice contracts on melting). This is the key to the regelation phenomenon — a wire under tension cuts through ice without breaking it.
10.8.2 Boiling and Condensation
The transition liquid → gas at the boiling point is called vaporisation (or boiling); the reverse is condensation. The boiling point depends strongly on pressure: at higher pressures, water boils at higher temperatures. This is why a pressure cooker (≈ 2 atm inside) raises the boiling point of water to about 120 °C and cooks food faster. Conversely, on Mount Everest (\(P\)≈ 0.3 atm) water boils at only 70 °C — and a normal cup of tea is too cool!
10.8.3 Sublimation
Some substances pass directly from solid to gas (or vice-versa) without becoming liquid. This is sublimation. Familiar examples: solid CO₂ ("dry ice"), iodine, naphthalene, camphor.
10.9 Latent Heat
During a phase change the absorbed heat does not raise temperature — it goes into rearranging the molecular structure (breaking bonds in melting, separating molecules in vaporisation). The hidden ("latent") heat absorbed per unit mass is the latent heat of that transition.
Latent heat (L):
\[\boxed{\,Q = m\, L\,}\]
where \(m\) is the mass undergoing phase change. SI unit: J/kg. Two specific cases dominate:
- Latent heat of fusion (Lf): for solid ↔ liquid transitions.
- Latent heat of vaporisation (Lv): for liquid ↔ gas transitions.
| Substance | Melting point (°C) | Lf (×10³ J/kg) | Boiling point (°C) | Lv (×10³ J/kg) |
|---|---|---|---|---|
| Water / Ice | 0 | 334 | 100 | 2256 |
| Ethanol | −114 | 104 | 78 | 854 |
| Mercury | −39 | 11.8 | 357 | 272 |
| Lead | 327 | 24.5 | 1750 | 871 |
| Copper | 1083 | 205 | 2570 | 4730 |
| Nitrogen | −210 | 25.7 | −196 | 199 |
10.9.1 Heating Curve of Water
Suppose 1 g of ice at −20 °C is heated steadily until it becomes steam at 120 °C. Plot temperature vs heat absorbed:
Adding up segment heats (in calories per gram):
A→B ice warms : m·s_ice·ΔT = 1·0.5·20 = 10 cal
B→C ice melts : m·L_f = 1·80 = 80 cal
C→D water warms : m·s_w·ΔT = 1·1·100 = 100 cal
D→E water vaporises: m·L_v = 1·540 = 540 cal
E→F steam warms : m·s_st·ΔT = 1·0.48·20 = 9.6 cal
TOTAL (1 g, −20 °C to 120 °C steam) ≈ 740 cal
Why is Lv so large for water? To convert liquid water into steam each H₂O molecule must completely escape the strong hydrogen-bond network — an enormous energy cost. Melting only loosens the network; the molecules still touch each other. That's why \(L_v \approx 7 L_f\) for water.
10.9.2 Evaporation vs Boiling
- Evaporation happens at any temperature, only at the free surface; molecules with enough kinetic energy leave the liquid.
- Boiling happens at one specific temperature (boiling point) throughout the bulk; bubbles of vapour form within the liquid.
Both involve absorption of latent heat, which is why evaporating sweat cools your skin and a wet earthen pot keeps water cool by surface evaporation — the principle behind the traditional Indian matka.
Worked Examples
Example 10.9: Heat to convert ice to steam (NCERT-style)
How much heat is required to convert 3.0 kg of ice at 0 °C kept in a calorimeter to steam at 100 °C at atmospheric pressure? Take \(s_w = 4186\) J/kg·K, \(L_f = 3.34\times 10^5\) J/kg, \(L_v = 22.6\times 10^5\) J/kg.
Step 1 — melt ice (0 °C → water 0 °C): \(Q_1 = m L_f = 3.0\times 3.34\times 10^5 = 1.002\times 10^6\) J.
Step 2 — warm water (0 → 100 °C): \(Q_2 = m s \Delta T = 3.0\times 4186\times 100 = 1.256\times 10^6\) J.
Step 3 — vaporise water (100 °C → steam 100 °C): \(Q_3 = m L_v = 3.0\times 22.6\times 10^5 = 6.78\times 10^6\) J.
Total: \(Q = Q_1 + Q_2 + Q_3 = (1.002 + 1.256 + 6.78)\times 10^6 \approx \boxed{9.04\times 10^6\text{ J}}\).
Notice how the vaporisation step alone consumes about 75 % of the total heat!
Step 2 — warm water (0 → 100 °C): \(Q_2 = m s \Delta T = 3.0\times 4186\times 100 = 1.256\times 10^6\) J.
Step 3 — vaporise water (100 °C → steam 100 °C): \(Q_3 = m L_v = 3.0\times 22.6\times 10^5 = 6.78\times 10^6\) J.
Total: \(Q = Q_1 + Q_2 + Q_3 = (1.002 + 1.256 + 6.78)\times 10^6 \approx \boxed{9.04\times 10^6\text{ J}}\).
Notice how the vaporisation step alone consumes about 75 % of the total heat!
Example 10.10: Steam burn vs water burn
Compare the heat released when (i) 5.0 g of steam at 100 °C cools and condenses to water at 37 °C (body temperature) on the skin, and (ii) 5.0 g of boiling water at 100 °C cools to 37 °C on the skin.
(i) Steam: condensation + cooling.
\(Q_{\text{steam}} = m L_v + m s_w \Delta T = 0.005(22.6\times 10^5) + 0.005(4186)(63)\)
\(= 11300 + 1319 = 12619\) J ≈ 12.6 kJ.
(ii) Boiling water: only cooling.
\(Q_{\text{water}} = m s_w \Delta T = 0.005(4186)(63) = 1319\) J ≈ 1.3 kJ.
Steam delivers nearly 10× more heat to the skin — confirming that steam burns are far more severe.
\(Q_{\text{steam}} = m L_v + m s_w \Delta T = 0.005(22.6\times 10^5) + 0.005(4186)(63)\)
\(= 11300 + 1319 = 12619\) J ≈ 12.6 kJ.
(ii) Boiling water: only cooling.
\(Q_{\text{water}} = m s_w \Delta T = 0.005(4186)(63) = 1319\) J ≈ 1.3 kJ.
Steam delivers nearly 10× more heat to the skin — confirming that steam burns are far more severe.
Example 10.11: Ice cube in tea
A 50 g ice cube at 0 °C is dropped into 250 g of tea at 80 °C in an insulated cup. Find the final temperature. (s of tea = s of water = 4186 J/kg·K, L_f = 3.34 × 10⁵ J/kg.)
First check whether all ice melts. Heat available from tea cooling to 0 °C = \(0.250 \times 4186 \times 80 = 83720\) J. Heat needed to melt the ice = \(0.050\times 3.34\times 10^5 = 16700\) J. Since 83 720 > 16 700, all ice melts and we proceed to find the final T.
Energy balance: Heat lost by tea = Heat to melt ice + Heat to warm melted-ice from 0 °C to T.
\(0.250 \times 4186 \times (80 - T) = 16700 + 0.050 \times 4186 \times T\)
\(1046.5(80 - T) = 16700 + 209.3 T\)
\(83720 - 1046.5 T = 16700 + 209.3 T\)
\(67020 = 1255.8 T \Rightarrow T = \boxed{53.4\,°\text{C}}\).
Energy balance: Heat lost by tea = Heat to melt ice + Heat to warm melted-ice from 0 °C to T.
\(0.250 \times 4186 \times (80 - T) = 16700 + 0.050 \times 4186 \times T\)
\(1046.5(80 - T) = 16700 + 209.3 T\)
\(83720 - 1046.5 T = 16700 + 209.3 T\)
\(67020 = 1255.8 T \Rightarrow T = \boxed{53.4\,°\text{C}}\).
Example 10.12: Energy to make ice in a freezer
A refrigerator must convert 1.0 kg of water at 25 °C into ice at −10 °C. Calculate the heat to be removed. (s_ice = 2090 J/kg·K.)
Cool water 25 → 0 °C: \(1.0\times 4186\times 25 = 1.046\times 10^5\) J.
Freeze water at 0 °C: \(1.0\times 3.34\times 10^5 = 3.34\times 10^5\) J.
Cool ice 0 → −10 °C: \(1.0\times 2090\times 10 = 2.09\times 10^4\) J.
Total: \(Q = (1.046 + 3.34 + 0.209)\times 10^5 = \boxed{4.60\times 10^5\text{ J}}\) — most of it is the latent-heat step.
Freeze water at 0 °C: \(1.0\times 3.34\times 10^5 = 3.34\times 10^5\) J.
Cool ice 0 → −10 °C: \(1.0\times 2090\times 10 = 2.09\times 10^4\) J.
Total: \(Q = (1.046 + 3.34 + 0.209)\times 10^5 = \boxed{4.60\times 10^5\text{ J}}\) — most of it is the latent-heat step.
Interactive: Ice → Steam Heat Calculator L3 Apply
Enter mass and starting/ending temperatures of water; the applet adds segment heats and shows where energy goes.
Activity — Boil Water in a Paper Cup
L4 Analyse
Predict: Will a paper cup catch fire if you heat water in it directly over a flame?
- Take a paper cup. Fill it three-quarters full with water.
- Hold it (with tongs) over a candle or small flame.
- Wait until the water boils.
- Observe the bottom of the cup — does the paper char?
Observation: The water boils, but the paper does not catch fire (it may brown slightly).
Explanation: Water is a strong absorber of heat (high specific heat) and, while boiling, holds the bottom temperature at 100 °C — well below paper's ignition temperature (~ 230 °C). The latent heat carried away by escaping steam keeps the cup cool.
Explanation: Water is a strong absorber of heat (high specific heat) and, while boiling, holds the bottom temperature at 100 °C — well below paper's ignition temperature (~ 230 °C). The latent heat carried away by escaping steam keeps the cup cool.
Competency-Based Questions
In a cold-chain pharmaceutical depot, vaccines must be kept between 2 °C and 8 °C. A worker places 4.0 kg of crushed ice at 0 °C inside an insulated container holding 12.0 kg of vaccines (assume average specific heat 3500 J/kg·K) initially at 25 °C.
Q1. L1 Remember Define the latent heat of fusion. State its SI unit.
It is the heat absorbed (per unit mass) when a substance changes from solid to liquid at constant temperature. SI unit: J/kg.
Q2. L3 Apply Calculate the heat the vaccines must lose to drop to 4 °C.
\(Q = m s \Delta T = 12.0 \times 3500 \times (25-4) = 8.82\times 10^5\) J.
Q3. L3 Apply How much ice will melt in absorbing this heat? Will any ice be left?
Heat absorbed by ice = \(m_\text{ice} L_f + m_\text{ice} s_w(4)\) (melted ice rises to 4 °C). Solving \(8.82\times 10^5 = m(3.34\times 10^5 + 4186\times 4)\) gives \(m = 8.82\times 10^5 / 3.508\times 10^5 = 2.51\) kg. So ~2.5 kg melts; about 1.5 kg of ice remains. ✓ Acceptable.
Q4. L4 Analyse Why is dry ice (solid CO₂) preferred over water-ice for shipping vaccines on long flights?
Dry ice sublimes (solid → gas) directly at −78 °C, leaving no liquid mess that could ruin packaging or short-circuit electronics. Its sublimation absorbs a huge amount of heat per kg and keeps temperatures well below freezing for many hours.
Q5. L5 Evaluate A mountaineer's dehydrated food sachet asks for "boiling water" but at altitude water boils at 80 °C. Will the food cook properly? Explain.
No, cooking requires temperatures around 95–100 °C for the chemistry of starch gelatinisation and protein denaturation. At 80 °C reactions proceed much slower, leaving food undercooked. Mountaineers use pressure cookers or higher-energy fuel cells to compensate.
Assertion-Reason Questions
Assertion (A): The temperature of a substance does not change while it is melting, even though heat is being added.
Reason (R): The added heat is used up to break the inter-molecular bonds of the solid lattice rather than increasing kinetic energy.
Answer: A. R correctly explains A — the latent heat is potential-energy of separation, not kinetic.
Assertion (A): A pressure cooker cooks food faster than an open pot.
Reason (R): The increased pressure inside the cooker raises water's boiling point above 100 °C, so food cooks at a higher temperature.
Answer: A. Cooking-rate roughly doubles for every 10 °C rise (Arrhenius). At ~120 °C, cooking is much faster.
Assertion (A): Steam at 100 °C is more harmful for the skin than water at 100 °C.
Reason (R): Steam carries an additional latent heat of vaporisation that is released on condensation onto the skin.
Answer: A. Steam delivers ~5× more heat per gram than equally hot boiling water, exactly because of the L_v term R describes.
Frequently Asked Questions - Change of State
What is the main concept covered in Change of State?
In NCERT Class 11 Physics Chapter 10 (Thermal Properties of Matter), "Change of State" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Change of State useful in real-life applications?
Real-life applications of Change of State from NCERT Class 11 Physics Chapter 10 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Change of State?
Key formulas in Change of State (NCERT Class 11 Physics Chapter 10 Thermal Properties of Matter) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 10?
NCERT Class 11 Physics Chapter 10 (Thermal Properties of Matter) is structured so each part builds on the previous one. Change of State connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Change of State?
CBSE board questions from Change of State typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Change of State lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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