This MCQ module is based on: Specific Heat Calorimetry
TOPIC 10 OF 33
Specific Heat Calorimetry
🎓 Class 11
Physics
CBSE
Theory
Ch 10 – Thermal Properties of Matter
⏱ ~14 min
🌐 Language: [gtranslate]
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Specific Heat Calorimetry
10.6 Heat Capacity & Specific Heat Capacity
If you place an iron rod and an aluminium rod of equal mass on identical hot-plates and supply them with the same amount of heat, the iron rod becomes considerably hotter than the aluminium rod. This shows that different substances need different amounts of heat for the same rise in temperature. The quantity that captures this property is called the heat capacity.
Heat capacity (S): the heat \(\Delta Q\) required to raise the temperature of a body by 1 K:
\[S = \frac{\Delta Q}{\Delta T} \qquad \text{SI unit: J K}^{-1}\]
Heat capacity depends on the size of the body. To get a property of the material alone, divide by mass:
Specific heat capacity (s or c):
\[s = \frac{S}{m} = \frac{1}{m}\frac{\Delta Q}{\Delta T} \qquad \text{SI unit: J kg}^{-1}\text{K}^{-1}\]
The heat absorbed by a body is therefore \(\boxed{\,\Delta Q = m\, s\, \Delta T\,}\).
For substances handled mole by mole (gases especially), it is more useful to define the molar specific heat capacity:
\[C = \frac{1}{\mu}\frac{\Delta Q}{\Delta T} \qquad \text{SI unit: J mol}^{-1}\text{K}^{-1}\]
where \(\mu = m/M\) is the number of moles (\(M\) = molar mass). For solids, \(C = M \times s\).
| Substance | s (J kg⁻¹ K⁻¹) | C (J mol⁻¹ K⁻¹) | Note |
|---|---|---|---|
| Water (liquid) | 4186 | 75.3 | Highest among common substances |
| Ice (0 °C) | 2090 | 37.7 | ~½ of liquid water |
| Steam (100 °C) | 2010 | 36.2 | Approx., at constant P |
| Aluminium | 900 | 24.4 | Light cookware metal |
| Iron / Steel | 460 | 25.7 | Dulong–Petit ≈ 25 J/mol·K |
| Copper | 387 | 24.5 | |
| Mercury | 140 | 28.0 | Liquid metal |
| Lead | 128 | 26.5 |
Why is water so special? The exceptionally high specific heat of water (≈ 5× that of iron) means oceans absorb large amounts of solar energy with little temperature rise. This moderates climate near coasts and stabilises the temperature of biological organisms (which are mostly water).
10.6.1 Molar Specific Heats of Gases — Cp and Cv
For gases the heat absorbed depends on whether the gas is held at constant volume or constant pressure during heating. So gases have two distinct molar specific heats:
- Cv = molar specific heat at constant volume. All heat goes into raising internal energy, since no work is done.
- Cp = molar specific heat at constant pressure. Heat raises the internal energy and does work pushing back the piston.
For an ideal gas (Mayer's relation, derived in Chapter 11):
\[\boxed{\,C_p - C_v = R\,}\]
For monatomic gases (He, Ar): \(C_v = \tfrac{3}{2}R\), \(C_p = \tfrac{5}{2}R\), \(\gamma = C_p/C_v = 5/3\).
For diatomic gases (H₂, N₂, O₂): \(C_v = \tfrac{5}{2}R\), \(C_p = \tfrac{7}{2}R\), \(\gamma = 7/5\).
For diatomic gases (H₂, N₂, O₂): \(C_v = \tfrac{5}{2}R\), \(C_p = \tfrac{7}{2}R\), \(\gamma = 7/5\).
10.7 Calorimetry — The Principle of Mixtures
Calorimetry measures heat. The instrument is the calorimeter — a polished metal cup, well-insulated, fitted with a stirrer and a thermometer. The technique relies on energy conservation:
Principle of Mixtures: When two bodies at different temperatures are placed in thermal contact in an insulated enclosure, the heat lost by the hotter body equals the heat gained by the colder body, until both reach a common equilibrium temperature.
\[\boxed{\,\text{Heat lost by hot body} = \text{Heat gained by cold body}\,}\]
10.7.1 Standard Worked Algorithm
m₁ s₁ (T₁ − T) = m₂ s₂ (T − T₂)
↑ hot body cools ↑ cold body warms
where T = final equilibrium temperature, T₁ > T > T₂. If the calorimeter itself absorbs heat, add a term \(W(T - T_2)\) where \(W\) is the calorimeter's water equivalent (mass of water that would have the same heat capacity as the calorimeter).
Water equivalent (W): if calorimeter has mass \(m_c\) and specific heat \(s_c\), then \(W = m_c s_c / s_w\) (in grams or kg, depending on units). Adding \(W\) is equivalent to imagining the calorimeter replaced by an additional \(W\) grams of water.
Worked Examples
Example 10.5: Heat needed to raise temperature
How much heat is required to raise the temperature of 5.0 kg of water from 25 °C to 90 °C? Take \(s_w = 4186\) J kg⁻¹ K⁻¹.
\(\Delta T = 90 - 25 = 65\) K.
\(\Delta Q = m s \Delta T = 5.0\times 4186\times 65 = 1.36\times 10^6\) J = 1.36 MJ.
This is roughly the energy of a 1.5-kW electric kettle running for 15 minutes.
\(\Delta Q = m s \Delta T = 5.0\times 4186\times 65 = 1.36\times 10^6\) J = 1.36 MJ.
This is roughly the energy of a 1.5-kW electric kettle running for 15 minutes.
Example 10.6: Determining specific heat by mixtures (NCERT-style)
A copper block of mass 0.20 kg is heated to 100 °C and dropped into 0.30 kg of water at 25 °C contained in a copper calorimeter of mass 0.10 kg. The final temperature settles at 28.4 °C. Find the specific heat of copper. (Take \(s_w = 4186\) J/kg·K.)
Let the specific heat of copper be \(s_c\).
Heat lost by copper block = heat gained by water + calorimeter.
\(0.20 \cdot s_c \cdot (100 - 28.4) = 0.30 \cdot 4186 \cdot (28.4 - 25) + 0.10 \cdot s_c \cdot (28.4 - 25)\)
\(0.20 s_c (71.6) = 0.30(4186)(3.4) + 0.10 s_c (3.4)\)
\(14.32 s_c = 4269.7 + 0.34 s_c\)
\(13.98 s_c = 4269.7 \Rightarrow s_c = \boxed{305\text{ J kg}^{-1}\text{K}^{-1}}\)
This is reasonably close to the standard value 387 J/kg·K (real experiments lose some heat to surroundings).
Heat lost by copper block = heat gained by water + calorimeter.
\(0.20 \cdot s_c \cdot (100 - 28.4) = 0.30 \cdot 4186 \cdot (28.4 - 25) + 0.10 \cdot s_c \cdot (28.4 - 25)\)
\(0.20 s_c (71.6) = 0.30(4186)(3.4) + 0.10 s_c (3.4)\)
\(14.32 s_c = 4269.7 + 0.34 s_c\)
\(13.98 s_c = 4269.7 \Rightarrow s_c = \boxed{305\text{ J kg}^{-1}\text{K}^{-1}}\)
This is reasonably close to the standard value 387 J/kg·K (real experiments lose some heat to surroundings).
Example 10.7: Final temperature of a mixture
Hot tea at 80 °C of mass 200 g is poured into a thick clay cup of mass 150 g initially at 25 °C. Specific heat of clay = 800 J/kg·K, of tea (assume = water) = 4186 J/kg·K. Neglect heat loss to surroundings. Find the final equilibrium temperature.
Let final temperature be \(T\).
Heat lost by tea = Heat gained by cup
\(0.200 \times 4186 \times (80 - T) = 0.150 \times 800 \times (T - 25)\)
\(837.2(80 - T) = 120(T - 25)\)
\(66976 - 837.2T = 120T - 3000\)
\(69976 = 957.2 T\)
\(T = 73.1\) °C.
The cup pulls the tea down by only ~7 °C — that's why earthen cups keep tea hot longer than thin steel cups (lower s × m of the cup).
Heat lost by tea = Heat gained by cup
\(0.200 \times 4186 \times (80 - T) = 0.150 \times 800 \times (T - 25)\)
\(837.2(80 - T) = 120(T - 25)\)
\(66976 - 837.2T = 120T - 3000\)
\(69976 = 957.2 T\)
\(T = 73.1\) °C.
The cup pulls the tea down by only ~7 °C — that's why earthen cups keep tea hot longer than thin steel cups (lower s × m of the cup).
Example 10.8: Power-rating problem
An electric immersion heater rated 1500 W is dipped into 2.0 kg of water at 30 °C. How long will it take to bring the water to a boil? Assume 90 % electric efficiency, no heat loss.
Heat needed: \(\Delta Q = m s \Delta T = 2.0 \times 4186 \times (100 - 30) = 5.86\times 10^5\) J.
Useful power = \(0.90 \times 1500 = 1350\) W.
Time \(t = \Delta Q / P = 5.86\times 10^5 / 1350 = 434\) s ≈ 7 min 14 s.
Useful power = \(0.90 \times 1500 = 1350\) W.
Time \(t = \Delta Q / P = 5.86\times 10^5 / 1350 = 434\) s ≈ 7 min 14 s.
Interactive: Mixture / Calorimetry Solver L3 Apply
Enter masses, specific heats and starting temperatures of two substances. The applet computes the final equilibrium temperature using the principle of mixtures.
Activity — Sand vs Water on a Sunny Day
L3 Apply
Predict: Why do beaches feel scorching hot at noon while the sea remains pleasantly cool?
- Take two identical wide bowls. Fill one with dry sand and the other with the same mass of water.
- Place a thermometer in each. Put both bowls in direct sunlight for 30 min.
- Record the rise in temperature in each bowl.
- Bring both indoors and again record temperatures every 10 min for half an hour.
Observation: Sand heats up much faster (and cools faster) than water.
Explanation: Specific heat of sand (~800 J/kg·K) is roughly five times less than that of water (4186 J/kg·K), so for the same heat input, sand temperature rises about five times more. The same physics is at work in coastal climates: the ocean acts as a thermal reservoir, smoothing day-night temperature swings.
Explanation: Specific heat of sand (~800 J/kg·K) is roughly five times less than that of water (4186 J/kg·K), so for the same heat input, sand temperature rises about five times more. The same physics is at work in coastal climates: the ocean acts as a thermal reservoir, smoothing day-night temperature swings.
Competency-Based Questions
A 50 g iron bullet (s = 460 J/kg·K) leaves a rifle barrel at 600 m/s. It strikes a wooden plank and embeds itself, instantly converting its kinetic energy into thermal energy of the bullet (assume all of it heats the bullet, none the plank).
Q1. L1 Remember Define molar specific heat capacity. State its SI unit.
Heat required to raise temperature of 1 mole of a substance by 1 K. SI unit: J mol⁻¹ K⁻¹.
Q2. L3 Apply Calculate the kinetic energy of the bullet.
\(KE = \tfrac{1}{2}m v^2 = 0.5\times 0.050\times (600)^2 = 9000\) J.
Q3. L3 Apply By what amount does the temperature of the bullet rise on impact?
\(\Delta T = \dfrac{KE}{m s} = \dfrac{9000}{0.050\times 460} = \boxed{391\text{ K (or °C)}}\). A nearly 400-degree jump — bullets can melt or deform on hard targets.
Q4. L4 Analyse If only 70 % of the kinetic energy were converted to bullet heat (the rest into the plank), recompute ΔT.
\(\Delta T = \dfrac{0.70\times 9000}{0.050\times 460} = 274\) K. Still huge — over 270 °C rise.
Q5. L5 Evaluate Cooling water in a car radiator is preferred over an oil with low specific heat. Why?
Water has a very high specific heat (4186 J/kg·K). For the same heat extracted from the engine, water's temperature rises only modestly, allowing it to absorb much more heat per unit mass and per cycle. It is also abundant, cheap, non-toxic and has good thermal conductivity. Anti-freeze is added only to prevent winter freezing and corrosion.
Assertion-Reason Questions
Assertion (A): The temperature of a body cannot rise without absorption of heat.
Reason (R): Heat is the only form of energy that can change the temperature of a body.
Answer: D. A is false. Mechanical work (e.g., rubbing) can also raise temperature without any external heat absorption (we will see this in Joule's experiment, Chapter 11). R alone is false too — but the assertion is the wrong one here.
Assertion (A): Two bodies of different mass and material can have the same heat capacity.
Reason (R): Heat capacity \(S = m s\), so a small mass with high \(s\) and a larger mass with low \(s\) can give the same product.
Answer: A. R correctly explains A. Heat capacity is the product m·s, allowing many combinations.
Assertion (A): For an ideal gas, Cp is always greater than Cv.
Reason (R): At constant pressure, part of the supplied heat does work pushing back the surroundings, so more heat is needed for the same temperature rise.
Answer: A. Cp − Cv = R > 0 (Mayer's relation), exactly because of the work term, as R correctly explains.
Frequently Asked Questions - Specific Heat Calorimetry
What is the main concept covered in Specific Heat Calorimetry?
In NCERT Class 11 Physics Chapter 10 (Thermal Properties of Matter), "Specific Heat Calorimetry" 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 Specific Heat Calorimetry useful in real-life applications?
Real-life applications of Specific Heat Calorimetry 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 Specific Heat Calorimetry?
Key formulas in Specific Heat Calorimetry (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. Specific Heat Calorimetry 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 Specific Heat Calorimetry?
CBSE board questions from Specific Heat Calorimetry 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 Specific Heat Calorimetry 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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