#calorimetry

Heat Transfer Work Distinction and Internal Energy Change

How much heat is required to raise the temperature of 0.2 kg of aluminium from 45^circ C to 75^circ C ? (Specific heat o

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. Δ Q = m s Δ T . m = 0.2 , s = 900 , Δ T = 75 - 45 = 30 . Δ Q = 0.2 × 900 × 30 = 5400 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ =

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

View Details

Cyclic Processes and Reversibility Concepts

How much heat is required to raise the temperature of 0.6 kg of silver from 20^circ C to 50^circ C ? (Specific heat of s

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. Δ Q = m s Δ T . m = 0.6 , s = 236.1 , Δ T = 50 - 20 = 30 . Δ Q = 0.6 × 236.1 × 30 = 4249.8 J ≈ 4250 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h,

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

View Details

Second Law Heat Engines and Kelvin-Planck

How much heat is required to raise the temperature of 0.35 kg of copper from 15^circ C to 45^circ C ? (Specific heat of

**Heat engine** absorbs Q_h from hot reservoir T_h, does work W, rejects Q_c to cold T_c, efficiency η = W/Q_h =1 - Q_c/Q_h, second law limits η

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

View Details

Second Law Heat Engines and Kelvin-Planck

How much heat is required to raise the temperature of 0.6 kg of copper from 20^circ C to 50^circ C ? (Specific heat of c

**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

View Details

Second Law Heat Engines and Kelvin-Planck

0.25 kg of a substance at 30°C absorbs 1500 J of heat, reaching 60°C. What is its specific heat capacity?

**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. Specific heat: s = (Δ Q)/(m Δ T) . Δ Q = 1500 J , m = 0.25 kg , Δ T = 60 - 30 = 30 K . s = (1500)/(0.25 × 30) = 200 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

View Details

Mean Free Path and Molecular Diameter

How much heat is required to raise the temperature of 0.2 moles of a diatomic gas by 10 K at constant volume, with no vi

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. Diatomic gas: 5 degrees of freedom, C_v = (5)/(2) R.Q = μ C_v Δ T = 0.2 × (5)/(2) × 8.31 × 10 = 83.1 J. Substituting values gives 83.1 J, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

View Details

CHEMISTRY

For the combustion of 1 g of graphite in a bomb calorimeter, Δ U = -20.7 kJ. What is Δ U for 1 mole of graphite (molar

Given: For the combustion of 1 g of graphite in a bomb calorimeter, Δ U = -20.7 kJ. What is Δ U for 1 mole of graphite (molar mass = 12 g/mol)? These values define the system as per NCERT data. Formula: Δ U per mole = Δ U per gram × molar mass = -20.7 kJ/g × 12 g/mol = -248.4 kJ/mol.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the cor

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details

PHYSICS

A 0.2 kg aluminium block at 120° C is placed in 0.8 kg of water at 20° C in a 0.1 kg copper calorimeter at 20° C . Wh

Given: A 0.2 kg aluminium block at 120° C is placed in 0.8 kg of water at 20° C in a 0.1 kg copper calorimeter at 20° C . What is the final temperature? (Specific heat of aluminium = 900 J kg^{-1 K^{-1, water = 4186 J kg^{-1 K^{-1, copper = 386 J kg^{-1 K^{-1 ) These values define the system as per NCERT data. Formula: Heat lost = Heat gained. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 0.2 × 900 × (120 - T) = (0.8 × 4186 + 0.1 × 386) × (T - 20) . 21600 - 180 T = (3348.8 + 38.6) × (T - 20) = 3387.4 T - 67748 . 21600 + 67748 = 3387.4

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details

CHEMISTRY

For a bomb calorimeter, which thermodynamic quantity is directly measured?

A bomb calorimeter operates at constant volume, so Δ U = q_v is measured. This follows from NCERT principle where relation explains outcome clearly for students.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

View Details

Ideal Gas Laws and Thermal Properties

How much heat is required to convert 0.2kg of ice at −22∘C to steam at 110∘C in a 0.05kg aluminium calorimeter initially

Q1 = 0.2×2100×22 = 9240J (ice to 0°C). Q2 = 0.2×3.35×105 = 67000J (melting). Q3 = (0.2×4186+0.05×900)×100 = (837.2+45)×100 = 88220J (to 100°C). Q4 = 0.2×2.256×106 = 451200J (vaporization). Q5 = 0.2×4186×10 = 8372J (steam to 110°C). Calorimeter cools: Q6 = 0.05×900×(30−0) = 1350J (assume it cools to 0°C). Total: Q = 9240+67000+88220+451200+8372−1350 = 622682J = 622.68kJ.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

View Details

Ideal Gas Laws and Thermal Properties

A 0.5kg silver block at 240∘C is placed in 1.5kg water at 21∘C in a 0.25kg copper calorimeter at 21∘C. What is the final

0.5×236×(240−T) = (1.5×4186+0.25×386)×(T−21). 28320−118T = (6279+96.5)×(T−21) = 6375.5T−133885.5. 28320+133885.5 = 6375.5T+118T. 162205.5 = 6493.5T⇒T≈24.97∘C≈25∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 25°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

View Details

Ideal Gas Laws and Thermal Properties

A 0.1kg iron block at 200∘C is dropped into 0.4kg water at 30∘C in a 0.05kg copper calorimeter at 30∘C. Find the final t

0.1×450×(200−T) = (0.4×4186+0.05×386)×(T−30). 9000−45T = (1674.4+19.3)×(T−30) = 1693.7T−50811. 9000+50811 = 1693.7T+45T. 59811 = 1738.7T⇒T≈34.4∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 34.4°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

View Details