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#thermal equilibrium

8 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

Which of the following statements is correct regarding the Zeroth Law of Thermodynamics?

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. The Zeroth Law states that if two systems are in thermal equilibrium with a third, they are in equilibrium with each other, establishing temperature as a measurable property. It does not involve heat flow direction or work, which are addressed by other laws. Using first law ΔU = Q -

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

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Heat Transfer Work Distinction and Internal Energy Change

Why is the concept of thermal equilibrium essential to the Zeroth Law of Thermodynamics?

**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. The Zeroth Law relies on thermal equilibrium, where no net heat flows between systems in contact, establishing that systems sharing this property with a third system have a common property—temperature—thus defining temperature measurement. 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 =

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

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Adiabatic Processes and Gamma Determination

Which of the following statements correctly defines thermal equilibrium?

**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. Thermal equilibrium occurs when two systems in contact have no net heat flow between them, indicating equal temperatures, as per the Zeroth Law. Option C is correct. 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

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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PHYSICS

Why is the concept of thermal equilibrium essential to the Zeroth Law of Thermodynamics?

The Zeroth Law relies on thermal equilibrium, where no net heat flows between systems in contact, establishing that systems sharing this property with a third system have a common property—temperature—thus defining temperature measurement.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Kinetic Theory (Latest NCERT 2026-27), Topic: RMS speed v_rms ∝ √T, temperature dependence, ratio v₂/v₁ = √(T₂/T₁) and calculation. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and illustrative examples. Page.

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Thermal Expansion of Solids and Liquids

Why does water in a calorimeter reach a steady temperature when mixed with a hot object?

In calorimetry (Section 10.7), heat lost by the hot object equals heat gained by the water and calorimeter at thermal equilibrium, where no further heat transfer occurs. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Heat lost equals heat gained. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Thermal Expansion of Solids and Liquids

A 0.25kg iron block at 150∘C is dropped into 1.5kg of water at 30∘C. What is the final temperature? (Specific heat of ir

Heat lost = Heat gained. 0.25×450×(150−T) = 1.5×4186×(T−30). 16875−112.5T = 6279T−188370. 16875+188370 = 6279T+112.5T. 205245 = 6391.5T⇒T≈32.11∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 32.11°C. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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Temperature Scales and Triple Point

A 0.15kg mercury block at 400∘C is placed in 0.5kg water at 20∘C in a 0.05kg aluminium calorimeter at 20∘C. Find the fin

0.15×140×(400−T) = (0.5×4186+0.05×900)×(T−20). 8400−21T = (2093+45)×(T−20) = 2138T−42760. 8400+42760 = 2138T+21T. 51160 = 2159T⇒T≈23.7∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 23.7°C. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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