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#heat capacity

40 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

0.5 kg of a substance at 15°C absorbs 1800 J of heat at constant volume, reaching 45°C. What is its specific heat capaci

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. Specific heat: s = (Δ Q)/(m Δ T) . Δ Q = 1800 J , m = 0.5 kg , Δ T = 45 - 15 = 30 K . s = (1800)/(0.5 × 30) = 120 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W,

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

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Heat Transfer Work Distinction and Internal Energy Change

What is the change in internal energy for 0.8 moles of an ideal gas heated from 310 K to 370 K at constant volume? ( C_v

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. Δ U = μ C_v Δ T . μ = 0.8 , C_v = 20.8 , Δ T = 370 - 310 = 60 . Δ U = 0.8 × 20.8 × 60 = 998.4 J ≈ 998 J . Using first law ΔU = Q - W, W

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

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Heat Transfer Work Distinction and Internal Energy Change

Which of the following statements is correct about C_p and C_v for an ideal gas?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For an ideal gas, C_p > C_v because at constant pressure, heat supplies both internal energy increase and work ( C_p = C_v + R ), while at constant volume, heat only increases internal energy. Option A is correct. Using first law ΔU

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

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Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant volume for a monatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. For monatomic gas: C_v = (3)/(2) R . C_v = (3)/(2) × 8.3 = 12.45 J mol⁻¹ K⁻¹ . 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, evaluation yields 12.45 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

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

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Isochoric Processes and Pressure-Temperature

What is the change in internal energy for 0.6 moles of an ideal gas heated from 270 K to 320 K at constant volume? ( C_v

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. Δ U = μ C_v Δ T . μ = 0.6 , C_v = 20.8 , Δ T = 320 - 270 = 50 . Δ U = 0.6 × 20.8 × 50 = 624 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Specific Heat Capacity and Latent Heat

An ideal gas absorbs 1000 J of heat in an isochoric process, increasing its temperature by 20 K . What is the number of

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. For isochoric process: Δ Q = μ C_v Δ T . Δ Q = 1000 , C_v = 20 , Δ T = 20 . 1000 = μ × 20 × 20 ⇒ μ = (1000)/(400) = 2.5 moles . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the change in internal energy when 1 mole of an ideal gas is heated from 300 K to 350 K at constant volume? ( C_

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. Δ U = μ C_v Δ T . μ = 1 , C_v = 20.8 , Δ T = 350 - 300 = 50 . Δ U = 1 × 20.8 × 50 = 1040 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

Which of the following statements is correct about specific heat capacity?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Specific heat capacity ( s = (Δ Q)/(m Δ T) ) measures heat required per unit mass to raise temperature, depending on the substance and conditions (e.g., C_p vs. C_v ). Option B 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 = 1 -

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

What characteristic of a solid’s molecular structure leads to a molar specific heat capacity of approximately 3R ?

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. In solids, each atom vibrates with 3 degrees of freedom, contributing 3 k_B T (kinetic and potential) per atom. For a mole, this totals 3 R T , so C = (Δ U)/(Δ T) = 3R , due to vibrational motion. Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the molar specific heat capacity at constant pressure for a monatomic gas if C_v = 12.45 J mol⁻¹ K⁻¹ and R = 8.3

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. C_p - C_v = R . C_p = C_v + R = 12.45 + 8.3 = 20.75 J mol⁻¹ K⁻¹ . 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 > Zeroth Law Thermal Equilibrium and Quasi-static

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Second Law Heat Engines and Kelvin-Planck

What is the relationship between C_p and C_v for an ideal gas?

**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. For an ideal gas, the molar specific heat at constant pressure ( C_p ) exceeds that at constant volume ( C_v ) by the gas constant ( R ), due to the work done during expansion at constant pressure: C_p - C_v = R . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

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

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First Law of Thermodynamics Applications

0.1 kg of a substance absorbs 1200 J of heat, increasing its temperature from 20°C to 50°C. What is its specific heat ca

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Specific heat: s = (Δ Q)/(m Δ T) . Given Δ Q = 1200 J , m = 0.1 kg , Δ T = 50 - 20 = 30 K . s = (1200)/(0.1 × 30) = 400 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV,

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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