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#heat absorption

18 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A system absorbs 730 J of heat and performs 190 J of work. What is the change in internal energy?

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). First Law: Δ Q = Δ U + Δ W . Δ Q = 730 , Δ W = 190 (work by system). 730 = Δ U + 190 ⇒ Δ U = 730 - 190 = 540 J . Using first law ΔU = Q - W, W = ∫ P

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

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Heat Transfer Work Distinction and Internal Energy Change

A system absorbs 850 J of heat and has 300 J of work done on it. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = 850 J , Δ W = -300 J (work on system). 850 = Δ U - 300 ⇒ Δ U = 850 + 300 = 1150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

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

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Isochoric Processes and Pressure-Temperature

A system in a cyclic process absorbs 1020 J of heat and rejects 380 J . What is the net work done?

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 1020 - 380 = 640 J . W = 640 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

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

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Isochoric Processes and Pressure-Temperature

A gas undergoes a cyclic process where 600 J of heat is absorbed. What is the net work done by the gas?

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. For a cyclic process, Δ U = 0 . Δ Q = Δ U + Δ W ⇒ 600 = 0 + Δ W . Δ W = 600 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 =

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

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Specific Heat Capacity and Latent Heat

A system absorbs 830 J of heat and has 290 J of work done on it. What is the change in internal energy?

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. First Law: Δ Q = Δ U + Δ W . Δ Q = 830 , Δ W = -290 (work on system). 830 = Δ U - 290 ⇒ Δ U = 830 + 290 = 1120 J . Using first law ΔU = Q - W, W

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Zeroth Law Thermal Equilibrium and Quasi-static

A system absorbs 920 J of heat and performs 280 J of work. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = 920 , Δ W = 280 (work by system). 920 = Δ U + 280 ⇒ Δ U = 920 - 280 = 640 J . Using first law ΔU = Q - W, W = ∫ P dV,

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

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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

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First Law of Thermodynamics Applications

0.2 moles of an ideal gas expand isothermally at 350 K from 4 L to 10 L. What is the heat absorbed? ( R = 8.3 J mol⁻¹ K⁻

**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. Isothermal: Δ U = 0 , Δ Q = Δ W = μ R T ln((V₂)/(V₁)) . μ = 0.2 , T = 350 , V₂ = 10 , V₁ = 4 . Δ Q = 0.2 × 8.3 × 350 × ln((10)/(4)) = 581 × 0.916 ≈ 532 J . Using first law ΔU

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

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First Law of Thermodynamics Applications

A system absorbs 450 J of heat and performs 150 J of work. What is the change in internal energy?

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. First Law: Δ Q = Δ U + Δ W . Δ Q = 450 , Δ W = 150 . 450 = Δ U + 150 ⇒ Δ U = 450 - 150 = 300 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

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

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Adiabatic Processes and Gamma Determination

A system absorbs 710 J of heat and performs 260 J of work. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = 710 , Δ W = 260 (work by system). 710 = Δ U + 260 ⇒ Δ U = 710 - 260 = 450 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

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

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Isobaric and Isothermal Processes Work Calculation

A system absorbs 700 J of heat while 300 J of work is done on it. What is the change in internal energy?

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. First Law: Δ Q = Δ U + Δ W . Δ Q = 700 , Δ W = -300 (work done on system, negative work by system). 700 = Δ U - 300 ⇒ Δ U = 700 + 300 = 1000 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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