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#energy change

7 public questions tagged with this topic.

Cyclic Processes and Reversibility Concepts

In a cyclic process, what is true about the change in internal energy?

**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. In a cyclic process, the system returns to its initial state. Since internal energy ( U ) is a state variable, its change ( Δ U ) is zero over a complete cycle, regardless of the path taken. 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

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

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Specific Heat Capacity and Latent Heat

A system releases 600 J of heat and performs 250 J of work. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = -600 J (heat released), Δ W = 250 J (work by system). -600 = Δ U + 250 ⇒ Δ U = -600 - 250 = -850 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

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

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Specific Heat Capacity and Latent Heat

A gas undergoes an adiabatic expansion, doing 450 J of work. What is the change in its internal energy?

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. Adiabatic: Δ Q = 0 , Δ U = -Δ W . Work by gas: Δ W = 450 J . Δ U = -450 J (internal energy decreases). 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 > Specific Heat Capacity and Latent Heat

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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 800 J of heat and has 350 J of work done on it. 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 = 800 , Δ W = -350 (work done on system). 800 = Δ U - 350 ⇒ Δ U = 800 + 350 = 1150 J . 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

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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Thermodynamics

A system at constant pressure undergoes an enthalpy change of -200 kJ. This means:

The system releases heat is the scientifically accurate answer to this question. Within the study of Thermodynamics, this concept is well-established through extensive research and is documented in standard scientific literature. The specific properties, mechanisms, or characteristics of The system releases heat directly address what is being asked. Among the other options, The system absorbs heat, The system does no work, and The system remains unchanged do not correctly answer this question because they either refer to different concepts, describe properties of other molecules or processes, or represent common misconceptions about this topic.

Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 1

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