#first law of thermodynamics

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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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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Zeroth Law Thermal Equilibrium and Quasi-static

A system releases 790 J of heat and has 310 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 = -790 (heat released), Δ W = -310 (work on system). -790 = Δ U - 310 ⇒ Δ U = -790 + 310 = -480 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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First Law of Thermodynamics Applications

A system absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. First Law: Δ Q = Δ U + Δ W .Given: Δ Q = 500 J , Δ W = 200 J . 500 = Δ U + 200 . Δ U = 500 - 200 = 300 J . Using first law ΔU = Q - W,

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

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