#heat energy

Cyclic Processes and Reversibility Concepts

How many joules are equivalent to 500 cal of heat? (1 cal = 4.186 J )

**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. Heat in J = Heat in cal × 4.186 . 500 × 4.186 = 2093 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 = 1 - T_c/T_h, evaluation yields 2093 J, consistent with thermodynamic laws and energy conservation.

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

View Details

Zeroth Law Thermal Equilibrium and Quasi-static

How many calories are equivalent to 2093 J of heat? (1 cal = 4.186 J )

**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. Heat in cal = Heat in J4.186 . (2093)/(4.186) ≈ 500 cal . 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 500 cal, consistent with thermodynamic laws and energy conservation.

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

View Details

First Law of Thermodynamics Applications

How many joules are equivalent to 300 cal of heat? (1 cal = 4.186 J )

**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. Heat in J = Heat in cal × 4.186 . 300 × 4.186 = 1255.8 J ≈ 1256 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

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

View Details

First Law of Thermodynamics Applications

How much heat is required to raise the temperature of 0.45 kg of aluminium from 30^circ C to 60^circ C ? (Specific heat

**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. Δ Q = m s Δ T . m = 0.45 , s = 900 , Δ T = 60 - 30 = 30 . Δ Q = 0.45 × 900 × 30 = 12150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

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

View Details

Temperature Scales and Triple Point

During vaporization, what does the supplied heat primarily do to the liquid?

During vaporization, the supplied heat (latent heat of vaporization) breaks intermolecular bonds to convert the liquid into vapor, without changing its temperature. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Converts it to vapor without temperature change. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

View Details