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#calorie conversion

2 public questions tagged with this topic.

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

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