How much heat is required to vaporize 0.5 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )
**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. Δ Q = m L . m = 0.5 , L = 2256 . Δ Q = 0.5 × 2256 = 1128 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T
Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications