Govt Exams
Maximum efficiency occurs in a Carnot engine: η = 1 - (T_cold/T_hot) = 1 - (300/500) = 1 - 0.6 = 0.4 or 40%
Efficiency η = W/Q_in = (Q_in - Q_out)/Q_in = (2000-1200)/2000 = 800/2000 = 0.4 = 40%. Work output W = 800 J
In constant volume process: W = ∫PdV = 0. From first law: ΔU = Q - W = -300 - 0 = -300 J
Water has exceptionally high specific heat (~4.18 kJ/kg·K), much higher than metals (iron ~0.46) and air (~1.01). This is due to hydrogen bonding in water.
During phase transitions (like vaporization at constant T and P), infinite heat can be absorbed without temperature change, making C_p → ∞.
From first law: ΔU = Q - W = 500 - 200 = 300 J. Internal energy increases by 300 J.
Otto cycle efficiency = 1 - (1/r^(γ-1)) = 1 - (1/10^0.4) = 1 - (1/2.512) = 1 - 0.3981 = 0.602 ≈ 60.2%
Heat (Q) and work (W) are path-dependent quantities, not state functions. Internal energy, enthalpy, and entropy are state functions depending only on initial and final states.
COP_heating = T_H/(T_H - T_C) = 330/(330-270) = 330/60 = 5.5. Q_H = W × COP = 1000 × 5.5 = 5500 J. Adding work input: Total = 5500 + 500 = 6000 J
For polytropic process with n=1: PV = constant, which is the ideal gas law at constant temperature, making it isothermal (T = constant).