Govt. Exams
Entrance Exams
At constant P, heat goes into both increasing internal energy and doing work: Q_p = ΔU + W. At constant V, heat only increases internal energy: Q_v = ΔU. Thus Cp > Cv by R.
Carnot (reversible) engine has maximum efficiency between any two temperatures. All real irreversible engines are less efficient.
According to third law of thermodynamics, entropy of a perfect crystal at 0 K is zero.
COP = Q_removed/W_input, so W = Q/COP = 500/5 = 100 J
Efficiency η = (Q_input - Q_output)/Q_input = (1000 - 600)/1000 = 400/1000 = 0.4 = 40%
For a cyclic process, ΔU = 0, so Q = W. The work equals the area enclosed in P-V diagram.
In adiabatic process, Q = 0. By first law, ΔU = -W. Since gas does work (W > 0), ΔU < 0, so internal energy decreases and temperature drops.
Carnot efficiency η = 1 - (T_cold/T_hot) = 1 - (300/400) = 1 - 0.75 = 0.25 = 25%
Using conservation of momentum: m(3v) + m(-v) = 2m(v_final). Therefore, 2mv = 2m(v_final), giving v_final = v. Taking rightward as positive direction.
Thrust force = rate of mass ejection × relative velocity = 2 kg/s × 2000 m/s = 4000 N. This applies Newton's third law to rocket propulsion.
