Chapter 15
----------


7) T_H = 950K, T_C = 300K, |Q_C| = 400J
       epsilon_Carnot = 1 - T_C/T_H = 0.684  .... (i)
   From the definition of efficiency
       epsilon = |W| / |Q_H| = |W| / ( |Q_C| + |W| ) = |W| / ( 4000 + |W| ) .. (ii)
   From (i) = (ii),
       |W| = 867J


14) (a) PV = nRT  ->  V = nRT/P = 2.41 x 10^(-2) [m^3]
    (b) At the beginning (point 1) and the end (point 2)
	       P_1 * V = nRT_1  and  P_2 * V = nRT_2  (V does not change)
		   ->  P_1 / P_2 = T_1 / T_2  ->  P_2 = 1.46 x 10^5 Pa
    (c) For an adiabatic process, PV^gamma = const
	      P_2 * V_2^gamma = P_3 * V_3^gamma   ->  V_3 = 3.12 x 10^(-2) [m^3]
    (d) PV = nRT  ->  T_3 = P_3 * V_3 / (nR) = 379K
    (e) W_1 = 0 for path 1 (isochoric process)
        W_2 = - n * c_V * delT   (adiabatic, delU + W = 0, W = - delU)
		    = - 1.0 * (5R/2) * (379-423)
			= 915J
        W_3 = P * delV = -7.3 x 10^2 J
		Therefore, W_total = W_1 + W_2 + W_3 = 1.9 x 10^2 J
    (f) Path 1: constant volume, Q_1 = n * c_V * delT = 2.70 x 10^3 J
        Path 2: Q_2 = 0J (adiabatic)
		Path 3: constant pressure, Q_3 = n * c_P * delT = -2.5 x 10^3 J
		Therefore, Q_total = Q_1 + Q_2 + Q_3 = 2 x 10^2 J
		------------------------------------------------------
		delT = T_f - T_i
		------------------------------------------------------
	(g) epsilon = |W| / |Q_H| = 0.07


18) epsilon_Carnot = 1 - T_C/T_H = (T_H - T_C)/T_H
    K_Carnot = T_C/(T_H - T_C)
	Therefore, epsilon_Carnot * K_Carnot = T_C / T_H


30) (a) W = mgh = 5 * 9.81 * 1.5 = 73.6 J
    (b) Engine does not hold any themal energy. delU = 0J
    (c) Q = delU + W = 0 + 73.6 = 73.6J
	(d) |Q_H| = |Q_C| + |W| = 90 + 73.6 = 164J
	(e) epsilon = |W| / |Q_H| = 0.449


44) |Q_H| - |Q_C| is less than |W| .... violates 1st and 2nd laws


52) T_H = 373K, T_C = 293K, delQ = 4500J
    (a) delS_H = -4500/T_H = -12.1 J/K
	(b) delS_C = 4500/T_C = 15.4 J/K
	(c) delS = delS_H + delS_C = 3.3 J/K    ( > 0 )
	(d) epsilon_Carnot = 1 - T_C/T_H = 0.214


54) V1 = 1 liter = 1 * 10^(-3) [m^3], 
    density of water rho = 1.0 x 10^3 Kg/m^3
	-> m1 = rho * V1 = 1Kg,  m2 = 2Kg
	To find out the equilibrium final t (in celsius),
	   delQ1 + delQ2 = 0
	   ->  m1 * c_water * (t-20) + m2 * c_water * (t-90) = 0
	   ->  t = 66 degrees in celsius  => T = 339K
	Therefore,
	  delS = Integrate(i,f) dQ/T =  Integrate(i,f) mcdT/T = mc*ln(T_f/T_i)
	  -> delS_total = delS_1 + delS_2 = 610 - 573 = 37 [J/K]    (>0)


55) (a) W = |Q_H| - |Q_C| = 1500J
    (b) epsilon = |W| / |Q_H| = 0.30
	(c) epsilon_Carnot = 1 - T_C/T_H = 0.621
	(d) delS = -|Q_H|/T_H + |Q_C|/T_C + 0 = 5.4J/K


58) Like a free expansion for each gas,
    delS = Integrate(i,f) dQ/T = Integrate(i,f) PdV/T = nR * Integrate(i,f) dV/V 
	     = nR*ln(V_f/V_i)
	delS_total = nR*ln(V/V1) + nR*ln(V/V2) = nR*ln(V^2/V1/V2), where V=V1+V2
	Since V^2/V1/V2 > 1, delS_total > 0


59) The final T is given by
    mc*(T-T1) + mc*(T-T2) = 0  ->  T = (T1+T2)/2
	delS = mc*ln(T/T1) + mc*ln(T/T2) = mc*ln(T^2/T1/T2) 
	     = mc*ln( (T1+T2)^2/(4*T1*T2) )
    However, delS >= 0 J/K
	since (T1+T2)^2 > 4*T1*T2 <-----------> (T1-T2)^2 >= 0
	





------------------------------------------------
   A^B = A to the Bth power (2^3 = 8)
   ln(x) = natural logarithm of x
   v_rms = rms velocity ... _rms stands for subscript rms