SM_2A - Chapter 2 - Section A - Mathcad Solutions 2.1 (a)...

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t 1 20 degC = C P 4.18 kJ kg degC = M H2O 25 kg = t 2 t 1 U total M H2O C P + = t 2 20.02degC = Ans. (d) For the restoration process, the change in internal energy is equal but of opposite sign to that of the initial process. Thus Q U total - = Q 1.715 - kJ = Ans. (e) In all cases the total internal energy change of the universe is zero. 2.2 Similar to Pb. 2.1 with mass of water = 30 kg. Answers are: (a) W = 3.43 kJ (b) Internal energy change of the water = 2.86 kJ (c) Final temp. = 20.027 deg C (d) Q = -2.86 kJ Chapter 2 - Section A - Mathcad Solutions 2.1 (a) M wt 35 kg = g 9.8 m s 2 = z 5 m = Work M wt g z = Work 1.715kJ = Ans. (b) U total Work = U total 1.715kJ = Ans. (c) By Eqs. (2.14) and (2.21): dU d PV ( ) + C P dT = Since P is constant, this can be written: M H2O C P dT M H2O dU M H2O P dV + = Take Cp and V constant and integrate: M H2O C P t 2 t 1 - ( 29 U total = 5
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Q 34 800 - J = W 34 300J = Ut 34 Q 34 W 34 + = Ut 34 500 - J = Ans. Step 1 to 2 to 3 to 4 to 1: Since U t is a state function, U t for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the U t values for all of the steps must sum to zero. Ut 41 4700J = Ut 23 Ut 12 - Ut 34 - Ut 41 - = Ut 23 4000 - J = Ans. Step 2 to 3: Ut 23 4 - 10 3 × J = Q 23 3800 - J = W 23 Ut 23 Q 23 - = W 23 200 - J = Ans. For a series of steps, the total work done is the sum of the work done for each step. W 12341 1400 - J = 2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor. i 9.7amp = E 110V = Wdot mech 1.25hp = Wdot elect i E = Wdot elect 1.067 10 3 × W = Qdot Wdot elect Wdot mech - = Qdot 134.875W = Ans. 2.5 Eq. (2.3): U t Q W + = Step 1 to 2: Ut 12 200 - J = W 12 6000 - J = Q 12 Ut 12 W 12 - = Q 12 5.8 10 3 × J = Ans. Step 3 to 4: 6
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U 12 - kJ = Q U = Q 12 - kJ = Ans. 2.13
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SM_2A - Chapter 2 - Section A - Mathcad Solutions 2.1 (a)...

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