Chapter2_B - Chapter 2 Section B Non-Numerical Solutions...

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Chapter 2 - Section B - Non-Numerical Solutions 2.3 Equation (2.2) is here written: 1 U t + 1 E P + 1 E K = Q + W ( a ) In this equation W does not include work done by the force of gravity on the system. This is accounted for by the 1 E K term. Thus, W = 0. ( b ) Since the elevation of the egg decreases, sign( 1 E P )is( ). ( c ) The egg is at rest both in its initial and final states; whence 1 E K = 0. ( d ) Assuming the egg does not get scrambled, its internal energy does not change; thus 1 U t = 0. ( e ) The given equation, with 1 U t = 1 E K = W = 0, shows that sign(Q) is ( ). A detailed exam- ination of the process indicates that the kinetic energy of the egg just before it strikes the surface appears instantly as internal energy of the egg, thus raising its temperature. Heat transfer to the surroundings then returns the internal energy of the egg to its initial value. 2.6 If the refrigerator is entirely contained within the kitchen, then the electrical energy entering the re- frigerator must inevitably appear in the kitchen. The only mechanism is by heat transfer (from the condenser of the refrigerator, usually located behind the unit or in its walls). This raises, rather than lowers, the temperature of the kitchen. The only way to make the refrigerator double as an air condi- tioner is to place the condenser of the refrigerator outside the kitchen (outdoors). 2.7 According to the phase rule [Eq. (2.7)], F = 2 π + N . According to the laboratory report a pure material ( N = 1) is in 4-phase ( π = 4) equilibrium. If this is true, then F = 2 4 + 1 =− 1. This is not possible; the claim is invalid. 2.8 The phase rule [Eq. (2.7)] yields: F = 2 π + N = 2 2 + 2 = 2. Specification of T and P fixes the intensive state, and thus the phase compositions, of the system. Since the liquid phase is pure species 1, addition of species 2 to the system increases its amount in the vapor phase. If the composition of the vapor phase is to be unchanged, some of species 1 must evaporate from the liquid phase, thus decreasing the moles of liquid present. 2.9 The phase rule [Eq. (2.7)] yields: F = 2 π + N = 2 2 + 3 = 3. With only T and P fixed, one degree of freedom remains. Thus changes in the phase compositions are possible for the given T and P . If ethanol is added in a quantity that allows T and P to be restored to their initial values, the ethanol distributes itself between the phases so as to form new equilibrium phase compostions and altered amounts of the vapor and liquid phases. Nothing remains the same except T and P .
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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter2_B - Chapter 2 Section B Non-Numerical Solutions...

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