606_PartUniversity Physics Solution

# 606_PartUniversity Physics Solution - The Second Law of...

This preview shows pages 1–3. Sign up to view the full content.

The Second Law of Thermodynamics 20-21 20.60. IDENTIFY: For a reversible isothermal process, . Q S T Δ= For a reversible adiabatic process, 0 Q = and 0. S Δ = The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic processes. SET UP: Use the results for the Stirling cycle from Problem 20.50. EXECUTE: (a) The graph is given in Figure 20.60. (b) For a reversible process, , and so , and dQ dS dQ T dS T == , Qd QT d S which is the area under the curve in the TS plane. (c) H Q is the area under the rectangle bounded by the horizontal part of the rectangle at H T and the verticals. C || Q is the area bounded by the horizontal part of the rectangle at C T and the verticals. The net work is then HC , QQ the area bounded by the rectangle that represents the process. The ratio of the areas is the ratio of the lengths of the vertical sides of the respective rectangles, and the efficiency is HH . WTT e (d) As explained in Problem 20.50, the substance that mediates the heat exchange during the isochoric expansion and compression does not leave the system, and the diagram is the same as in part (a). As found in that problem, the ideal efficiency is the same as for a Carnot-cycle engine. EVALUATE: The derivation of e Carnot using the concept of entropy is much simpler than the derivation in Section 20.6, but yields the same result. Figure 20.60 20.61. IDENTIFY: The temperatures of the ice-water mixture and of the boiling water are constant, so . Q S T The heat flow for the melting phase transition of the ice is f . Qm L = + SET UP: For water, 5 f 3.34 10 J/kg. L EXECUTE: (a) The heat that goes into the ice-water mixture is 54 f (0.160 kg)(3.34 10 J/kg) 5.34 10 J. L × This is same amount of heat leaves the boiling water, so 4 5.34 10 J 143 J/K. 373 K Q S T −× Δ= = = (b) 4 196 J/K 273 K Q S T × = + (c) For any segment of the rod, the net heat flow is zero, so 0. S Δ = (d) tot 143 J/K 196 J/K 53 J/K. S + = + EVALUATE: The heat flow is irreversible, since the system is isolated and the total entropy change is positive. 20.62. IDENTIFY: Use the expression derived in Example 20.6 for the entropy change in a temperature change. SET UP: For water, 4190 J/kg K. c =⋅ 20 C 293.15 K, = ° 65 C 338.15 K = ° and 120 C 393.15 K. = ° EXECUTE: (a) 3 21 ln( ) (250 10 kg)(4190 J kg K)ln(338.15 K 293.15 K) 150 J K. Sm cTT = × = (b) 3 element (250 10 kg)(4190 J kg K)(338.15 K 293.15 K) 120 J/K. 393.15 K mc T S T −Δ − × = = (c) The sum of the result of parts (a) and (b) is system 30 J/K. S Δ =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
20-22 Chapter 20 EVALUATE: (d) Heating a liquid is not reversible. Whatever the energy source for the heating element, heat is being delivered at a higher temperature than that of the water, and the entropy loss of the source will be less in magnitude than the entropy gain of the water. The net entropy change is positive. 20.63. IDENTIFY: Use the expression derived in Example 20.6 for the entropy change in a temperature change. For the value of T for which S Δ is a maximum, () / 0 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

606_PartUniversity Physics Solution - The Second Law of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online