This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions for Homework 2 Chapter 19 Problems 17. We assume that all of the kinetic energy of the hammer goes into heating the nail. KE = Q ! 10 1 2 m hammer v hammer 2 ( ) = m nail c Fe " T ! " T = 10 1 2 m hammer v hammer 2 ( ) m nail c Fe = 5 1.20 kg ( ) 7.5m s ( ) 2 0.014 kg ( ) 450J kg & C ° ( ) = 53.57C ° # 54C ° 40. ( a ) Leg ba is an isobaric expansion, and so the work done is positive. Leg ad is an isovolumetric reduction in pressure, and so the work done on that leg is 0. Leg dc is an isobaric compression, and so the work done is negative. Leg cb is an isovolumetric expansion in pressure, and so the work done on that leg is 0. ( b ) From problem 38, W cda = W cd + W da = 38J, so W adc = W ad + W dc = ! 38J. Also from problem 38, W abc = ! 84J, and so W Cba = W Cb + W ba = 84J. So the net work done during the cycle is as follows. W net = W ba + W ad + W dc + W cb = 84J ! 38J = 46J ( c ) Since the process is a cycle, the initial and final states are the same, and so the internal energy does not change. ! E int = ( d ) Use the first law of thermodynamics, applied to the entire cycle. ! E int tot = Q net " W net # Q net = ! E int + W net = + 46J = 46J ( e ) From problem 39( b ), we have Q adc = ! 68J. This is the exhaust heat. So the input heat is found as follows....
View
Full
Document
This note was uploaded on 11/19/2010 for the course LECTURE 1 taught by Professor Yildiz during the Fall '10 term at University of California, Berkeley.
 Fall '10
 yildiz
 Physics, Work, Heat

Click to edit the document details