.MCAD 303010000 1 74 81 0
.CMD PLOTFORMAT
0 0 1 1 0 0 1 
0 0 1 1 0 0 1 
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15 0 3 
.CMD FORMAT  rd=d ct=10 im=i et=3 zt=15 pr=3 mass length time charge temperature tr=0 vm=0
.CMD SET ORIGIN 0
.CMD SET TOL 0.001000000000000
.CMD SET PRNCOLWIDTH 8
.CMD SET PRNPRECISION 4
.CMD PRINT_SETUP 1.200000 1.218750 1.200000 1.200000 0
.CMD HEADER_FOOTER 1 1 *empty* *empty* *empty* 0 1 *empty* *empty* *empty*
.CMD HEADER_FOOTER_FONT fontID=14 family=Arial points=10 bold=0 italic=0 underline=0
.CMD HEADER_FOOTER_FONT fontID=15 family=Arial points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE_NAME fontID=0 name=Variables
.CMD DEFINE_FONTSTYLE_NAME fontID=1 name=Constants
.CMD DEFINE_FONTSTYLE_NAME fontID=2 name=Text
.CMD DEFINE_FONTSTYLE_NAME fontID=4 name=User^1
.CMD DEFINE_FONTSTYLE_NAME fontID=5 name=User^2
.CMD DEFINE_FONTSTYLE_NAME fontID=6 name=User^3
.CMD DEFINE_FONTSTYLE_NAME fontID=7 name=User^4
.CMD DEFINE_FONTSTYLE_NAME fontID=8 name=User^5
.CMD DEFINE_FONTSTYLE_NAME fontID=9 name=User^6
.CMD DEFINE_FONTSTYLE_NAME fontID=10 name=User^7
.CMD DEFINE_FONTSTYLE fontID=0 family=Times^New^Roman points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=1 family=Times^New^Roman points=10 bold=0 italic=0 underline=0
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.CMD UNITS U=1
.CMD DIMENSIONS_ANALYSIS 0 0
.TXT 10 1 65 0
Cg a47.375000,47.000000,619
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain {[Adapted from Bromberg 8.14] The
 following table gives the heat capacity
 of silver at temperatures from 15K to 300K.
 (a) Construct two graphs, one showing C
 versus T, and the other showing C/T versus
 T. (b) Assume the cubic dependence of heat
 capacity below 15K: C=AT}{\fs16 \up6 3}{.
 Determine A and compute the absolute entropy
 of silver at 15K. (c) Approximate the integral
 at higher temperature in a reasonable way
 and compute the entropy of silver at 100K,
 at 200K, and at 298.1K. (d) Suppose that
 the error in C at 15K is 10%. How great
 an error does this introduce into the absolute
 entropy at 298.1K?}}
}
.EQN 25 47 67 0
{0:T}NAME:({21,1}ö300ö298.1ö280ö260ö240ö220ö200ö180ö160ö140ö120ö100ö90ö80ö70ö60ö50ö40ö30ö20ö15)*{0:K}NAME
.EQN 0 12 70 0
{0:C}NAME:({21,1}ö25.5ö25.49ö25.39ö25.21ö24.89ö24.57ö24.27ö23.9ö23.3ö22.59ö21.6ö20.96ö19.13ö17.89ö16.33ö14.31ö11.65ö8.39ö4.77ö1.72ö0.67)*({0:joule}NAME)/({0:K}NAME)
.TXT 28 -59 75 0
Cg a17.250000,73.000000,44
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain {(a) Plots of C versus T\par and
 C/T versus T}}
}
.EQN 0 24 15 0
{0:i}NAME:0,1;20
.EQN 3 -24 14 0
&&(_n_u_l_l_&_n_u_l_l_)&({0:C}NAME)[({0:i}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&({0:T}NAME)[({0:i}NAME)
0 0 1 1 0 0 1 
0 0 1 1 0 0 1 
2 0 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15 0 3 
.EQN 1 35 16 0
&&(_n_u_l_l_&_n_u_l_l_)&(({0:C}NAME)[({0:i}NAME))/(({0:T}NAME)[({0:i}NAME))@&&(_n_u_l_l_&_n_u_l_l_)&({0:T}NAME)[({0:i}NAME)
0 0 1 1 0 0 1 
0 0 1 1 0 0 1 
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15 0 3 
.TXT 24 9 80 0
Cg a18.625000,21.500000,71
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain The area under this curve represents
 the absolute entropy of silver.}
}
.TXT 1 -44 79 0
Cg a25.750000,25.625000,80
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain The area under this curve represents
 the enthalpy to heat the silver to 300K.}
}
.TXT 25 1 19 0
Cg a27.500000,72.000000,37
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain {(b) Assume Cp=AT^3 below 15K, then}}
}
.EQN 1 29 20 0
{0:A}NAME:(({0:C}NAME)[(0))/(((15*{0:K}NAME))^(3))
.EQN 0 14 21 0
{0:A}NAME={0}?{0:joule}NAME
.EQN 5 -2 23 0
{0:A}NAME*((15*{0:K}NAME))^(3)={0}?({0:joule}NAME)/({0:K}NAME)
.TXT 1 21 24 0
Cg a5.000000,12.125000,9
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
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{\plain {Check!}}
}
.EQN 3 -50 25 0
{0:C0}NAME({0:T}NAME):{0:A}NAME*({0:T}NAME)^(3)
.EQN 6 -13 26 0
({0:\DS}NAME)[(1):(0*{0:K}NAME&15*{0:K}NAME`({0:C0}NAME({0:T}NAME))/({0:T}NAME)&{0:T}NAME)
.EQN 0 20 27 0
({0:\DS}NAME)[(1)={0}?({0:joule}NAME)/({0:K}NAME)
.TXT 0 16 81 0
Cg a14.250000,37.000000,20
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain The entropy at 15K.}
}
.TXT 7 -36 58 0
Cg a36.625000,40.000000,55
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain {(c) Take the average heat capacity
 in each interval.}}
}
.TXT 0 39 63 0
Cg a9.875000,42.000000,17
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
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{\plain {integrate on T}}
}
.EQN 0 12 59 0
({0:Cav}NAME)/({0:T}NAME)
.TXT 0 4 64 0
Cg a3.625000,24.000000,8
{\rtf1\ansi \deff0
{\fonttbl
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{\f1\fnil Symbol;}
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{\plain {gives}}
}
.EQN 0 5 62 0
{0:Cav}NAME*{0:ln}NAME({0:T}NAME)
.EQN 4 -32 76 0
({0:T}NAME)[(8+1)={0}?_n_u_l_l_
.EQN 2 -28 28 0
({0:\DS}NAME)[(2):((0,8,{0:i}NAME,(({0:C}NAME)[({0:i}NAME+1)+({0:C}NAME)[({0:i}NAME))/(2)*{0:ln}NAME((({0:T}NAME)[({0:i}NAME+1))/(({0:T}NAME)[({0:i}NAME)))){64})
.EQN 0 41 31 0
({0:\DS}NAME)[(2)={0}?({0:joule}NAME)/({0:K}NAME)
.TXT 3 -11 32 0
Cg a6.125000,42.000000,11
{\rtf1\ansi \deff0
{\fonttbl
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{\f1\fnil Symbol;}
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{\plain {at 100K,}}
}
.EQN 0 11 34 0
{0:S}NAME:({0:\DS}NAME)[(1)+({0:\DS}NAME)[(2)
.EQN 0 14 43 0
{0:S}NAME={0}?({0:joule}NAME)/({0:K}NAME)
.EQN 6 -28 77 0
({0:T}NAME)[(13+1)={0}?_n_u_l_l_
.EQN 1 -27 35 0
({0:\DS}NAME)[(3):((9,13,{0:i}NAME,(({0:C}NAME)[({0:i}NAME+1)+({0:C}NAME)[({0:i}NAME))/(2)*{0:ln}NAME((({0:T}NAME)[({0:i}NAME+1))/(({0:T}NAME)[({0:i}NAME)))){64})
.EQN 1 40 38 0
({0:\DS}NAME)[(3)={0}?({0:joule}NAME)/({0:K}NAME)
.TXT 6 -13 39 0
Cg a6.625000,41.000000,12
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
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{\plain {at 200K, }}
}
.EQN 0 11 40 0
{0:S}NAME:({0:\DS}NAME)[(1)+({0:\DS}NAME)[(2)+({0:\DS}NAME)[(3)
.EQN 0 18 44 0
{0:S}NAME={0}?({0:joule}NAME)/({0:K}NAME)
.EQN 4 -29 78 0
({0:T}NAME)[(18+1)={0}?_n_u_l_l_
.EQN 2 -27 45 0
({0:\DS}NAME)[(4):((14,18,{0:i}NAME,(({0:C}NAME)[({0:i}NAME+1)+({0:C}NAME)[({0:i}NAME))/(2)*{0:ln}NAME((({0:T}NAME)[({0:i}NAME+1))/(({0:T}NAME)[({0:i}NAME)))){64})
.EQN 0 41 48 0
({0:\DS}NAME)[(4)={0}?({0:joule}NAME)/({0:K}NAME)
.TXT 5 -14 49 0
Cg a6.125000,40.000000,11
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain {at 298K,}}
}
.EQN 0 7 50 0
{0:S}NAME:({0:\DS}NAME)[(1)+({0:\DS}NAME)[(2)+({0:\DS}NAME)[(3)+({0:\DS}NAME)[(4)
.EQN 0 22 53 0
{0:S}NAME={0}?({0:joule}NAME)/({0:K}NAME)
.TXT 6 -56 72 0
Cg a67.625000,73.000000,212
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
{\f1\fnil Symbol;}
}
{\plain {(d) If the heat capacity measurement
 is off by 10% at 15K, then also A is off
 by 10% so that\par }{\f1 D}{S1 is in error
 by 10%. This means an error in all higher
 temperature absolute entropy values amounting
 to}}
}
.EQN 8 11 73 0
0.1*({0:\DS}NAME)[(1)={0}?({0:joule}NAME)/({0:K}NAME)