.MCAD 304020000 1 76 156 0 .CMD PLOTFORMAT 0 0 1 1 1 0 0 1 1 0 0 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 0 0 0 0 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 21 15 0 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 colrid=-1 .CMD HEADER_FOOTER_FONT fontID=15 family=Arial points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFAULT_TEXT_PARPROPS 0 0 2 .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 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=1 family=Times^New^Roman points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=2 family=Times^New^Roman points=14 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=4 family=Arial points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=5 family=Courier^New points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=6 family=System points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=7 family=Script points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=8 family=Roman points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=9 family=Modern points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=10 family=Times^New^Roman points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD UNITS U=1 .CMD DIMENSIONS_ANALYSIS 0 0 .CMD COLORTAB_ENTRY 0 0 0 .CMD COLORTAB_ENTRY 128 0 0 .CMD COLORTAB_ENTRY 0 128 0 .CMD COLORTAB_ENTRY 128 128 0 .CMD COLORTAB_ENTRY 0 0 128 .CMD COLORTAB_ENTRY 128 0 128 .CMD COLORTAB_ENTRY 0 128 128 .CMD COLORTAB_ENTRY 128 128 128 .CMD COLORTAB_ENTRY 192 192 192 .CMD COLORTAB_ENTRY 255 0 0 .CMD COLORTAB_ENTRY 0 255 0 .CMD COLORTAB_ENTRY 255 255 0 .CMD COLORTAB_ENTRY 0 0 255 .CMD COLORTAB_ENTRY 255 0 255 .CMD COLORTAB_ENTRY 0 255 255 .CMD COLORTAB_ENTRY 255 255 255 .CMD COLORTAB_ENTRY 0 64 128 .TXT 3 1 1 0 0 0 Cg a59.125000,59.125000,115 {\rtf1\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue0;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1\fs20 }{\cf2 \fs32\b THERMAL PHYSICS COMPUTER LAB. #0: \tab \tab TUTORIAL}{\cf2\fs32 \par }{\cf1\fs20 \par }{\cf2 First we learn how to do simple computations with MathCad.}{\cf1\fs20 \par }} .TXT 15 -1 67 0 0 0 Cg b73.000000,73.000000,32 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue0;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset0\fnil Arial;}}{ \stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 \f1 (1) Type:3*4= The answer is: 12.}} .EQN 5 1 2 0 0 0 3*4={0}?_n_u_l_l_ .TXT 3 -1 68 0 0 0 Cg b73.000000,73.000000,34 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue0;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset0\fnil Arial;}}{ \stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 \f1 (2) Type 24-35= The answer is -11}} .EQN 4 2 3 0 0 0 24-35={0}?_n_u_l_l_ .TXT 4 -2 69 0 0 0 Cg b73.000000,73.000000,60 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue0;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset0\fnil Arial;}}{ \stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 \f1 (3) Type: 3\\= (this is square root of 3). The answer is 1.73}} .EQN 5 4 4 0 0 0 \(3)={0}?_n_u_l_l_ .TXT 6 -4 64 0 0 0 Cg b73.000000,73.000000,70 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue0;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset0\fnil Arial;}}{ \stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 \f1 (4) Type 2.3^4.5= (this is 2.3 to the power 4.5). The answer is 42.44}} .EQN 6 7 5 0 0 0 (2.3)^(4.5)={0}?_n_u_l_l_ .TXT 3 -7 65 0 0 0 Cg b73.000000,73.000000,39 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue0;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset0\fnil Arial;}}{ \stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 \f1 (5) Type: (3-4)*4-6= The answer is -10}} .EQN 6 10 6 0 0 0 (3-4)*4-6={0}?_n_u_l_l_ .TXT 3 -10 66 0 0 0 Cg b73.000000,73.000000,57 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue0;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset0\fnil Arial;}}{ \stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 \f1 (6) Type: 7/(2.3-8.6)space bar+3=. The answer is:1.889 }} .EQN 5 13 7 0 0 0 (7)/((2.3-8.6))+3={0}?_n_u_l_l_ .TXT 34 -12 54 0 0 0 Cg a72.000000,72.000000,29 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 copyright Miron Kaufman, 1997}} .TXT 4 1 71 0 0 0 Cg a71.000000,71.000000,534 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 We learn next to do simple calculations and to graph with MathCad.\par Let us consider two objects of temperatures T}{\cf1\dn 1}{\cf1 and T}{\cf1\dn 2}{\cf1 and heat capacities C}{\cf1\dn 1}{\cf1 and C}{\cf1\dn 2}{\cf1 . The equilibrium temperature is obtained by equating the heat leaving the hot object 2 to the heat entering the cold object 1. \tab \tab \par Q = C}{\cf1\dn 1}{\cf1 (T-T}{\cf1\dn 1}{\cf1 )= C}{\cf1\dn 2}{\cf1 (T}{ \cf1\dn 2}{\cf1 -T). Solve for: T = (C}{\cf1\dn 1}{\cf1 T}{\cf1\dn 1 }{ \cf1 + C}{\cf1\dn 2}{\cf1 T}{\cf1\dn 2}{\cf1 )/(C}{\cf1\dn 1}{\cf1 + C}{ \cf1\dn 2}{\cf1 ). \tab Note: T}{\cf1\dn 1}{\cf1 < T 0, }{ \cf1\f1 D}{\cf1 S}{\cf1\dn 2}{\cf1 < 0. The total entropy change is positive acording to the 2'nd law: }{\cf1\f1 D}{\cf1 S}{\cf1\dn 1}{\cf1 \f1 + D}{\cf1 S}{\cf1\dn 2 }{\cf1 > 0. \par \par The fact that the two objects equilibrate at the same temperature is required by the 2'nd law of thermodynamics: entropy is at its largest possible value consistent with constraints. We are going to demonstrate this statement. Let us denote T_1 the final temperature of 1 and T_2 final temperature of 2. From: Q = C}{\cf1\dn 1}{\cf1 (T_1-T}{\cf1\dn 1}{\cf1 ) = C}{\cf1\dn 2}{\cf1 (T}{\cf1\dn 2}{\cf1 -T_2) we get: T_2 = T}{\cf1 \dn 2}{\cf1 + (C}{\cf1\dn 1}{\cf1 /C}{\cf1\dn 2}{\cf1 )(T}{\cf1\dn 1}{ \cf1 -T_1). We then graph }{\cf1\f1 D}{\cf1 S(T_1) and show that the entropy change is largest when:\tab T_1 = T_2 = (C}{\cf1\dn 1}{\cf1 T}{ \cf1\dn 1 }{\cf1 + C}{\cf1\dn 2}{\cf1 T}{\cf1\dn 2}{\cf1 )/(C}{\cf1\dn 1}{ \cf1 + C}{\cf1\dn 2}{\cf1 ).\par \par Object #1 is a porcelain piece of mass 500g, specific heat 0.22cal/g*K and temperature 20C.\par Object #2 is water of mass 1000g, specific heat 1cal/g*K and temperature 100C.}} .EQN 42 23 123 0 0 0 {0:C.2}NAME:1000*1 .EQN 0 15 126 0 0 0 {0:T.1}NAME:20+273 .EQN 0 16 128 0 0 0 {0:T.2}NAME:100+273 .EQN 1 -53 88 0 0 0 {0:C.1}NAME:500*.22 .EQN 3 0 118 0 0 0 {0:C.1}NAME={0}?_n_u_l_l_ .EQN 0 22 138 0 0 0 {0:C.2}NAME={0}?_n_u_l_l_ .EQN 0 15 139 0 0 0 {0:T.1}NAME={0}?_n_u_l_l_ .EQN 0 16 141 0 0 0 {0:T.2}NAME={0}?_n_u_l_l_ .EQN 10 -53 87 0 0 0 {0:\DS}NAME({0:T_1}NAME):{0:C.1}NAME*{0:ln}NAME(({0:T_1}NAME)/({0:T.1}NAME))+{0:C.2}NAME*{0:ln}NAME((({0:C.1}NAME)/({0:C.2}NAME)*({0:T.1}NAME-{0:T_1}NAME)+{0:T.2}NAME)/({0:T.2}NAME)) .EQN 5 3 93 0 0 0 {0:T_1}NAME:50,100;1000 .TXT 5 -4 154 0 0 0 Cg a29.250000,29.250000,29 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 copyright Miron Kaufman,1997 }} .EQN 8 0 94 0 0 0 &&(_n_u_l_l_&_n_u_l_l_)&{0:\DS}NAME({0:T_1}NAME)@&&(({0:C.1}NAME*{0:T.1}NAME+{0:C.2}NAME*{0:T.2}NAME)/({0:C.1}NAME+{0:C.2}NAME)&_n_u_l_l_)&{0:T_1}NAME 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 1 1 1 1 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 57 27 10 0 3 .TXT 45 0 137 0 0 0 Cg a70.000000,70.000000,273 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 Note that at the equilibrium temperature: }{\object{\*\objclass \eqn}\rsltpict{\*\objdata .EQN 255 39 133 0 0 0 {0:T}NAME:({0:C.1}NAME*{0:T.1}NAME+{0:C.2}NAME*{0:T.2}NAME)/({0:C.1}NAME+{0:C.2}NAME) }}{\cf1 , }{\object{\*\objclass \eqn}\rsltpict{\*\objdata .EQN 255 56 143 0 0 0 {0:T}NAME={0}?_n_u_l_l_ }}{\cf1 Kelvins the entropy increase is largest. \par Note that the entropy change for the object #1 (porcelain) is }{\object{ \*\objclass \eqn}\rsltpict{\*\objdata .EQN 264 2 142 0 0 0 {0:C.1}NAME*{0:ln}NAME(({0:T}NAME)/({0:T.1}NAME))={0}?_n_u_l_l_ }}{\cf1 calories/Kelvin and for object #2 (water) is }{\object{ \*\objclass \eqn}\rsltpict{\*\objdata .EQN 268 2 145 0 0 0 {0:C.2}NAME*{0:ln}NAME(({0:T}NAME)/({0:T.2}NAME))={0}?_n_u_l_l_ }}{\cf1 calories/Kelvin. The total entropy change is positive: }{ \object{\*\objclass \eqn}\rsltpict{\*\objdata .EQN 272 2 152 0 0 0 24.191-21.484={0}?_n_u_l_l_ }}{\cf1 calories/Kelvin.}} .TXT 50 1 156 0 0 0 Cg a70.000000,70.000000,29 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 copyright Miron Kaufman, 1997}} .TXT 10 0 148 0 0 0 Cg a67.000000,67.000000,160 {\rtf1\ansi \deff0{\colortbl;\red0\green64\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}{ \stylesheet{\f0\fs28 Normal;}}\plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 We can show the fact that }{\cf1\f1 D}{\cf1 S(T_1) is maximum when the temperatures are the same}{\cf1 by using the symbolic processor. First copy the expression for }{\cf1\f1 D}{\cf1 S(T_1).}} .EQN 13 14 105 0 0 0 {0:C.1}NAME*{0:ln}NAME(({0:T_1}NAME)/({0:T.1}NAME))+{0:C.2}NAME*{0:ln}NAME((({0:C.1}NAME)/({0:C.2}NAME)*({0:T.1}NAME-{0:T_1}NAME)+{0:T.2}NAME)/({0:T.2}NAME)) .TXT 10 -12 110 0 0 0 Cg a68.000000,68.000000,86 {\rtf1\ansi \deff0{\colortbl;\red0\green64\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 Then set the blue marker on T_1 and use }{\cf1\i Differentiate on Variable}{\cf1 from under }{\cf1 \i Symbolic}{\cf1 .}} .EQN 7 12 106 0 0 0 ({0:C.1}NAME)/({0:T_1}NAME)-({0:C.1}NAME)/((({0:C.1}NAME)/({0:C.2}NAME)*({0:T.1}NAME-{0:T_1}NAME)+{0:T.2}NAME)) .TXT 8 -11 112 0 0 0 Cg a67.000000,67.000000,111 {\rtf1\ansi \deff0{\colortbl;\red0\green64\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 Next set the above expression equal to zero by using the }{\cf1\i Boolean Equal}{\cf1 from the }{ \cf1\i Evaluation and Boolean}{\cf1 palette.}} .EQN 7 18 111 0 0 0 ({0:C.1}NAME)/({0:T_1}NAME)-({0:C.1}NAME)/((({0:C.1}NAME)/({0:C.2}NAME)*({0:T.1}NAME-{0:T_1}NAME)+{0:T.2}NAME))÷0 .TXT 8 -18 114 0 0 0 Cg a67.000000,67.000000,79 {\rtf1\ansi \deff0{\colortbl;\red0\green64\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 Then set the blue marker on T_1 and use }{\cf1\i Solve for Variable}{\cf1 from under }{\cf1\i Symbolic}{ \cf1 .}} .EQN 7 11 107 0 0 0 (-({0:C.1}NAME*{0:T.1}NAME+{0:T.2}NAME*{0:C.2}NAME))/((-{0:C.1}NAME-{0:C.2}NAME)) .TXT 7 -12 115 0 0 0 Cg a68.000000,68.000000,40 {\rtf1\ansi \deff0{\colortbl;\red0\green64\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 Finlly use }{\cf1\i Simplify }{ \cf1 from under }{\cf1\i Symbolic}{\cf1 .}} .EQN 6 12 108 0 0 0 (({0:C.1}NAME*{0:T.1}NAME+{0:T.2}NAME*{0:C.2}NAME))/(({0:C.1}NAME+{0:C.2}NAME)) .TXT 22 -14 117 0 0 0 Cg a72.000000,72.000000,29 {\rtf1\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}{\stylesheet{\f0\fs28 Normal;}} \plain\fs28 \pard\plain\s0\f0\fs28 {\cf1 copyright Miron Kaufman, 1997}}