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author | Trupti Kini | 2016-08-24 23:30:24 +0600 |
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committer | Trupti Kini | 2016-08-24 23:30:24 +0600 |
commit | 2e6cacc645c7aaaf848ea854843461f5bd589766 (patch) | |
tree | e348cd1aa9d4235edb12a6357a24b6295308bfe4 /Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb | |
parent | 6d761d4b1820e2b0e271f7ff8ee1e5da5814b07b (diff) | |
download | Python-Textbook-Companions-2e6cacc645c7aaaf848ea854843461f5bd589766.tar.gz Python-Textbook-Companions-2e6cacc645c7aaaf848ea854843461f5bd589766.tar.bz2 Python-Textbook-Companions-2e6cacc645c7aaaf848ea854843461f5bd589766.zip |
Added(A)/Deleted(D) following books
A Principles_of_Physics_by_F.J.Bueche/Chapter10_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter10_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter13_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter14_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter15_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter16_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter17_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter18_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter19_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter1_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter1_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter20_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter21_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter22_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter23_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter24_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter25_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter26_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter27_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter2_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter2_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter3_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter3_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter4_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter4_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter5_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter5_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter6_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter6_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter7_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter7_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter8_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter9_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter9_2.ipynb
A Principles_of_Physics_by_F.J.Bueche/chapter12_1.ipynb
A Principles_of_Physics_by_F.J.Bueche/screenshots/11.2_1.png
A Principles_of_Physics_by_F.J.Bueche/screenshots/24.4_1.png
A Principles_of_Physics_by_F.J.Bueche/screenshots/8.4_1.png
Diffstat (limited to 'Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb')
-rw-r--r-- | Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb | 493 |
1 files changed, 493 insertions, 0 deletions
diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb new file mode 100644 index 00000000..cfca5d7a --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter11_1.ipynb @@ -0,0 +1,493 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11:Thermal Properties of Matter" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.1:pg-303" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat required for 400 gm of water is Q= 2000.0 Cal\n", + "\n", + "The heat required for 400 gm of copper is Q= -186.0 Cal\n", + "\n" + ] + } + ], + "source": [ + " import math #Example 11_1\n", + " \n", + " \n", + " #To find out how much heat is required to change the temperature\n", + " #With 400 Grams of water\n", + "c=1 #units in cal/g Centigrade\n", + "m=400 #Units in gm\n", + "t=5 #Units in centigrade\n", + "q=c*m*t #Units in Cal\n", + "print \"The heat required for 400 gm of water is Q=\",round(q),\" Cal\\n\"\n", + " #With 400 grams of copper\n", + "c=0.093 #units in cal/g Centigrade\n", + "m=400 #Units in gm\n", + "t=-5 #Units in centigrade\n", + "q=c*m*t #Units in Cal\n", + "print \"The heat required for 400 gm of copper is Q=\",round(q),\" Cal\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.2:pg-303" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When it crystallizes heat required is Q= 4000.0 Cal\n", + "\n", + "When it condenses heat required is Q= 26950.0 Cal\n", + "\n" + ] + } + ], + "source": [ + " import math #Example 11_2\n", + " \n", + " \n", + " #To findout how much water is released\n", + " #When it crystallizes\n", + "m=50 #Units in gm\n", + "h=80 #Units in Cal/gm\n", + "q=m*h #Units in Cal\n", + "print \"When it crystallizes heat required is Q=\",round(q),\" Cal\\n\"\n", + " #When it Condenses\n", + "m=50 #Units in gm\n", + "h=539 #Units in Cal/gm\n", + "q=m*h #Units in Cal\n", + "print \"When it condenses heat required is Q=\",round(q),\" Cal\\n\"\n", + " #In textbook answer is printed wrong as Q=27000 cal but the correct answer is Q=26950 Cal\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.3:pg-304" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amount of ice that has to be added is M= 54.0 gm\n" + ] + } + ], + "source": [ + " import math #Example 11_3\n", + " \n", + " \n", + " #To findout the amount of Ice that has to be added\n", + "m=200 #Units in gm\n", + "c=1 #Units in Cal/gm Centigrade\n", + "tf=60 #Units in Centigrade\n", + "to=98 #Units in Centigrade\n", + "change=m*c*(tf-to) #units in Cal\n", + "tf=60 #Units in centigrade\n", + "to=0 #Units in centigrade\n", + "Hf=80 #Units in Cal/gm\n", + "change1=Hf+c*(tf-to) #Units in Cal/gm\n", + "M=change/-(change1)\n", + "print \"The amount of ice that has to be added is M=\",round(M,1),\" gm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.4:pg-305" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The specific heat of metal is Cm= 0.216 cal/gm C\n" + ] + } + ], + "source": [ + " import math #Example 11_4\n", + " \n", + " \n", + " #To findout the specific heat capacity of the metal\n", + "m=400 #Units in gm\n", + "c=0.65 #Units in Cal/gm Centigrade\n", + "tf=23.1 #Units in Centigrade\n", + "to=18 #Units in Centigrade\n", + "oil=m*c*(tf-to) #units in cal\n", + "m1=80 #Units in gm\n", + "tf=23.1 #Units in Centigrade\n", + "to=100 #Units in Centigrade\n", + "cm=m1*(tf-to) #units in in terms of cm and gm Centigrade\n", + "cmm=oil/-cm #Units in Cal/gm Centigrade\n", + "print \"The specific heat of metal is Cm=\",round(cmm,3),\" cal/gm C\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.5:pg-305" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time taken is t= 449.0 sec\n" + ] + } + ], + "source": [ + " import math #Example 11_5\n", + " \n", + " \n", + " #To findout how long does the heater takes to heat\n", + "m=500 #Units in gm\n", + "c=0.033 #Units in Cal/gm Centigrade\n", + "tf=357 #Units in Centigrade\n", + "to=20.0 #Units in Centigrade\n", + "m1=30 #Units in gm\n", + "hv=65 #Units in cal/gm\n", + "Hg=((m*c*(tf-to))+(m1*hv))*4.1808135 #units in Joules\n", + "delivered=70 #Units in Joule/Sec\n", + "t=Hg/delivered #Units in sec\n", + "print \"The time taken is t=\",round(t),\" sec\"\n", + " #In textbook answer printed wrong as t=450 sec correct answer is t=448 sec\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.6:pg-306" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rise in temperature is DeltaT= 38.6 C\n" + ] + } + ], + "source": [ + " import math #Example 11_6\n", + " \n", + " \n", + " #To findout the rise in temperature\n", + "m=0.01 #Units in Kg\n", + "v=100 #Units in meters/sec\n", + "KE=(0.5*m*v**2)/4.1808135 #units in Cal\n", + "m=10 #units in gm\n", + "c=0.031 #units in cal/gm Centigrade\n", + "t=KE/(m*c)\n", + "print \"the rise in temperature is DeltaT=\",round(t,1),\" C\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.8:pg-307" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The slab is longer by= 0.01 meters\n" + ] + } + ], + "source": [ + " import math #Example 11_8\n", + " \n", + " \n", + " #To findout how much longer is at 35 degrees\n", + "alpha=10*10**-6 #Units in Centigrade\n", + "dist=20.0 #Unis in meters\n", + "t=50 #Units in centigrade\n", + "L=alpha*dist*t #Units in meters\n", + "print \"The slab is longer by=\",round(L,3),\" meters\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.9:pg-308" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The new diameter of the hole is= 2.0076 cm\n" + ] + } + ], + "source": [ + " import math #Example 11_9\n", + " \n", + " \n", + " #To findout how large a diameter when the sheet is heated\n", + "dist=2 #Units in cm\n", + "delta=19*10**-6 #Units in Centigrade**-1\n", + "t=200 #Units in centigrade\n", + "L=dist*delta*t #Units in cm\n", + "print \"The new diameter of the hole is=\",round(2+L,4),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.10:pg-309" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The change in benzene volume is V30= 101.253 cm**3\n" + ] + } + ], + "source": [ + " import math #Example 11_10\n", + " \n", + " \n", + " #To findout the change in benzene volume\n", + "delta=1.24*10**-3 #Units in Centigrade**-1\n", + "t=10 #Units in Centigrade\n", + "v10=100.0 #Units in cm**3 \n", + "v20=delta*t+v10 #Units in cm**3\n", + "V=v20*delta*t #Units in cm**3\n", + "v30=V+v20 #Units in cm**3\n", + "print \"The change in benzene volume is V30=\",round(v30,3),\" cm**3\"\n", + " #In textbook the answer is printed wrng as V3=0102.5 cm**3 the correct answer is V3=101.253 cm**3 \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.11:pg-309" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ice melts by 116.0 gm\n" + ] + } + ], + "source": [ + " import math #Example 11_11\n", + " \n", + " \n", + " #To findout how much ice melts each hour\n", + "s=30 #Units in cm\n", + "a=s*s*10**-4 #units in meter**2\n", + "k=0.032 #Units in W/K meter\n", + "t=25 #Units in K\n", + "l=0.040 #Units in meters\n", + "q_t=(6*k*((a*t)/l))/4.1808135 #Units in cal/sec\n", + "Q=3600*q_t #Units in cal\n", + "qq=80 #Units in cal/gm\n", + "melted=Q/qq #Units in gm\n", + "print \"The ice melts by \",round(melted),\" gm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.12:pg-310" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The radiation defers by 34.0 percent\n" + ] + } + ], + "source": [ + " import math #Example 11_12\n", + " \n", + " \n", + " #To compare the energy emitted per unit area of our body to with the same emissivity\n", + "t1=37.0 #Units in Centigrade\n", + "t1=273+t1 #Units in K\n", + "t2=15 #Units in Centigrade\n", + "t2=273+t2 #Units in K\n", + "tb_tc=(t1/t2)**4 #Units in terms of (Tb/Tc)**4\n", + "tb_tc=tb_tc*100 #In terms of percentage\n", + "print \"The radiation defers by \",round(tb_tc-100),\" percent\"\n", + "\n", + " #In textbook answer is printed wrong as 40% the correct answer is 34%\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.13:pg-310" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amount of heat lost is Q= 736363.6 J\n" + ] + } + ], + "source": [ + " import math #Example 11_13\n", + " \n", + " \n", + " #To findout how much heat is lost through it\n", + "a=15 #Unis in meter**2\n", + "t=30.0 #Units in K\n", + "R=2.2 #Units in Meter**2 K/W\n", + "q_t=(a*t)/R #Units in W\n", + "T=3600.0 #Units in sec\n", + "Q=q_t*T #Units in J\n", + "print \"The amount of heat lost is Q=\",round(Q,1),\" J\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |