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author | kinitrupti | 2017-05-12 18:40:35 +0530 |
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committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb | |
parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
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diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb deleted file mode 100755 index 0781dd9b..00000000 --- a/Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb +++ /dev/null @@ -1,493 +0,0 @@ -{ - "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": 3, - "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": [ - " #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": 4, - "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": [ - " #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": 5, - "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": [ - " #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": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The specific heat of metal is Cm= 0.216 cal/gm C\n" - ] - } - ], - "source": [ - " #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": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The time taken is t= 449.0 sec\n" - ] - } - ], - "source": [ - " #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": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "the rise in temperature is DeltaT= 38.6 C\n" - ] - } - ], - "source": [ - " #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": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The slab is longer by= 0.01 meters\n" - ] - } - ], - "source": [ - " #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": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The new diameter of the hole is= 2.0076 cm\n" - ] - } - ], - "source": [ - " #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": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The change in benzene volume is V30= 101.253 cm**3\n" - ] - } - ], - "source": [ - " #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": 13, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The ice melts by 116.0 gm\n" - ] - } - ], - "source": [ - " #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": 15, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The radiation defers by 34.0 percent\n" - ] - } - ], - "source": [ - " #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": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The amount of heat lost is Q= 736363.6 J\n" - ] - } - ], - "source": [ - " #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 -} |