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-{
- "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
-}