path: root/Heat_And_Thermodynamics/ch3.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3 : The mechanical equivalent of  heat"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1 pageno : 44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "m = 20;\t\t\t#calorimeter of water equivalent in gm\n",
      "n = 1030;\t\t\t#weight of water in gm\n",
      "p = 2;\t\t\t#no.of paddles\n",
      "a = 10;\t\t\t#weight of each paddle in kg\n",
      "s = 80;\t\t\t#dismath.tance between paddles in m\n",
      "g = 980;\t\t\t#accelaration due to gravity in cm/sec**2\n",
      "\n",
      "# Calculations\n",
      "E = (p*a*1000*g*s*100);\t\t\t#potential energy in dyne cm\n",
      "T = (E)/(1050*4.18*10**7);\t\t\t#rise in temperature in deg.C\n",
      "\n",
      "# Result\n",
      "print 'the rise in temperature of water is %3.2f deg.C'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rise in temperature of water is 3.57 deg.C\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.2 pageno : 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "cp = 0.1;\t\t\t#specific heat of copper in kj/kg-K\n",
      "w = 120;\t\t\t#weight of copper calorimeter in gm\n",
      "a = 1400;\t\t\t#weight of paraffin oil in gm\n",
      "cp1 = 0.6;\t\t\t#specific of parafin oil in kj/kg-K\n",
      "b = 10**8;\t\t\t#force to rotate the paddle in dynes\n",
      "T = 16;\t\t\t#rise in temperature in deg.C\n",
      "n = 900;\t\t\t#no.of revolutions stirred \n",
      "pi = 3.14;\t\t\t#value of pi\n",
      "\n",
      "# Calculations\n",
      "c = 2*pi*b;\t\t\t#work done by a rotating paddle per rotation in dyne cm per rotation\n",
      "d = c*n;\t\t\t#total work done in dyne cm \n",
      "hc = w*cp*16;\t\t\t#heat gained by calorimeter in calories\n",
      "hp = a*cp1*16;\t\t\t#heat gaained by paraffin oil in calories \n",
      "J = d/(hc+hp);\t\t\t#mecanical equivalent of heat in erg/cal\n",
      "\n",
      "# Result\n",
      "print 'mecanical equivalent of heat is %.2e erg/cal'%(J)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "mecanical equivalent of heat is 4.15e+07 erg/cal\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3 pageno : 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables \n",
      "cp = 0.12;\t\t\t#specific heat of iron in kj/kg-K\n",
      "m = 25;\t\t\t#mass of iron in lb\n",
      "h = 0.4;\t\t\t#horse power developed in 3 min\n",
      "t = 3;\t\t\t#time taken to develop the horse power in min\n",
      "T = 17;\t\t\t#raise in temp in deg.C\n",
      "\n",
      "# Calculations\n",
      "w = h*33000*t;\t\t\t#total work done in ft-lb\n",
      "H = m*cp*T;\t\t\t#aount of heat developed in B.Th.U\n",
      "J = (w)/H;\t\t\t#the value of mechanical equivalent of heat\n",
      "\n",
      "# Result\n",
      "print 'the mechanical equivalent of water is %3.1f ft-lb/B.Th.U'%(J)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mechanical equivalent of water is 776.5 ft-lb/B.Th.U\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4 pageno : 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables \n",
      "n = 2.;\t\t\t#no.of lead blocks\n",
      "m = 210.;\t\t\t#mass of each lead block in gm\n",
      "v = 20000.;\t\t\t#velocity of block relative to earth in cm/sec\n",
      "J = 4.2*10**7;\t\t\t#mechanical equivalent of heat in ergs/calorie\n",
      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg-K\n",
      "\n",
      "# Calculations\n",
      "E = (m*v**2)/2;\t\t\t#kinetic energy of each block in ergs\n",
      "E2 = n*E;\t\t\t#total kinetic energy in ergs\n",
      "T = E2/(J*m*n*cp);\t\t\t#mean rise in temperature in T\n",
      "\n",
      "# Result\n",
      "print 'the mean rise in temperature is %3.1f deg.C'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mean rise in temperature is 158.7 deg.C\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5 pageno : 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables \n",
      "h = 150;\t\t\t#height froom which ball fallen in ft\n",
      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg-K\n",
      "J = 778;\t\t\t#mechanical equivalent of heat in ft lb/B.Th.U\n",
      "\n",
      "# Calculations\n",
      "#work done in falling is equal to heat absorbed by the ball\n",
      "T = 160./(J*cp)*(5./9);\t\t\t#the raise in temperature in T\n",
      "\n",
      "# Result\n",
      "print 'the raise in temperature is %3.1f deg.C'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the raise in temperature is 3.8 deg.C\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.6 pageno : 46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "# Variables \n",
      "w = 26.6;\t\t\t#work done one horse in to raise the temperature in lb\n",
      "T1 = 32.;\t\t\t#temperature at initial in deg.F\n",
      "T2 = 212.;\t\t\t#temperature at final in deg.F\n",
      "t = 2.5;\t\t\t#time to raise the tmperature in hrs\n",
      "p = 25.;\t\t\t#percentage of heat lossed \n",
      "\n",
      "# Calculations\n",
      "#only 75% of heat is utillised\n",
      "x = w*180.*100.*778./((100-p)*150);\t\t\t#the rate at which horse worked\n",
      "\n",
      "# Result\n",
      "print 'the rate at which horse worked is %3.0f ft-lb wt/min'%(x)\n",
      "print \"Note : Answer in book is rounded off, Please calculate manually. This answer is accurate.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rate at which horse worked is 33112 ft-lb wt/min\n",
        "Note : Answer in book is rounded off, Please calculate manually. This answer is accurate.\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.7 pageno : 46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables \n",
      "l = 100.;\t\t\t#length of glass tube in cm\n",
      "m = 500.;\t\t\t#mass of mercury in glass tube in gm\n",
      "n = 20.;\t\t\t#number of times inverted i succession\n",
      "cp = 0.03;\t\t\t#specific heat of mercury in cal/gm/deg.C\n",
      "J = 4.2;\t\t\t#joule's equivalent in j/cal\n",
      "g = 981.;\t\t\t#accelaration due to gravity in cm/s**2\n",
      "\n",
      "# Calculations\n",
      "PE = m*g*l;\t\t\t#potential energy for each time in ergs\n",
      "TE = PE*n;\t\t\t#total loss in ergs\n",
      "T = TE/(m*cp*J*10**7);\t\t\t#rise in temperature in deg.C\n",
      "#if T is the rise in temperature,then heat devoloped is m*cp*T\n",
      "\n",
      "# Result\n",
      "print 'the rise in temperature is %3.2f deg.C'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rise in temperature is 1.56 deg.C\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.8 page no : 46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# Variables \n",
      "d = 0.02;\t\t\t#diameter of the copper wire in cm\n",
      "i = 1;\t\t\t#current in amp\n",
      "T = 100;\t\t\t#maximum steady temperature in deg.C\n",
      "r = 2.1;\t\t\t#resistance of the wire in ohm cm\n",
      "J = 4.2;\t\t\t#mechanical equivalent of heat in j/cal\n",
      "a = 3.14*d**2/4;\t\t\t#area of the copper wire in sq.cm\n",
      "a2 = 1;\t\t\t#area of the copper surface in sq.cm\n",
      "\n",
      "# Calculations \n",
      "l = 1/(2*3.14*d/2);\t\t\t#length corresponding to the area in cm\n",
      "R = r*l/a;\t\t\t#resistance of the copper wire in ohm\n",
      "w = R*a2**2;\t\t\t#work done in joule\n",
      "h = w/J;\t\t\t#heat devoleped in cal\n",
      "\n",
      "# Result\n",
      "print 'the heat developed is %.f calories'%(round(h,-1))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the heat developed is 25360 calories\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.9 pageno: 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "h = 10000;\t\t\t#vertical height of water fall in cm\n",
      "v = 5;\t\t\t    #volume disharged per sec in litres\n",
      "J = 4.18;\t\t\t#joule's constant in j/cal\n",
      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
      "\n",
      "# Calculations\n",
      "m = v*1000;\t\t\t#mass of water disharged per sec in gm\n",
      "w = m*h*g;\t\t\t#work done in falling through 100m in erg\n",
      "H = (v*10**7 *g)/(J*10**7);\t#quantity of heat produced in cal\n",
      "T = H/m;\t\t\t#rise in temperature in deg.C\n",
      "\n",
      "# Result\n",
      "print 'the quantity of heat produced is %3f cal  \\\n",
      "\\nthe rise in temperature is %3.2f deg.C'%(H,T)\n",
      "\n",
      "print \"Note : Answer for part A in book is wrong. Please calculate manually.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the quantity of heat produced is 1173.444976 cal  \n",
        "the rise in temperature is 0.23 deg.C\n",
        "Note : Answer for part A in book is wrong. Please calculate manually.\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10 page no : 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# Variables \n",
      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg.k\n",
      "v = 10000;\t\t\t#initial velocity of bullet in cm/sec\n",
      "J = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
      "\n",
      "# Calculations\n",
      "ke = (v**2)/2;\t\t\t#kinetic energy of the bullet per unit mass in (cm/sec)**2\n",
      "T = ke*95/(cp*J*100);\t\t\t#rise in temperature in deg.C\n",
      "\n",
      "# Result\n",
      "print 'the rise in temperature is %3.1f deg.C'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rise in temperature is 37.7 deg.C\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.11 page no : 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables \n",
      "h = 5000.;\t\t\t#height of the niagara falls in cm\n",
      "J = 4.2*10**7;\t\t#joules constant in ergs per cal\n",
      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
      "\n",
      "#CALCULATIONS\n",
      "w = h*g;\t\t\t#work done per unit mass in ergs/gn\n",
      "T = w/J;\t\t\t#rise in temperature in deg.C\n",
      "\n",
      "# Result\n",
      "print 'the rise in temperature is %3.2f deg.C'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rise in temperature is 0.12 deg.C\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.12 page no : 48\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "# Variables \n",
      "E1 = 3.75;\t\t\t#potential difference in v\n",
      "E2 = 3.;\t\t\t#potential differnce in v\n",
      "i1 = 2.5;\t\t\t#current in amp\n",
      "i2 = 2;\t\t\t    #current in amp\n",
      "T = 2.7;\t\t\t#the rise in temperature of the water in deg.C\n",
      "m1 = 48.;\t\t\t#water flow rate at 3 volts in gm/min\n",
      "m2 = 30.;\t\t\t#water flow rate at 3.75volts in gm/min\n",
      "s = 1;\t\t\t    #specific heat of the water kj/kg-K\n",
      "\n",
      "# Calculations\n",
      "J = (E1*i1-E2*i2)/(s*T*(m1-m2)/60);\t\t\t#the mechanical equivalent in j/cal\n",
      "\n",
      "# Result\n",
      "print 'the mechanical equivalent is %3.3f j/cal'%(J)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mechanical equivalent is 4.167 j/cal\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.13 page no : 48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# Variables \n",
      "R = 64*10**7;\t\t\t#mean radius of the earth in cm\n",
      "cp = 0.15;\t\t\t#specific heat of earth in kj/kg-K\n",
      "J = 4.2*10**7;\t\t\t#joules consmath.tant in erg/cal\n",
      "\n",
      "# Calculations\n",
      "i = 2./5*R**2;\t\t\t#moment of inertia of the earth per unit mass in joules\n",
      "w = (2*3.14)/(24*60*60);\t\t\t#angular velocity of the earth in rad/sec\n",
      "T = (i*w**2)/(2*J*cp);\t\t\t#rise in temperature in deg.C\n",
      "\n",
      "# Result\n",
      "print 'the rise in the temperature is %.1f deg C'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rise in the temperature is 68.7 deg C\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.14 page no : 49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables \n",
      "cp = 1.25;\t\t\t#specific heat of helium inkj/kg-K\n",
      "v = 1000;\t\t\t#volume of the gas in ml\n",
      "w = 0.1785;\t\t\t#mass of the gas at N.T.P in gm\n",
      "p = 76*13.6*981;\t#pressure of the gas at N.T.P in dynes\n",
      "T = 273;\t\t\t#temperature at N.T.P in K\n",
      "\n",
      "# Calculations\n",
      "V = 1000/w;\t\t\t#volume occupied by the 1gm of helium gas in cc\n",
      "cv = cp/1.66;\t\t#specific heat at constant volume it is monatomuc gas kj/kg-K\n",
      "r = p*V/T;\t\t\t#gas constant in cm**3.atm./K.mol\n",
      "J = r/(cp-cv);\t\t#mechanical equivalent of heat in erg/cal\n",
      "\n",
      "# Result\n",
      "print 'the mechanical equivalent of heat is %.2e ergs/calories'%(J)\n",
      "print \"Note: answer slightly different because of rounding error.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mechanical equivalent of heat is 4.19e+07 ergs/calories\n",
        "Note: answer slightly different because of rounding error.\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "\n",
      "Example 3.15 pageno : 49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables \n",
      "n = 1./273;     \t\t\t#coefficent of expaaansion of air\n",
      "a = 0.001293;\t    \t\t#density of air in gm/cc\n",
      "cp = 0.2389;\t\t    \t#specific heat at consmath.tant pressure in kj/kg.K\n",
      "p = 76*13.6*981;\t\t\t#pressure at 0 deg.C in dynes\n",
      "\n",
      "# Calculations\n",
      "J = (p*n)/(a*(cp-(cp/1.405)));\t\t\t#mechanical equivalent of heat\n",
      "\n",
      "# Result\n",
      "print 'mechanical equivalent of heat is %.2e ergs/cal'%(J)\n",
      "print \"Note: answer slightly different because of rounding error.\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "mechanical equivalent of heat is 4.17e+07 ergs/cal\n",
        "Note: answer slightly different because of rounding error.\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.16 pageno : 49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "# Variables \n",
      "r = 120./60;\t\t\t#rate of flow of water in gm/sec\n",
      "T1 = 27.30;\t\t\t#temperature at initial in deg.C\n",
      "T2 = 33.75;\t\t\t#temperature at final in deg.C\n",
      "v = 12.64;\t\t\t#potential drop in volts\n",
      "s = 1.; \t\t\t#specific heat of water in kj/kg-K\n",
      "i = 4.35;\t\t\t#current through the heating element in amp\n",
      "\n",
      "# Calculations\n",
      "J = (v*i)/(r*s*(T2-T1));\t\t\t#the mechanical equivalent of heat in joule/calorie\n",
      "\n",
      "# Result\n",
      "print 'the mechanical equivalent of heat is %3.2f j/cal'%(J)\n",
      "print \"Note: answer slightly different because of rounding error.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mechanical equivalent of heat is 4.26 j/cal\n",
        "Note: answer slightly different because of rounding error.\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.17 page no : 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# Variables \n",
      "cp = 6.865;\t\t\t#molar specific heat of hydrogen at consmath.tant pressure in kj/kg-K\n",
      "cv = 4.880;\t\t\t#molar specific heat of hydrogen at consmath.tant volume in kj/kg-K\n",
      "p = 1.013*10**6;\t\t\t#atmospheric pressure in dynes/cm**2\n",
      "v = 22.4*10**3;\t\t\t#gram molar volume in ml\n",
      "T = 273;\t\t\t#temperature at N.T.P in kelvins\n",
      "\n",
      "# Calculations\n",
      "J = (p*v)/(T*(cp-cv));\t\t\t#mechanical equivalent of heat\n",
      "\n",
      "# Result\n",
      "print 'the mechanical equivalent of heat is %.2e ergs/cal'%(J)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mechanical equivalent of heat is 4.19e+07 ergs/cal\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.18 page no : 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "# Variables\n",
      "v = 1000.;\t\t\t#volume of hydrogen in ml\n",
      "t = 273.;\t\t\t#tempature of hydrogen in kelvin\n",
      "p = 76.;\t\t\t#pressure of hydrogen in mm of hg\n",
      "w = 0.0896;\t\t\t#weigh of hydrogen in gm\n",
      "cp = 3.409;\t\t\t#specific heat of hydogen in kj/kg-K\n",
      "cv = 2.411;\t\t\t#specific heat of hydrogen in kj/kg-K\n",
      "g = 981.;\t\t\t#accelaration due to gravity in cm/sec**2\n",
      "a = 13.6;\t\t\t#density of mercury in gm/cm**2\n",
      "\n",
      "# Calculations\n",
      "J = (p*v*g*a)/(w*t*(cp-cv));\t\t\t#mechanical equivalent of heat in ergs/cals\n",
      "\n",
      "# Result\n",
      "print 'mechanical equivalent of heat is %.2e ergs/calorie'%(J)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "mechanical equivalent of heat is 4.15e+07 ergs/calorie\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.19 page no : 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "cp = 0.23;\t\t\t#specific heat at constant pressure in kj/kg-K\n",
      "a = 1.18;\t\t\t#density of air in gm/lit\n",
      "J = 4.2*10**7;\t\t\t#mechanical equivalent of heat in ergs/cal\n",
      "t = 300;\t\t\t#temperature of air in kelvin\n",
      "p = 73*13.6*981;\t\t\t#pressure of air in dynes\n",
      "\t\t\t#cp-cv = (r/J) = pv/(tj)\n",
      "\n",
      "#CALCULATON\n",
      "cv = cp-(p*1000/(a*t*J));\t\t\t#specific heat at constant volume in calories\n",
      "\n",
      "# Result\n",
      "print 'the specific heat at constant volume is %.4f calories'%(cv)\n",
      "print \"Note: answer slightly different because of rounding error.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the specific heat at constant volume is 0.1645 calories\n",
        "Note: answer slightly different because of rounding error.\n"
       ]
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.20 pageno : 51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "t1 = 0;\t\t\t#temperature of water in deg.C\n",
      "t2 = 0;\t\t\t#temperature of ice in deg.C\n",
      "J = 4.18*10**7;\t\t\t#the joules thomson coefficent in erg/cal\n",
      "l = 80;\t\t\t#latent heat og fusion kj/kg\n",
      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
      " \n",
      "# Calculations\n",
      "h = l*J/(15*g);\t\t\t#height from which ice has fallen\n",
      "\n",
      "# Result\n",
      "print 'the height from which ice has fallen is %.2e cm'%(h)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the height from which ice has fallen is 2.27e+05 cm\n"
       ]
      }
     ],
     "prompt_number": 30
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.21 page no : 51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "# Variables\n",
      "T = 80;\t\t\t#temperature of bullet in deg.C\n",
      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg-K\n",
      "J = 4.2;\t\t\t#mechanical equivalent of heat in j/cal\n",
      "\n",
      "# Calculations\n",
      "h = T*cp;\t\t\t#heat developed per unit mass in calorie\n",
      "v = (J*10**7*h*2/0.9)**0.5;\t\t\t#velocity of bullet in cm/sec\n",
      "\n",
      "# Result\n",
      "print 'the velocity of bullet is %.1e cm/sec'%(v)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the velocity of bullet is 1.5e+04 cm/sec\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.22 pageno : 51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "w = 5.0;\t\t\t#weight of lead ball in lb\n",
      "cp = 0.032;\t\t\t#specific heat of lead in Btu/lbdeg.F\n",
      "h = 50;\t\t\t#height at which ball thrown in feets\n",
      "v = 20;\t\t\t#vertical speed in ft/sec\n",
      "g = 32;\t\t\t#accelararion due to gravity in ft/sec**2\n",
      "\n",
      "# Calculations\n",
      "u = (v**2)+2*g*h\n",
      "ke = (w/2*(u));\t\t\t#kinetic energy of the ball at ground\n",
      "T = ke/(2*32*778*w*cp);\t\t\t#rise of temperature in deg.F\n",
      "\n",
      "# Result\n",
      "print 'the rise in temperature is %.1f deg.F'%(T)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rise in temperature is 1.1 deg.F\n"
       ]
      }
     ],
     "prompt_number": 32
    }
   ],
   "metadata": {}
  }
 ]
}