{"id":390,"date":"2017-08-21T06:47:23","date_gmt":"2017-08-21T06:47:23","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=390"},"modified":"2020-02-03T17:49:18","modified_gmt":"2020-02-03T17:49:18","slug":"energy-work-and-heat-flow","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/energy-work-and-heat-flow\/","title":{"rendered":"Energy, Work and Heat Flow"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/cell-potential-electric-work-and-free-energy\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/hesss-law-and-gibbs-free-energy\">Next Lesson\u00a0\u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Energy, Work, and Heat Flow<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will review the equations for internal energy, and work relevant to a system and its surroundings.\u00a0We will also calculate heat flow and the energy of phase changes. Finally, we will perform calculations using the\u00a0Clausius-Clapeyron equation.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Introducing cell potential and tables of standard reduction potentials.<\/li>\n<li>Calculating emf using tables of standard reduction potentials and the Nernst Equation.<\/li>\n<li>Determining the relationship between cell potential, work and free energy.<\/li>\n<li>Using the equation <em>w = -qE<\/em>.<\/li>\n<li>Using the equation <em>\u0394G = -nFE<\/em>\u00b0.<\/li>\n<\/ul>\n<section>\n<h3>Overview<\/h3>\n<p>When looking at equilibrium,\u00a0the system is that part\u00a0of the universe\u00a0being studied. The surroundings describe\u00a0everything\u00a0else around the system being studied. The universe is defined as the\u00a0system and\u00a0its surroundings. Enthalpy\u00a0is the internal energy of a system plus the\u00a0pressure of the system\u00a0multiplied by the volume of the system. Entropy can be considered as the measure of disorder\u00a0or randomness\u00a0of a system. The free energy of a system is the minimum amount of\u00a0energy available when a\u00a0system is in equilibrium with its surroundings.<\/p>\n<h3>Internal Energy<\/h3>\n<p>The internal energy of a system is the sum of its\u00a0kinetic and potential\u00a0energies. It can be changed by work being done on or by the system and by heat being added or removed from the system. When considering such changes we\u00a0always assume the frame of reference of the system.<\/p>\n<p class=\"center\">\u0394<em>E<\/em> = <em>q<\/em> + <em>w<\/em><\/p>\n<p>\u0394<em>E<\/em> represents change\u00a0in the internal\u00a0energy of a system, <em>q<\/em> represents\u00a0heat, and <em>w <\/em>represents work.<\/p>\n<h4>Sample Problem<\/h4>\n<p>When 12.4 kJ of heat are added to a system on which 3.5\u00a0kJ of work is done,\u00a0what is the change in the internal energy of that system?<\/p>\n<p>Since \u0394<em>E<\/em> = <em>q<\/em> + <em>w, <\/em>the\u00a0change in the\u00a0internal energy of the system, is equal to 12.4 kJ + 3.5 kJ, or 15.9 kJ.<\/p>\n<h3>Work<\/h3>\n<p>A gas, by virtue of its compressibility, can have work done on it when the gas is compressed by an increase in\u00a0pressure. Work can defined as the pressure multiplied by a change\u00a0in volume, or <em>w<\/em> = <em>P<\/em>\u0394<em>V<\/em>. <em>w<\/em> and <em>P<\/em>\u0394<em>V\u00a0<\/em>will have opposite signs since\u00a0when a gas expands, work\u00a0is done by the gas on\u00a0the surroundings, and when a gas\u00a0contracts\u00a0work\u00a0is done on\u00a0the system. The equation becomes:<\/p>\n<p class=\"center\"><em>w<\/em> = \u2013<em>P<\/em>\u0394<em>V<\/em><\/p>\n<h4>Sample Problem<\/h4>\n<p>How much work is done when a gas expands from 23 L to 34\u00a0L at a constant\u00a0pressure of 3.2 atm?<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>w<\/em> = \u2013<em>P<\/em>\u0394<em>V<br \/>\n<\/em><em>w<\/em> = \u2013(3.2 atm) (34 L \u2013 23 L)<br \/>\n<em>w<\/em> = \u201335 atm <sup>.<\/sup> L<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Calculate \u0394<em>E<\/em> when 84.2 kJ\u00a0of work are\u00a0done on a system when 321 J of heat are added.<\/p>\n<ol>\n<li>405.2 kJ<\/li>\n<li>405.2 J<\/li>\n<li>84.5 kJ<\/li>\n<li>84.5 J<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The\u00a0correct answer is C. \u0394E is the sum of heat and work. The\u00a0units need to be the same, so the joules of heat needs to be divided by\u00a01000 to obtain kilojoules.<\/p>\n<p class=\"q-reveal center\" style=\"text-align: center;\">\u0394<em>E<\/em> = <em>q<\/em> + <em>w<\/em><br \/>\n\u0394<em>E<\/em> = [(321 J)\/1000] + 84.2 kJ<br \/>\n\u0394<em>E<\/em> = 84.5 kJ<\/p>\n<\/section>\n<p>&nbsp;<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Calculate the work done on a system when its volume\u00a0increases from 32.3\u00a0L to\u00a048.2 L at a constant pressure of 0.931 atm.<\/p>\n<ol>\n<li>14.8 atm<sup>.<\/sup>L<\/li>\n<li>-14.8 atm<sup>.<\/sup>L<\/li>\n<li>15.9 atm<sup>.<\/sup>L<\/li>\n<li>-15.9 atm<sup>.<\/sup>L<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The\u00a0correct answer is B. The change in volume is the final volume minus the\u00a0initial volume and then multiplied by the pressure, not divided.\u00a0Remember that:<\/p>\n<p class=\"center q-reveal\">w = \u2013<em>P<\/em>\u0394<em>V<\/em><strong>\u00b7<\/strong><em>w<\/em> = \u2013(0.931 atm)(48.2 L \u201332.3 L)w = \u201314.8 atm<strong>\u00b7<\/strong>L<\/p>\n<\/section>\n<h3>Calorimetry<\/h3>\n<p>Calorimetry is the science of measuring heat transfer from one object to another. The quantity of heat energy\u00a0transferred may be calculated using the equation, <em>Q<\/em> = <em>mC<sub>p<\/sub><\/em>\u0394<em>T<\/em>, where <em>Q\u00a0<\/em>is heat, <em>m<\/em> is the mass\u00a0of the substance, <em>C<sub>p<\/sub><\/em> the specific\u00a0heat\u00a0of the substance and \u0394<em>T<\/em> is\u00a0the\u00a0difference in temperature (final \u2013 initial).<\/p>\n<h4>Sample Problem 1<\/h4>\n<p>Calculate the final temperature\u00a0of a sample of Te (<em>C<sub>p<\/sub><\/em> = 0.201 J\/g\u00b0C) if 123.4 g\u00a0if its initial temperature is 56.7 \u00b0C and it\u00a0absorbs 8901 J of heat.<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>Q<\/em> = <em>m<\/em>C<sub>p<\/sub>(<em>T<\/em><sub>final\u00a0<\/sub>\u2212 <em>T<\/em><sub>initial<\/sub>)<\/p>\n<p class=\"center\" style=\"text-align: center;\">8901 J = (123.4 g)(0.201 J\/g C\u00b0) (<em>T<\/em><sub>final<\/sub> \u2212 56.7\u00b0C)<\/p>\n<p class=\"center\" style=\"text-align: center;\">8901 J = (24.8034 J\/\u00b0C)(<em>T<\/em><sub>final<\/sub> \u2212 56.7\u00b0C)<\/p>\n<p class=\"center\" style=\"text-align: center;\">358.9 \u00b0C\u00a0= (<em>T<\/em><sub>final<\/sub> \u2212 56.7\u00b0C)<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>T<\/em><sub>final<\/sub> = 415.6\u00b0C<\/p>\n<p>&nbsp;<\/p>\n<h4>Sample Problem 2<\/h4>\n<p>How much thermal energy is\u00a0released by 123 g of H<sub>2<\/sub>O when its\u00a0temperature decreases from 45.6\u00b0C to 7.89\u00b0C?<br \/>\nThe specific\u00a0heat of water is 4.184 J\/g\u00b0C.<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>Q<\/em> = <em>m<\/em>C<sub>p<\/sub>(<em>T<\/em><sub>final\u00a0<\/sub>\u2212 <em>T<\/em><sub>initial<\/sub>)<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>Q\u00a0<\/em>= (123 g)(4.184 J\/g\u00b0C)(7.89\u00a0\u00b0C \u2212 45.6 \u00b0C)<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>Q\u00a0<\/em>= (123 g)(4.184 J\/g\u00b0C)(-37.71\u00a0C\u00b0)<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>Q\u00a0<\/em>= \u201319406 J<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>Q<\/em>= \u201319.4 kJ<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>How much thermal energy is absorbed by 123.4 g H<sub>2<\/sub>O when\u00a0its temperature\u00a0increases from 56.7\u00b0C to 89.0\u00b0C?<\/p>\n<ol>\n<li>-46.0 kJ<\/li>\n<li>-16.7 kJ<\/li>\n<li>16.7 kJ<\/li>\n<li>46.0 kJ<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The\u00a0correct answer is C. The heat will be positive since the temperature\u00a0change is positive, and one must remember to calculate the change in\u00a0temperature and not use just the final temperature.<\/p>\n<p class=\"center q-reveal\"><em>Q<\/em> = <em>mc <\/em>(<em>T<\/em><sub>final <\/sub>\u2212 <em>T<\/em><sub>initial<\/sub>)<\/p>\n<p class=\"center q-reveal\">Q = (123.4 g)(4.184 J\/g\u00b0C)(89.0 C\u00b0 \u2212 56.7 C\u00b0)<br \/>\nQ = 16.7 kJ.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Calculate the specific heat of\u00a0a metal if 293 grams of the\u00a0metal release 282 J when it is cooled from 45.6\u00b0C\u00a0to 38.2\u00b0C.<\/p>\n<ol>\n<li>-0.130 J\/g\u00b0C<\/li>\n<li>0.130 J\/g\u00b0C<\/li>\n<li>0.0 &#8211; 252 J\/g\u00b0C<\/li>\n<li>-0.0252 J\/g\u00b0C<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The\u00a0correct answer is B. Specific heat for an object is always positive.\u00a0One must remember to use the change in temperature and not just the\u00a0final temperature.<\/p>\n<p class=\"q-reveal center\"><em>Q<\/em> = <em>mC<sub>p<\/sub><\/em>\u0394<em>T<\/em>,<br \/>\n28200 J = (293 g)(<em>C<sub>p<\/sub><\/em>)(38.2 \u00b0C \u2013 45.6 \u00b0C)<\/p>\n<p class=\"q-reveal\">Solving for <em>C<sub>p<\/sub><\/em>, the specific heat of the metal is: 0.130 J\/g\u00b0C<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>If 1234 J are absorbed by 567 g\u00a0of H<sub>2<\/sub>O at an initial\u00a0temperature of 8.90\u00b0C, what is the final\u00a0temperature?<\/p>\n<ol>\n<li>-8.38 \u00b0C<\/li>\n<li>-6.72 \u00b0C<\/li>\n<li>9.42 \u00b0C<\/li>\n<li>11.1 \u00b0C<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is C.\u00a0Remember the specific heat for water is 4.184 J\/g\u00b0C. The final\u00a0temperature will be positive:<\/p>\n<p class=\"q-reveal center\"><em>Q<\/em> = <em>mc<\/em>\u0394<em>T<\/em><br \/>\n1234 J = (567 g)(4.184 J\/g\u00b0C)(T<sub>final<\/sub> \u2013 8.90 \u00b0C)<\/p>\n<p class=\"q-reveal center\">T<sub>final<\/sub> = 9.42 \u00b0C<\/p>\n<\/section>\n<hr \/>\n<h3>Energy in Phase Changes<\/h3>\n<p>During a phase change,\u00a0the temperature of a substance does not change. The amount of heat transferred to the substance depends on the mass\u00a0of the material and what it is made of. Because there is no temperature change,\u00a0the heat equation becomes:<\/p>\n<p class=\"center\"><em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>fus<\/sub><\/p>\n<p><em>q<\/em> is heat measured in Joules, <em>m<\/em> is mass measured in grams, and \u0394<em>H<\/em><sub>fus <\/sub>is\u00a0the enthalpy of\u00a0fusion (if vaporization, it will read \u0394<em>H<\/em><sub>vap<\/sub>)\u00a0measured in kJ\/mol. This means the mass must be expressed in moles. If\u00a0the\u00a0value is in J\/g or kJ\/g the value need not be converted into moles.<\/p>\n<p>To determine the amount of energy required to change a given mass of solid to a gas, four separate heat transfer\u00a0calculations must be considered. They are:<\/p>\n<p class=\"center\"><em>q <\/em>= <em>mC<sub>p<\/sub><\/em>\u0394<em>T<\/em> for heating the solid<\/p>\n<p class=\"center\"><em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>fus <\/sub>from the\u00a0solid to liquid\u00a0phase<\/p>\n<p class=\"center\"><em>q <\/em>= <em>mC<sub>p<\/sub><\/em>\u0394<em>T<\/em> for heating the liquid, and<\/p>\n<p class=\"center\"><em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>vap<\/sub> for the phase\u00a0change\u00a0from liquid to gas<\/p>\n<h4>Sample Problem 1<\/h4>\n<p>Calculate the heat needed to melt 19.28 g sodium chloride (NaCl), if its enthalpy of fusion is 30.2 kJ\/mol.<\/p>\n<p class=\"center\"><em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>fus<\/sub><\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s8_004.gif\" width=\"384\" height=\"44\" \/><\/center><\/p>\n<p class=\"center\"><em>q<\/em> = (0.3299 mol NaCl)(30.2 J\/mol)<br \/>\n<em>q<\/em> = 9.96 kJ<\/p>\n<h4>Sample Problem 2<\/h4>\n<p>Determine the amount of energy required to convert 193.2 g ice at \u201325.0\u00b0C to steam at 125\u00b0C.<br \/>\nThe specific heat of ice is 2.06 J\/g\u00b0C.<br \/>\nThe specific heat for water is 4.184 J\/g\u00b0C.<br \/>\nThe specific heat for steam is 2.02 J\/g\u00b0C.<br \/>\nThe enthalpy of fusion for ice is 334 J\/g.<br \/>\nThe enthalpy of vaporization for water is 2260 J\/g.<\/p>\n<p class=\"center\">Heating of ice:<br \/>\n<em>q <\/em>= <em>m<em>C<sub>p<\/sub><\/em><\/em>\u0394<em>T<\/em><br \/>\n<em>q<\/em> = (193.2 g)(2.06 J\/g\u00b0C)(0\u00b0C\u2013 (-25.0\u00b0C))<br \/>\n<em>q<\/em> = 9949.8 J<\/p>\n<p class=\"center\">Ice \u2192 water<br \/>\n<em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>fus<br \/>\n<\/sub><em>q<\/em> = (193.2 g)(334 J\/g)<br \/>\n<em>q<\/em> = 64528.8 J<\/p>\n<p class=\"center\">Heating of water<br \/>\n<em>q<\/em>= <em>mc<\/em>\u0394<em>T<\/em><br \/>\n<em>q<\/em> = (193.2 g)(4.184 J\/g\u00b0C)(100\u00b0C\u2013 0\u00b0C)<br \/>\n<em>q<\/em> = 80834.88 J<\/p>\n<p class=\"center\">Water \u2192 steam<br \/>\n<em>q <\/em>= <em>m<\/em>\u0394<em>H<\/em> vap<br \/>\n<em>q<\/em> = (193.2 g)(2260 J\/g)<br \/>\n<em>q<\/em> = 436632 J<\/p>\n<p class=\"center\">Heating the steam<br \/>\n<em>q <\/em>= <em>m<em>C<sub>p<\/sub><\/em><\/em>\u0394<em>T<\/em><br \/>\n<em>q<\/em> = (193.2 g)(2.02 J\/g\u00b0C)(125\u00b0C \u2013 100\u00b0C)<br \/>\n<em>q<\/em> = 9756.6 J<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 1<\/h4>\n<p>How much energy is needed to\u00a0melt 13.57 g sodium fluoride\u00a0(NaF)? The enthalpy of fusion for sodium fluoride is 29.3 kJ\/mol.<\/p>\n<ol>\n<li>9.47 kJ<\/li>\n<li>17.3 kJ<\/li>\n<li>20.9 kJ<\/li>\n<li>397.0 kJ<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is A. Choices B and C show incorrect mole\u00a0conversions, while choice D is obtained by failure to convert to moles.<\/p>\n<p class=\"center q-reveal\"><em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>fus<br \/>\n<\/sub><em>q<\/em> = (0.3232 mol)(29.3 kJ\/mol)<br \/>\n<em>q<\/em> = 9.47 kJ<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question 2<\/h4>\n<p>How much energy is needed to change 4392 g\u00a0ice at \u20137.5\u00b0C\u00a0into steam at 103.4\u00b0C?<\/p>\n<p>The specific heat of ice is 2.06 J\/g\u00b0C.<br \/>\nThe\u00a0specific heat for water is\u00a04.184 J\/g\u00b0C.<br \/>\nThe specific heat for steam is 2.02 J\/g\u00b0C.<br \/>\nThe enthalpy of\u00a0fusion for ice is 334 J\/g.<br \/>\nThe enthalpy of vaporization for water is\u00a02260 J\/g.<\/p>\n<ol>\n<li>9.62\u00d710<sup>3<\/sup> kJ<\/li>\n<li>1.04\u00d710<sup>4<\/sup> kJ<\/li>\n<li>1.33\u00d710<sup>4<\/sup> kJ<\/li>\n<li>1.46\u00d710<sup>4<\/sup> kJ<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is C. The others reflect erroneous conversions to kJ or incorrect multiplications.<\/p>\n<p class=\"center q-reveal\">Heating of ice<em><br \/>\nq <\/em>= <em>mc<\/em>\u0394<em>T<\/em><br \/>\n<em>q <\/em>= (4392 g)(2.06 J\/g \u00b0C)(0 \u00b0C\u2013 (-7.5 \u00b0C))<br \/>\n<em>q <\/em>= 67856.4 J<\/p>\n<p class=\"center q-reveal\">Ice \u2192 water<br \/>\n<em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>fus<\/sub><br \/>\n<em>q<\/em> = (4392 g)(334 J\/g)<br \/>\n<em>q<\/em> = 1466928 J<\/p>\n<p class=\"center q-reveal\">Heating of water<br \/>\n<em>q <\/em>= <em>mc<\/em>\u0394<em>T<\/em><br \/>\n<em>q <\/em>= (4392 g)(4.184 J\/g \u00b0C)(100 \u00b0C\u2013 0 \u00b0C)<br \/>\n<em>q<\/em> = 1837612.8 J<\/p>\n<p class=\"center q-reveal\">Water \u2192 steam<\/p>\n<p class=\"center q-reveal\"><em>q<\/em> = <em>m<\/em>\u0394<em>H<\/em><sub>vap<\/sub><br \/>\n<em>q<\/em> = (4392 g)(2260 J\/g)<br \/>\n<em>q<\/em> = 9,925,920 J<\/p>\n<p class=\"center q-reveal\">Heating the steam<br \/>\n<em>q <\/em>= <em>mc<\/em>\u0394<em>T<\/em><br \/>\n<em>q <\/em>= (4392 g)( 2.02 J\/g \u00b0C)(103.4 \u00b0C \u2013 100 \u00b0C)<br \/>\n<em>q<\/em> = 30164.256 J<\/p>\n<p class=\"center q-reveal\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s8_005.gif\" width=\"540\" height=\"66\" \/><\/p>\n<\/section>\n<h3>Clausius-Clapeyron Equation<\/h3>\n<p>The Clausius-Clapeyron equation provides a method to\u00a0determine vapor\u00a0pressure or temperature using the heat of vaporization. It can also be\u00a0used to\u00a0obtain values for \u0394<em>H<\/em><sub>vap<\/sub>.<\/p>\n<p>The equation is:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s8_015.gif\" width=\"164\" height=\"50\" \/><\/center><em> R<\/em> is\u00a0the ideal gas law\u00a0constant 8.314 J\/mol<strong><sup>.<\/sup><\/strong>K. If results are to be\u00a0expressed in kJ, the\u00a0value will need to be divided by 1000. <em>P<\/em><sub>1\u00a0<\/sub>and <em>P<\/em><sub>2\u00a0<\/sub>are the pressures at two temperatures <em>T<\/em><sub>1\u00a0<\/sub>and <em>T<\/em><sub>2<\/sub>.\u00a0Those temperatures must be expressed in Kelvins.<\/p>\n<p>&nbsp;<\/p>\n<h4>Sample Problem<\/h4>\n<p>Determine the normal boiling point for a substance with\u00a0a vapor pressure of\u00a01.02 atm at a temperature of 3.5\u00b0C. The heat of\u00a0vaporization for the\u00a0substance, methane, is 8.2 kJ\/mol.<\/p>\n<p><em>Hint: Remember the normal boiling point occurs at 1 atm.<\/em><\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s8_015.gif\" width=\"164\" height=\"50\" \/><\/center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s8_017.gif\" width=\"374\" height=\"176\" \/><\/p>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/cell-potential-electric-work-and-free-energy\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/hesss-law-and-gibbs-free-energy\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson\u00a0\u27a1 Energy, Work, and Heat Flow Objective In this lesson we will review the equations for internal energy, and work relevant to a system and its surroundings.\u00a0We will also calculate heat flow and the energy of phase changes. Finally, we will perform calculations using the\u00a0Clausius-Clapeyron equation. Previously we covered&#8230; Introducing cell [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-390","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/390","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=390"}],"version-history":[{"count":30,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/390\/revisions"}],"predecessor-version":[{"id":415,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/390\/revisions\/415"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=390"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}