{"id":7384,"date":"2021-06-08T21:56:01","date_gmt":"2021-06-08T21:56:01","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/work-and-heat\/"},"modified":"2021-09-24T20:59:24","modified_gmt":"2021-09-24T20:59:24","slug":"work-and-heat","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/work-and-heat\/","title":{"raw":"Work and Heat","rendered":"Work and Heat"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Define a type of work in terms of pressure and volume.<\/li>\r\n \t
  2. Define heat<\/em>.<\/li>\r\n \t
  3. Relate the amount of heat to a temperature change.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nWe have already defined work as a force acting through a distance. It turns out that there are other equivalent definitions of work that are also important in chemistry.\r\n\r\nWhen a certain volume of a gas expands, it works against an external pressure to expand (see Figure 7.5 \"Volume versus Pressure\"). That is, the gas must perform work. Assuming that the external pressure P<\/i>ext<\/sub> is constant, the amount of work done by the gas is given by the following equation:\r\n

    [latex]w=-P_{\\text{ext}}\\times \\Delta V[\/latex]<\/p>\r\nIn which \u0394V<\/i> is the change in volume of the gas. This term is always the final volume minus the initial volume, as shown here:\r\n

    [latex]\\Delta V=V_{\\text{final}}-V_{\\text{initial}}[\/latex]<\/p>\r\n\u0394V<\/i> can be positive or negative, depending on whether V<\/i>final<\/sub> is larger (is expanding) or smaller (is contracting) than V<\/i>initial<\/sub>. The negative sign in the equation for work is important and implies that as volume expands (\u0394V<\/i> is positive), the gas in the system is losing<\/em> energy as work. On the other hand, if the gas is contracting, \u0394V<\/i> is negative, and the two negative signs make the work positive, so energy is being added to the system.\r\n\r\n[caption id=\"attachment_349\" align=\"aligncenter\" width=\"300\"]\"\" Figure 7.5 \"Volume versus Pressure.\" When a gas expands against an external pressure, the gas does work.[\/caption]\r\n\r\nFinally, let us consider units. Volume changes are usually expressed in units like litres, while pressures are usually expressed in atmospheres. When we use the equation to determine work, the unit for work comes out as litre\u00b7atmospheres, or L\u22c5atm. This is not a very common unit for work. However, there is a conversion factor between L\u22c5atm and the common unit of work, joules:\r\n

    [latex]1\\text{ L}\\cdot \\text{atm}=101.32\\text{ J}[\/latex]<\/p>\r\nUsing this conversion factor and the previous equation for work, we can calculate the work performed when a gas expands or contracts.\r\n

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    Example 7.11<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nWhat is the work performed by a gas if it expands from 3.44 L to 6.19 L against a constant external pressure of 1.26 atm? Express the final answer in joules.\r\n

    Solution<\/h2>\r\nFirst we need to determine the change in volume, \u0394V<\/i>. A change is always the final value minus the initial value:\r\n

    [latex]\\Delta V = V_{\\text{final}}-V_{\\text{initial}}=6.19\\text{ L}-3.44\\text{ L}=2.75\\text{ L}[\/latex]<\/p>\r\nNow we can use the definition of work to determine the work done:\r\n

    [latex]w=-P_{\\text{ext}}\\cdot \\Delta V=-(1.26\\text{ atm})(2.75\\text{ L})=-3.47\\text{ L}\\cdot \\text{atm}[\/latex]<\/p>\r\nNow we construct a conversion factor from the relationship between litre\u00b7atmospheres and joules:\r\n

    [latex]-3.47\\text{ }\\cancel{\\text{L}\\cdot\\text{atm}}\\times \\dfrac{101.32\\text{ J}}{1\\text{ }\\cancel{\\text{L}\\cdot\\text{atm}}}=-351\\text{ J}[\/latex]<\/p>\r\nWe limit the final answer to three significant figures, as appropriate.\r\n

    Test Yourself<\/h1>\r\nWhat is the work performed when a gas expands from 0.66 L to 1.33 L against an external pressure of 0.775 atm?\r\n

    Answer<\/h2>\r\n\u221253 J\r\n\r\n<\/div>\r\n<\/div>\r\nHeat is another aspect of energy. [pb_glossary id=\"8110\"]Heat[\/pb_glossary]\u00a0is the transfer of energy from one body to another due to a difference in temperature. For example, when we touch something with our hands, we interpret that object as either hot or cold depending on how energy is transferred: If energy is transferred into your hands, the object feels hot. If energy is transferred from your hands to the object, your hands feel cold. Because heat is a measure of energy transfer, heat is also measured in joules.\r\n\r\nFor a given object, the amount of heat (q<\/i>) involved is proportional to two things: the mass of the object (m<\/i>) and the temperature change (\u0394T<\/i>) evoked by the energy transfer. We can write this mathematically as:\r\n

    [latex]q\\propto m\\times \\Delta T[\/latex]<\/p>\r\nwhere \u221d means \u201cis proportional to.\u201d To make a proportionality an equality, we include a proportionality constant. In this case, the proportionality constant is labelled c<\/i> and is called the [pb_glossary id=\"8111\"]specific heat capacity[\/pb_glossary], or, more succinctly, specific heat:\r\n

    [latex]q=mc\\Delta T[\/latex]<\/p>\r\nwhere the mass, specific heat, and change in temperature are multiplied together. Specific heat is a measure of how much energy is needed to change the temperature of a substance; the larger the specific heat, the more energy is needed to change the temperature. The units for specific heat are [latex]\\dfrac{\\text{J}}{\\text{g}\\cdot \\celsius}[\/latex] or [latex]\\dfrac{\\text{J}}{\\text{g}\\cdot \\text{K}}[\/latex], depending on what the unit of \u0394T<\/i> is. You may note a departure from the insistence that temperature be expressed in Kelvin. That is because a change in temperature has the same value, whether the temperatures are expressed in degrees Celsius or kelvins.\r\n

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    Example 7.12<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nCalculate the heat involved when 25.0 g of Fe increase temperature from 22\u00b0C to 76\u00b0C. The specific heat of Fe is 0.449 [latex]\\dfrac{\\text{J}}{\\text{g}\\cdot \\celsius}[\/latex].\r\n

    Solution<\/h2>\r\nFirst we need to determine \u0394T<\/i>. A change is always the final value minus the initial value:\r\n

    [latex]\\Delta T = 76\\celsius -22\\celsius=54\\celsius[\/latex]<\/p>\r\nNow we can use the expression for q<\/i>, substitute for all variables, and solve for heat:\r\n

    [latex]q=(25.0\\text{ \\cancel{g}})\\left(0.449\\cdot \\dfrac{\\text{J}}{\\cancel{\\text{g}}\\cdot\\cancel{\\celsius}}}\\right)(54\\cancel{\\celsius})=610\\text{ J}[\/latex]<\/p>\r\nNote how the g and \u00b0C units cancel, leaving J, a unit of heat. Also note that this value of q<\/i> is inherently positive, meaning that energy is going into the system.\r\n

    Test Yourself<\/h1>\r\nCalculate the heat involved when 76.5 g of Ag increase temperature from 17.8\u00b0C to 144.5\u00b0C. The specific heat of Ag is 0.233 [latex]\\dfrac{\\text{J}}{\\text{g}\\cdot \\celsius}[\/latex].\r\n

    Answer<\/h2>\r\n2,260 J\r\n\r\n<\/div>\r\n<\/div>\r\nAs with any equation, when you know all but one variable in the expression for q<\/i>, you can determine the remaining variable by using algebra.\r\n
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    Example 7.13<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nIt takes 5,408 J of heat to raise the temperature of 373 g of Hg by 104\u00b0C. What is the specific heat of Hg?\r\n

    Solution<\/h2>\r\nWe can start with the equation for q<\/i>, but now different values are given, and we need to solve for specific heat. Note that \u0394T<\/i> is given directly as 104\u00b0C. Substituting:\r\n

    [latex]5,408\\text{ J}=(373\\text{ g})c(104\\celsius)[\/latex]<\/p>\r\nWe divide both sides of the equation by 373 g and 104\u00b0C:\r\n

    [latex]c=\\dfrac{5408\\text{ J}}{(373\\text{ g})(104\\celsius)}[\/latex]<\/p>\r\nCombining the numbers and bringing together all the units, we get:\r\n

    [latex]c=0.139\\cdot \\dfrac{\\text{J}}{\\text{g}\\cdot\\celsius}}[\/latex]<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nGold has a specific heat of 0.129 [latex]\\dfrac{\\text{J}}{\\text{g}\\cdot\\celsius}[\/latex]. If 1,377 J are needed to increase the temperature of a sample of gold by 99.9\u00b0C, what is the mass of the gold?\r\n

    Answer<\/h2>\r\n107 g\r\n\r\n<\/div>\r\n<\/div>\r\nTable 7.3 \"Specific Heats of Various Substances\" lists the specific heats of some substances. Specific heat is a physical property of substances, so it is a characteristic of the substance. The general idea is that the lower the specific heat, the less energy is required to change the temperature of the substance by a certain amount.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Table 7.3 Specific Heats of Various Substances<\/caption>\r\n
    Substance<\/th>\r\nSpecific Heat [latex]\\dfrac{\\text{J}}{\\text{g}\\cdot \\celsius}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    water<\/td>\r\n4.184<\/td>\r\n<\/tr>\r\n
    iron<\/td>\r\n0.449<\/td>\r\n<\/tr>\r\n
    gold<\/td>\r\n0.129<\/td>\r\n<\/tr>\r\n
    mercury<\/td>\r\n0.139<\/td>\r\n<\/tr>\r\n
    aluminum<\/td>\r\n0.900<\/td>\r\n<\/tr>\r\n
    ethyl alcohol<\/td>\r\n2.419<\/td>\r\n<\/tr>\r\n
    magnesium<\/td>\r\n1.03<\/td>\r\n<\/tr>\r\n
    helium<\/td>\r\n5.171<\/td>\r\n<\/tr>\r\n
    oxygen<\/td>\r\n0.918<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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