What Is Heat? Help (page 2)
When a confined sample of gas gets hotter, its pressure increases. The converse of this is also true: When a gas is put under increasing pressure, it gets hotter. However, what do we mean when we talk about heat and temperature? What effects do heat and temperature have on matter? In this chapter we will find out. We’ll also see how matter can change state with changes in temperature or pressure.
What Is Heat?
Heat is a special kind of energy transfer that can take place from one material object, place, or region to another. For example, if you place a kettle of water on a hot stove, heat is transferred from the burner to the water. This is conductive heat, also called conduction (Fig. 11-1a). When an infrared lamp, sometimes called a heat lamp , shines on your sore shoulder, energy is transferred to your skin surface from the filament of the lamp; this is radiative heat , also called radiation (see Fig. 11-1b). When a blower-type electric heater warms up a room, air passes through the heating elements and is blown by a fan into the room, where the heated air rises and mixes with the rest of the air in the room. This is convective heat , also called convection (see Fig. 11-1c).
Heat is not quite the same thing as energy, although the units of heat and energy are defined in the same physical dimensions. Heat is the transfer of energy that occurs when conduction, radiation, and/or convection take place. Sometimes the energy transfer takes place in only one of these three modes, but sometimes it occurs in two or all three.
The unit of heat used by physicists is the calorie . You’ve heard and read this word many times (probably too often, but that’s a subject for another book). The calorie that scientists use is a much smaller unit than the calorie used by nutritionists—only 1/1,000 as large—and the scientific use of the term usually refers to inanimate things, whereas the nutritional term involves biologic processes.
The calorie (cal) in which we, as physicists, are interested is the amount of energy transfer that raises or lowers the temperature of exactly one gram (1 g) of pure liquid water by exactly one degree Celsius (1°C). The kilocalorie (kcal), equivalent to the nutritionist’s calorie , is the amount of energy transfer that will raise or lower the temperature of 1 kg, or 1,000 g, of pure liquid water by 1°C. This holds true only as long as the water is liquid during the entire process. If any of the water freezes, thaws, boils, or condenses, this definition falls apart. At standard atmospheric pressure at Earth’s surface, in general, this definition holds for temperatures between approximately 0°C (the freezing point of water) and 100°C (the boiling point).
Pure liquid water requires 1 calorie per gram (1 cal/g) to warm it up or cool it down by 1°C (provided it is not at the melting/freezing temperature or the vaporization/condensation temperature, as we shall shortly see.) However, what about oil, alcohol, or salt water? What about solids such as steel or wood? What about gases such as air? It is not so simple then. A certain, fixed amount of heat energy will raise or lower the temperatures of fixed masses of some substances more than others. Some matter takes more than 1 cal/g to get hotter or cooler by 1°C; some matter takes less. Pure liquid water takes exactly 1 cal/g to warm up or cool down by 1°C simply because this is the substance on which the definition of the calorie is based. It is one of those things scientists call a convention .
Suppose that we have a sample of some mysterious liquid. Call it substance X . We measure out 1 gram (1.00 g), accurate to three significant figures, of this liquid by pouring some of it into a test tube placed on a laboratory balance. Then we transfer 1 calorie (1.00 cal) of energy to substance X. Suppose that, as a result of this energy transfer, substance X increases in temperature by 1.20°C? Obviously, substance X is not water because it behaves differently from water when it receives a transfer of energy. In order to raise the temperature of 1.00 g of this stuff by 1.00°C, it takes somewhat less than 1.00 cal of heat. To be exact, at least insofar as we are allowed by the rules of significant figures, it will take 1.00/1.20 = 0.833 cal to raise the temperature of this material by 1.00°C.
Now suppose that we have a sample of another material, this time a solid. Let’s call it substance Y . We carve a chunk of it down until we have a piece that masses 1.0000 g, accurate to five significant figures. Again, we can use our trusty laboratory balance for this purpose. We transfer 1.0000 cal of energy to substance Y. Suppose that the temperature of this solid goes up by 0.80000°C? This material accepts heat energy in a manner different from either liquid water or substance X. It takes a little more than 1.0000 cal of heat to raise the temperature of 1.0000 g of this material by 1.0000°C. Calculating to the allowed number of significant figures, we can determine that it takes 1.0000/0.80000 = 1.2500 cal to raise the temperature of this material by 1.0000°C.
We’re onto something here: a special property of matter called the specific heat , defined in units of calories per gram per degree Celsius (cal/g/°C). Let’s say that it takes c calories of heat to raise the temperature of exactly 1 gram of a substance by exactly 1°C. For water, we already know that c = 1 cal/g/°C, to however many significant figures we want. For substance X , c = 0.833 cal/g/°C (to three significant figures), and for substance Y, c = 1.2500 cal/g/°C (to five significant figures).
Alternatively, c can be expressed in kilocalories per kilogram per degree Celsius (kcal/kg/°C), and the value for any given substance will be the same. Thus, for water, c = 1 kcal/kg/°C, to however many significant figures we want. For substance X , c = 0.833 kcal/kg/°C (to three significant figures), and for substance Y, c = 1.2500 kcal/kg/°C (to five significant figures).
The British Thermal Unit (btu)
In some applications, a completely different unit of heat is used: the British thermal unit (Btu). You’ve heard this unit mentioned in advertisements for furnaces and air conditioners. If someone talks about Btus alone in regard to the heating or cooling capacity of a furnace or air conditioner, this is an improper use of the term. They really mean to quote the rate of energy transfer in Btus per hour, not the total amount of energy transfer in Btus.
The Btu is defined as the amount of heat that will raise or lower the temperature of exactly one pound (1 lb) of pure liquid water by one degree Fahrenheit (1°F). Does something seem flawed about this definition? If you’re uneasy about it, you have a good reason. What is a pound? It depends where you are. How much water weighs 1 lb? On the Earth’s surface, it’s approximately 0.454 kg or 454 g. On Mars, however, it takes about 1.23 kg of liquid water to weigh 1 lb. In a weightless environment, such as on board a space vessel orbiting the Earth or coasting through deep space, the definition of Btu is meaningless because there is no such thing as a pound at all.
Despite these flaws, the Btu is still used once in awhile, so you should be acquainted with it. Specific heat is occasionally specified in Btus per pound per degree Fahrenheit (Btu/lb/°F). In general, this is not the same number, for any given substance, as the specific heat in cal/g/°C.
Heat Practice Problem
Suppose that you have 3.00 g of a certain substance. You transfer 5.0000 cal of energy to it, and the temperature goes up uniformly throughout the sample by 1.1234°C. It does not boil, condense, freeze, or thaw during this process. What is the specific heat of this stuff?
Let’s find out how much energy is accepted by 1.00 g of the matter in question. We have 3.00 g of the material, and it gets 5.0000 cai, so we can conclude that each gram gets 1 /3 of this 5.0000 cal, or 1.6667 cal.
We’re told that the temperature rises uniformly throughout the sample. This is to say, it doesn’t heat up more in some places than in other places. It gets hotter to exactly the same extent everywhere. Therefore, 1.00 g of this stuff goes up in temperature by 1.1234°C when 1.6667 cal of energy is transferred to it. How much heat is required to raise the temperature by 1.0000°C? This is the number c we seek, the specific heat. To get c , we must divide 1.6667 cal/g by 1.1234°C. This gives us c = 1.4836 cal/g/°C. Because we are given the mass of the sample to only three significant figures, we must round this off to 1.48 cal/g/°C.
Practice problems of these concepts can be found at: Temperature, Pressure, And Changes Of State Practice Test
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