Specific Heat CapacitySuppose that several objects composed of different materials are heated in the same manner. Will the objects warm up at equal rates? The answer: most likely not. Different materials would warm up at different rates because each material has its own specific heat capacity. The specific heat capacity refers to the amount of heat required to cause a unit of mass (say a gram or a kilogram) to change its temperature by 1°C. Specific heat capacities of various materials are often listed in textbooks. Standard metric units are Joules/kilogram/Kelvin (J/kg/K). More commonly used units are J/g/°C.
For example, the specific heat capacity of solid aluminum (0.904 J/g/°C) is different than the specific heat capacity of solid iron (0.449 J/g/°C). This means that it would require more heat to increase the temperature of a given mass of aluminum by 1°C compared to the amount of heat required to increase the temperature of the same mass of iron by 1°C. In fact, it would take about twice as much heat to increase the temperature of a sample of aluminum a given amount compared to the same temperature change of the same amount of iron. This is because the specific heat capacity of aluminum is nearly twice the value of iron. The fact that they are listed on a per amount basis is an indication that the quantity of heat required to raise the temperature of a substance depends on how much substance there is. Any person who has boiled a pot of water on a stove, undoubtedly know this truth. Water boils at 100°C at sea level and at slightly lowered temperatures at higher elevations. To bring a pot of water to a boil, its temperature must first be raised to 100°C. This temperature change is achieved by the absorption of heat from the stove burner. One quickly notices that it takes considerably more time to bring a full pot of water to a boil than to bring a half-full of water to a boil. This is because the full pot of water must absorb more heat to result in the same temperature change. In fact, it requires twice as much heat to cause the same temperature change in twice the mass of water. Specific heat capacities are also listed on a per K or a per °C basis. The fact that the specific heat capacity is listed on a per degree basis is an indication that the quantity of heat required to raise a given mass of substance to a specific temperature depends upon the change in temperature required to reach that final temperature. In other words, it is not the final temperature that is of importance, it is the overall temperature change. It takes more heat to change the temperature of water from 20°C to 100°C (a change of 80°C) than to increase the temperature of the same amount of water from 60°C to 100°C (a change of 40°C). In fact, it requires twice as much heat to change the temperature of a given mass of water by 80°C compared to the change of 40°C. A person who wishes to bring water to a boil on a stovetop more quickly should begin with warm tap water instead of cold tap water. This discussion of specific heat capacity deserves one final comment. The term specific heat capacity is somewhat of a misnomer. The term implies that substances may have the ability to contain a thing called heat. As has been previously discussed, heat is not something that is contained in an object. Heat is something that is transferred to or from an object. Objects contain energy in a variety of forms. When that energy is transferred to other objects of different temperatures, we refer to transferred energy as heat or thermal energy. While it's not likely to catch on, a more appropriate term would be specific energy capacity. Relating the Quantity of Heat to the Temperature Change Specific heat capacities provide a means of mathematically relating the amount of thermal energy gained (or lost) by a sample of any substance to the sample's mass and its resulting temperature change. The relationship between these four quantities is often expressed by the following equation. Q = m•C•ΔT where Q is the quantity of heat transferred to or from the object, m is the mass of the object, C is the specific heat capacity of the material the object is composed of, and ΔT is the resulting temperature change of the object. As in all situations in science, a delta (∆) value for any quantity is calculated by subtracting the initial value of the quantity from the final value of the quantity. In this case, ΔT is equal to Tfinal - Tinitial. When using the above equation, the Q value can turn out to be either positive or negative. As always, a positive and a negative result from a calculation has physical significance. A positive Q value indicates that the object gained thermal energy from its surroundings; this would correspond to an increase in temperature and a positive ΔT value. A negative Q value indicates that the object released thermal energy to its surroundings; this would correspond to a decrease in temperature and a negative ΔT value. Knowing any three of these four quantities allows an individual to calculate the fourth quantity. A common task in many physics/chemistry classes involves solving problems associated with the relationships between these four quantities. Example Problem What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? The specific heat capacity of water is 4.18 J/g/°C. Like any problem in chem, the solution begins by identifying known quantities and relating them to the symbols used in the relevant equation. In this problem, we know the following: Given: m = 450 g C = 4.18 J/g/°C Tinitial = 15°C Tfinal = 85°C We wish to determine the value of Q - the quantity of heat ( Required). To do so, we would use the equation Q = m•C•ΔT. The m and the C are known; the ΔT can be determined from the initial and final temperature. T = Tfinal - Tinitial = 85°C - 15°C = 70.°C With three of the four quantities of the relevant equation known, we can substitute and solve for Q. (Solution) Q = m•C•ΔT = (450 g)•(4.18 J/g/°C)•(70.°C) Q = 131670 J Q = 1.3x105 J = 130 kJ (rounded to two significant digits) [ Source : http://www.physicsclassroom.com/class/thermalP/Lesson-2/Measuring-the-Quantity-of-Heat ]
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. Archives
November 2017
Categories |