1. Measuring the Quantity of Heat Physics January 21, 2013 Coach Stephens

2. Review On the previous page, we learned what heat does to an object when it is gained or released.the previous page – Heat gains or losses result in changes in temperature, changes in state, or the performance of work. – Heat is a transfer of energy. – When gained or lost by an object, there will be corresponding energy changes within that object. – A change in temperature is associated with changes in the average kinetic energy of the particles within the object. – A change in state is associated with changes in the internal potential energy possessed by the object. – And when work is done, there is an overall transfer of energy to the object upon which the work is done. – In this part of Lesson 2, we will investigate the question How does one measure the quantity of heat gained or released by an object?

3. Specific Heat Capacity Suppose 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. http://www.wolframalpha.com/widgets/view.jsp?id =1e747ddbea997a1b933aaf58a7953c3c

4. 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. Standard metric units are Joules/kilogram/Kelvin (J/kg/K). More commonly used units are J/g/°C.

5. Examples 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.

6. Molar Heat Capacity Heat capacities are listed on a per gram or per kilogram basis. Occasionally, the value is listed on a per mole basis, in which case it is called the molar heat capacity. 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.

7. Continued… 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.

8. per K or per °C 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.

9. Specific ENERGY Capacity 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 have a capacity to contain 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.

10. 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 = mCΔT

11. 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 T final - T initial. When using the above equation, the Q value can turn out to be either positive or negative.

12. Positive vs. Negative Results 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.

13. Example Problems Knowing any three of these four quantities allows an individual to calculate the fourth quantity. A common task in many physics classes involves solving problems associated with the relationships between these four quantities. As examples, consider the two problems on the next slide. The solution to each problem is worked out for you.

14. Example #1 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 physics, 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: m = 450 g C = 4.18 J/g/°C T initial = 15°C T final = 85°C

15. Continued… We wish to determine the value of Q - the quantity of heat. To do so, we would use the equation Q = mCΔT. The m and the C are known; the ΔT can be determined from the initial and final temperature. T = T final - T initial = 85°C - 15°C = 70.°C With three of the four quantities of the relevant equation known, we can substitute and solve for Q. Q = mCΔT = (450 g)(4.18 J/g/°C)(70.°C) Q = 131670 J Q = 1.3x10 5 J = 130 kJ (rounded to two significant digits)

16. Example #2 A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.

17. Continued… Compared to the previous problem, this is a much more difficult problem. In fact, this problem is like two problems in one. At the center of the problem-solving strategy is the recognition that the quantity of heat lost by the water (Q water ) equals the quantity of heat gained by the metal (Q metal ). Since the m, C and ΔT values of the water are known, the Q water can be calculated. This Q water value equals the Q metal value. Once the Q metal value is known, it can be used with the m and ΔT value of the metal to calculate the Q metal. Use of this strategy leads to the following solution:

18. Continued… Part 1: Determine the Heat Lost by the Water Given: m = 50.0 g C = 4.18 J/g/°C T initial = 88.6°C T final = 87.1°C ΔT = -1.5°C (T final - T initial ) Solve for Q water : Q water = mCΔT = (50.0 g)(4.18 J/g/°C)(-1.5°C) Q water = -313.5 J (unrounded) (The - sign indicates that heat is lost by the water)

19. Continued… Part 2: Determine the value of C metal Given: Q metal = 313.5 J (use a + sign since the metal is gaining heat) m = 12.9 g T initial = 26.5°C T final = 87.1°C ΔT = (T final - T initial ) Solve for C metal : Rearrange Q metal = m metal C metalΔT metal to obtain C metal = Q metal / (m metalΔT metal ) C metal = Q metal / (m metalΔT metal ) = (313.5 J)/[(12.9 g)(60.6°C)] C metal = 0.40103 J/g/°C C metal = 0.40 J/g/°C (rounded to two significant digits)

20. Heat & Changes of State The discussion above and the accompanying equation relates the heat gained or lost by an object to the resulting temperature changes of that object. As we have learned, sometimes heat is gained or lost but there is no temperature change. This is the case when the substance is undergoing a state change. So now we must investigate the mathematics related to changes in state and the quantity of heat. To begin the discussion, let's consider the various state changes that could be observed for a sample of matter. The table below lists several state changes and identifies the name commonly associated with each process.

21. State Changes ProcessChange of State MeltingSolid to Liquid FreezingLiquid to Solid VaporizationLiquid to Gas CondensationGas to Liquid SublimationSolid to Gas DepositionGas to Solid

22. State Changes In the case of melting, boiling and sublimation, energy would have to be added to the sample of matter in order to cause the change of state. Such state changes are referred to as being endothermic. Freezing, condensation and deposition are exothermic; energy is released by the sample of matter when these state changes occur. So one might notice that a sample of ice (solid water) undergoes melting when it is placed on or near a burner. Heat is transferred from the burner to the sample of ice; energy is gained by the ice causing the change of state. But how much energy would be required to cause such a change of state? Is there a mathematical formula that might help in determining the answer to this question? There most certainly is.

23. Energy Required The amount of energy required to change the state of a sample of matter depends on three things. – It depends upon what the substance is, on how much substance is undergoing the state change, and upon what state change that is occurring. For instance, it requires a different amount of energy to melt ice (solid water) compared to melting iron. And it requires a different amount of energy to melt ice (solid water) as it does to vaporize the same amount of liquid water. And finally, it requires a different amount of energy to melt 10.0 grams of ice compared to melting 100.0 grams of ice. The substance, the process and the amount of substance are the three variables that affect the amount of energy required to cause a specific change in state. http://www.wolframalpha.com/widgets/view.jsp?id=48f7d3043b c03e6c48a6f0ebc0f258a8