Chapter 19 Heat and the First Law of Thermodynamics

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Presentation transcript:

Chapter 19 Heat and the First Law of Thermodynamics The Study Guide is posted online on my webpage under Homework Chapter opener. When it is cold, warm clothes act as insulators to reduce heat loss from the body to the environment by conduction and convection. Heat radiation from a campfire can warm you and your clothes. The fire can also transfer energy directly by heat convection and conduction to what you are cooking. Heat, like work, represents a transfer of energy. Heat is defined as a transfer of energy due to a difference of temperature. Work is a transfer of energy by mechanical means, not due to a temperature difference. The first law of thermodynamics links the two in a general statement of energy conservation: the heat Q added to a system minus the net work W done by the system equals the change in internal energy ΔEint of the system: ΔEint = Q – W. Internal energy Eint is the sum total of all the energy of the molecules of the system.

17-4 Thermal Expansion

17-4 Thermal Expansion Example 17-5: Ring on a rod. An iron ring is to fit snugly on a cylindrical iron rod. At 20°C, the diameter of the rod is 6.445 cm and the inside diameter of the ring is 6.420 cm. To slip over the rod, the ring must be slightly larger than the rod diameter by about 0.008 cm. To what temperature must the ring be brought if its hole is to be large enough so it will slip over the rod? Solution: The temperature needs to be raised by 430°C, to 450°C.

17-4 Thermal Expansion Conceptual Example 17-6: Opening a tight jar lid. When the lid of a glass jar is tight, holding the lid under hot water for a short time will often make it easier to open. Why? Solution: The lid will heat before the glass, and expand sooner. Also, metals generally expand more than glass for the same temperature difference.

17-4 Thermal Expansion Here, β is the coefficient of volume expansion. Volume expansion is similar, except that it is relevant for liquids and gases as well as solids: Here, β is the coefficient of volume expansion. For uniform solids, β ≈ 3α.

17-4 Thermal Expansion Example 17-7: Gas tank in the Sun. The 70-liter (L) steel gas tank of a car is filled to the top with gasoline at 20°C. The car sits in the Sun and the tank reaches a temperature of 40°C (104°F). How much gasoline do you expect to overflow from the tank? Solution: Both the tank and the gasoline expand; the amount that spills is the difference. However, the gasoline expands by about 1.3 L, whereas the tank expands by about 0.05 L – the expansion of the tank makes little difference.

17-4 Thermal Expansion Water behaves differently from most other solids—its minimum volume occurs when its temperature is 4°C, and has biggest density at 4°C. As it cools further, it expands, as anyone who leaves a bottle in the freezer to cool and then forgets about it can testify. Figure 17-12: Behavior of water as a function of temperature near 4°C. (a) Volume of 1.00000 g of water, as a function of temperature. (b) Density vs. temperature. [Note the break in each axis.]

19-1 Heat as Energy Transfer We often speak of heat as though it were a material that flows from one object to another; it is not. Rather, it is a form of energy. Unit of heat: calorie (cal) 1 cal is the amount of heat necessary to raise the temperature of 1 g of water by 1 Celsius degree. Don’t be fooled—the Calories on our food labels are really kilocalories (kcal or Calories), the heat necessary to raise 1 kg of water by 1 Celsius degree.

19-1 Heat as Energy Transfer If heat is a form of energy, it ought to be possible to equate it to other forms. The experiment below found the mechanical equivalent of heat by using the falling weight to heat the water: 4.186 J = 1 cal 4.186 kJ = 1 kcal Figure 19-1. Joule’s experiment on the mechanical equivalent of heat. Mechanical Joule experiment

19-1 Heat as Energy Transfer Definition of heat: Heat is energy transferred from one object to another because of a difference in temperature. Remember that the temperature of a gas is a measure of the kinetic energy of its molecules.

19-1 Heat as Energy Transfer Example 19-1: Working off the extra calories. Suppose you throw caution to the wind and eat too much ice cream and cake on the order of 500 Calories. To compensate, you want to do an equivalent amount of work climbing stairs or a mountain. How much total height must you climb? Solution: Remember food calories are kcal; letting W = mgh, approximate your mass as 60 kg, then h = 3600 m.

Problem 5 5.(II) How many joules and kilocalories are generated when the brakes are used to bring a 1200-kg car to rest from a speed of 95km/h

19-2 Internal Energy The total sum of all the energy of all the molecules in a substance is its internal (or thermal) energy. Temperature: measures molecules’ average kinetic energy Internal energy: total energy of all molecules Heat: transfer of energy due to difference in temperature

19-2 Internal Energy Internal energy of an ideal (atomic) gas: But since we know the average kinetic energy in terms of the temperature, we can write: N is the number of molecules, and k is the Boltzman constant=1.38X10-23J/K

19-2 Internal Energy If the gas is molecular rather than atomic, rotational and vibrational kinetic energy need to be taken into account as well. Figure 19-2. Besides translational kinetic energy, molecules can have (a) rotational kinetic energy, and (b) vibrational energy (both kinetic and potential).

19-3 Specific Heat The specific heat: The amount of heat required to change the temperature of a material and is proportional to the mass and to the temperature change: The specific heat,c, is characteristic of the material. Some material values are listed on the left.

19-3 Specific Heat Example 19-2: How heat transferred depends on specific heat. How much heat input is needed to raise the temperature of an empty 20-kg vat made of iron from 10°C to 90°C? (b) What if the vat is filled with 20 kg of water? Solution: a. Q = 720 kJ. b. For the water alone, Q = 6700 kJ, so the total is 7400 kJ.

19-4 Calorimetry—Solving Problems Closed system: no mass enters or leaves, but energy may be exchanged Open system: mass may transfer as well Isolated system: closed system in which no energy in any form is transferred For an isolated system, energy out of one part = energy into another part, or: heat lost = heat gained.

19-4 Calorimetry—Solving Problems Example 19-3: The cup cools the tea. If 200 cm3 of tea at 95°C is poured into a 150-g glass cup initially at 25°C, what will be the common final temperature T of the tea and cup when equilibrium is reached, assuming no heat flows to the surroundings? Solution: Assume all the heat that leaves the tea flows into the cup. Tea is mostly water; the common final temperature is 86 °C.

Problem 15 15.(II) When a 290-g piece of iron at 180°C is placed in a 95-g aluminum calorimeter cup containing 250 g of glycerin at 10°C, the final temperature is observed to be 38°C. ­Estimate the specific heat of glycerin.

19-4 Calorimetry—Solving Problems The instrument to the left is a calorimeter, which makes quantitative measurements of heat exchange. A sample is heated to a well-measured high temperature and plunged into the water, and the equilibrium temperature is measured. This gives the specific heat of the sample. Figure 19-4. Simple water calorimeter.