0 o K is called absolute zero. Water Freezes: 273 o K Water Boils: 373 o K

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1 Part I Notes Temperature and Heat The terms at the right all mean the same thing. The heat energy of a substance is the sum of the kinetic and potential energies of all of the atoms and molecules in the substance. Heat Heat energy Thermal energy Internal energy Kelvin Temperature Scale Symbol: T Units: degrees Kelvin ( o K) The average kinetic energy per atom of a monatomic gas such as argon is proportional to the Kelvin temperature. Example: If the average kinetic energy per argon atom is 2 x J at 300 K, then it would be twice that at twice the Kelvin temperature: at 600 K, the average kinetic energy per atom would be 4 x J. Celsius Temperature Scale Symbol: TC Units: degrees Celsius ( o C) TC = T The lowest-possible Kelvin temperature of a substance is T = 0 o K, called absolute zero. It s called absolute zero in the sense that there is absolutely no motion of the atoms whatsoever. Every atom has zero kinetic energy. There are no negative Kelvin temperatures. 0 o K is called absolute zero. The Kelvin temperature scale is also called the Absolute Temperature Scale. Water Freezes: 273 o K Water Boils: 373 o K Absolute Zero: TC = -273 o C Water Freezes: TC = 0 o C Water Boils: TC = 100 o C Because there are 100 degrees separating the freezing and boiling points of water, this scale is often called the Centigrade Temperature Scale. Fahrenheit Temperature Scale Absolute Zero: TF = 9/5 (-273) +32 = o F Symbol: TF Units: degrees Fahrenheit ( o F) TF = 9/5 TC + 32 Water Freezes: Water Boils: TF = 32 o F TF = 212 o F 1

2 Example A: A Fahrenheit thermometer and a Celsius thermometer are placed in a substance. The Fahrenheit reading is 20 degrees higher than the Celsius reading. What are the two temperatures? TF = TC + 20 (9/5) TC + 32 = TC + 20 Solve for TC: TC = -15 o C Now get TF: TF = (9/5)(-15) + 32 = 5 o F Kelvin Temperature Changes Symbol: ΔT Units: Kelvin degrees (K o ) (not degrees Kelvin) Note: superscript comes after the K Example B: To = 273 o K (0 o C) T = 373 o K (100 o C) ΔT = T - To = = 100 K o Same as ΔT = T - To = = 100 C o (Celsius degrees) Kelvin temperature change = Celsius temperature change Note: Kelvin degrees and Celsius degrees are not the same as degrees Kelvin and degrees Celsius. The former are units of temperature change, while the latter are units of temperature. For example, 40 C o is NOT a temperature; it is not the same as the temperature 40 o C Calories Non-Standard Alternative Energy Unit calorie lower-case c 1.0 calorie (cal) = 4.19 J Also called a small calorie, or the scientist s calorie. Example B: How much exercise would be necessary to burn off one chocolate-frosting cupcake (140 Cal). A 154-lb male walking for 30 minutes will burn off the cupcake. Calorie 1.0 Calorie (Cal) = 1000 Cal Also called a large Calorie, or the dieter s calorie. 1.0 Cal = 4190 J 2

3 Q symbolizes the quantity of heat entering or leaving a substance. The value of Q for heat entering a substance is positive. The value of Q for heat leaving a substance is negative. Phase Changes Three of the more common forms (states) of matter, called phases of matter, are solids, liquids, and gases. When a substance changes from one phase to another, a phase change has occurred. When a phase change is occurring, the temperature of the substance does not change. 3

4 Freezing Water Melting Ice The only substance we will study in phase changes is H2O, whose various phases include water, ice, and water vapor (steam). Water will not begin to become ice until its temperature is lowered to the freezing point of water, which is 0 o C. After that temperature is reached, further removal of heat from the water will causes ice crystals to form. To create one gram of ice from one gram of water at 0 o C, 80 cal of heat must be removed. This number--80 cal/gram--is called the latent heat of freezing. L = 80 cal/g Note: as the water at 0 o C is freezing, the temperature of the freezing ice and water mixture does not change: it remains at a temperature of 0 o C. Temperature doesn t change in a phase change. Ice will not begin to melt until its temperature has been raised to its melting point, which is 0 o C. After that temperature is reached, additional heat will cause the ice to melt. For each gram of ice at 0 o C to be melted, 80 cal of heat must be removed. This number, 80 cal/g, is called the latent heat of melting L = 80 cal/g This number is called the latent heat of melting, and is the same as the latent heat of freezing. Note: as the ice at 0 o C is melting, the temperature of the melting ice and water mixture remains at a temperature of 0 o C. Temperature doesn t change in a phase change. Example: What quantity Q of heat must be removed from 200 grams of water at 0 o C to convert it to 200 grams of ice at 0 o C? Answer: Q = -ml = -200 g (80 cal/g) = -16,000 cal = -16 kilocalories (kcal) Negative signs associated with Q s indicate that heat is removed. 4

5 Vaporization There are two types of vaporization of water. The first occurs at the surface of water when water molecules with sufficient kinetic energy and moving away from the surface escape into the surrounding air and becomes water vapor. This type of vaporization is called evaporation. Example A: How many calories of heat must be added to 200 grams of water at 100 o C to convert it to 200 grams of steam at 100 o C? Q = +ml = 200 (540) = 108,000 cal = 108 kcal The second type of vaporization is the one in which a phase change occurs: at the boiling point of water, 100 o C, water molecules are transformed into water vapor in a process called boiling. Example B: How many calories of heat must be removed from 50 g of steam at 100 o C to convert it to 50 g of water at 100 o C? Q = -ml = -50 (540) = -27,000 = -27 kcal (Negative, as expected, because heat is removed.) Example C: Once water s temperature has been raised to its boiling point of 100 o C, it will begin to vaporize. Each gram of water at 100 o C requires 540 cal of heat to undergo a phase change to water vapor (steam). This number is called the latent heat of vaporization. L = 540 cal/g The reverse process is condensation. To condense one gram of water vapor (steam) at 100 o C, 540 cal must be removed. When five grams of water produced by sweat glands evaporate from the skin of a human, how many calories of heat energy leaves the body? Answer: Q = - ml = -5 g (540 cal/g) = 2700 cal L = 540 cal/g This number is called the latent heat of condensation. 5

6 Example: How many calories of heat energy will be delivered to a soda can via phase transformation when 12 grams of water vapor condenses on it? Solution: Q = + ml = 12 (540) = 6480 cal Other heat energy would also be delivered to the can by a process called conduction, a topic to be discussed in Part J, when a mixture of substances at different temperatures occurs. See the section below called, Mixtures. Water vapor condenses on a colder beer can, which won t be cold for long. The water vapor s heat that s delivered to the can will raise the beer to a temperature to a point that is undrinkably high for most beer-lovers. Specific Heat Capacity If heat is added or removed from a substance whose phase is not changing, then a temperature change will occur. In what follows, temperature changes will be symbolized as ΔT, and will be measured in Celsius degrees (C o ). All substances have a property called specific heat capacity, symbolized as c, and measured in units of cal/g-c o. The quantity Q of heat that must be removed or added to m grams of a substance in order to cause a temperature change ΔT is Q = mcδt Substance c cal/g-c o Water 1.0 Ice 0.5 Steam 0.5 concrete 0.2 aluminum 0.2 bone 0.1 6

7 Re-write the equation above as: ΔT = Q/mc Because c is in the denominator, the larger the specific heat capacity c of a substance, the smaller will be the temperature change when heat is added or removed, compared to other substances with smaller heat capacities, and vice-versa. Example A: 900 cal of heat is added to 100 grams of water, and the same amount of heat is added to the same mass of concrete. Calculate and compare the temperature changes of each substance. Water: ΔT = 900 / (100 x 1.0) = 9 C o Concrete: ΔT = 900 / (100 x 0.2) = 45 C o The specific heat capacity of concrete is only one-fifth that of water, so concrete s temperature rise is five times greater. The greater the capacity of a substance to absorb heat, the less effect heat loss or gain has on it. Water has a comparatively high specific heat capacity; almost everything else heats up or cools down more dramatically than water does. Example B: How much heat must be added to 100 grams of water at 30 o C to raise its temperature to 70 o C? ΔT = = 40 C o Q = mcδt = (100 g) (1.0 cal/g-c o ) 40 C o = 4000 cal Example C: How much heat must be removed from 60 grams of ice at -30 o C to cool it down to -90 o C? ΔT = (-30) = -60 C o Q = mcδt = 60 (0.5) (-60) = cal Note that the sign of Q appears automatically; if ΔT had been negative, then Q would have been negative. Example D: See figure below. 100 grams of ice at -30 o C is to be converted to 100 grams of water at 40 o C. How much heat must be added to the ice to accomplish this? Q1 = 100 (0.5) (30) = 1500 Q2 = 100 (80) = 8000 Q3 = 100 (1.0) (40) = 4000 Q = Q1 + Q2 + Q3 = 13,500 cal = 13.5 kcal 7

8 Mixture Problems When two substances at different temperatures are mixed together, one of them gains heat, while the other loses heat. Whatever one gains, the other loses; one of the Q s will be positive, and the other will be negative, and the sum of the Q s will be zero: Q1 + Q2 = 0 When equilibrium is reached, both substances will be at the same temperature. Example A: Sixty grams of aluminum at 75 o C is added to 300 g of water at 30 o C. What is the equilibrium temperature? Q1 = 60 (0.20) (T -75) Q2 = 300 (1.00) (T - 30) Q1 + Q2 = 0 60 (0.20) (T -75) (1.00) (T - 30) = 0 Get T = 31.7 o C 8