Physics 142: Lecture 10 Today s Agenda. Review: Specific Heat. Review: Units of Heat. Review: Internal Energy of Ideal Gas
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1 Physics 142: Lecture 10 Today s Agenda Heat: Phase Change and Latent Heat Heat Transfer (time permitting) Examples Suggested problems for CH11: 23, 46, 49, 0, 1, 78 Review: Internal Energy of Ideal Gas Temperature is a measure of the average Kinetic Energy of molecules Average Kinetic Energy (per molecule) of any Ideal Gas 2 K = 12mv = 32k T rms Total Internal Energy of an Ideal Monatomic Gas 3 U = nrt 2 b Midterm: Friday, Feb 16, :2 13:2 THEATER AUDITORIUM Total Internal Energy of an Ideal Diatomic Gas U = nrt 2 Lecture 10, Pg 1 Lecture 10, Pg 2 Review: Units of Heat Review: Specific Heat Heat is FLOW of energy Flow of energy may increase temperature Specific Heat = c m T c c p V = n T p = n T V (molar specific heat at constant pressure) (molar specific heat at constant volume) Lecture 10, Pg 3 Lecture 10, Pg 4
2 Problem A student ate a Thanksgiving dinner that totaled 2800 Cal. He wants to use up all that energy by lifting a 20-kg mass up a distance of 1.0 m. (a) How many times must he lift the mass? (b) If he can lift the mass once every.0 s, how long does this exercise take (neglecting time taken to lower the mass)? (a) Work done in each lift: W = Fd = (20 kg)(9.80 m/s 2 )(1.0 m) = 196 J. Problem A.0-g pellet of aluminum at 20 C gains 200 J of heat. What is its final temperature? = c m T Energy input: E = 2800 Cal = cal = ( cal) (4.186 J/cal) = = J. So the number of lifts is J 196J (b) t = 60,000 (.0 s) = s = 83 h = 60, 000 T = cm = (200 J) [920 J/(kg C )]( kg) So the final temperature is 20 C + 43 C = 63 C = 43 C Lecture 10, Pg Lecture 10, Pg 6 Problem How many kilograms of aluminum will experience the same temperature rise as 3.00 kg of copper when the same amount of heat is added to each? = cm T m a m c = c a T c c T = c c c a m = c T = 390 J(/kg C ) 920 J/(kg C ) = Therefore: m a = 0.424(3.00 kg) = 1.27kg Problem A 0.20-kg coffee cup at is filled with 0.20 kg of boiling coffee. The cup and the coffee come to thermal equilibrium at 80 C. If no heat is lost to the environment, what is the specific heat of the cup material? [Hint: Consider the coffee essentially to be boiling water.] The heat gained by the cup is gained = c c m c T c = c c (0.20 kg)(80 C 20 C) = (1 kg C )c c The heat lost by the coffee is lost = [4186 J/(kg C )](0.20 kg)(80 C 100 C) = J. Since no heat is lost to the environment gained + lost = 0, lost = gained. So ( J) J = (1 kg C )c c, therefore c c = J/(kg C ) Lecture 10, Pg 7 Lecture 10, Pg 8
3 Phase Changes and Latent Heat solid phase Latent Heat L As you add heat to water, the temperature increases to 100 C, then it remains constant, despite the heat being added! liquid phase T 100 o C water temp rises water changes to steam (boils) Latent Heat steam temp rises added to water gaseous (vapor) phase Latent Heat L [J/kg] is heat which must be added to (or removed from) 1kg of substance to change its phase (liquid-gas or solid-liquid). Lecture 10, Pg 9 Lecture 10, Pg 10 Latent Heat L Latent Heat L =± ml The latent heat for a solid liquid phase change is called latent heat of fusion (L f ) The latent heat for a liquid-gas phase change is called latent heat of vaporization (L v ) Substance L f (J/kg) L v (J/kg) water 3.33 x x 10 Lecture 10, Pg 11 Lecture 10, Pg 12
4 Example Heat is added to 0.00 kg of water at room temperature (20 C). How much heat is required to change water to superheated steam at 110 C? heating water: 1 = mcw T1 = (0.00kg) 4186J /(kg C) (100 C 20 C) = J Lecture 10, Pg 13 Lecture 10, Pg 14 Example Example phase change: v v J ( ) =+ ml = (0.00kg) J / kg =+ heating steam: 2 = mcs T2 = (0.00kg) 2010J /(kg C) (110 C 100 C) = J Lecture 10, Pg 1 Lecture 10, Pg 16
5 Example Cooling During a tough work out, your body sweats (and evaporates) 1 liter of water to keep cool. How much water would you need to drink (at 2C) to achieve the same thermal cooling? 1) 0.1 liters 2) 1.0 liters 3) 1 liters 4) 10 liters Total heat: = i = ( ) J evaporative = L m = J c = c w m T = 4186 J/(kg C) (37 C-2 C) m = m evaporative = c J = m m = / = 1kg or 1 liters! Lecture 10, Pg 17 Lecture 10, Pg 18 Example Summers in Phoenix, Arizona are very hot ( C is not uncommon), and very dry. If you hop into an outdoor swimming pool on a summer day in Phoenix, you will probably find that the water is too warm to be very refreshing. However, when you get out of the pool and let the sun dry you off, you find that you are quite cold for a few minutes (yes...you will have goose-bumps on a day when the air temperature is over C). How can you explain this? The water is evaporating off of your skin. This means that enough heat (or energy) is entering the water drops to break bonds between water and allow them to evaporate. Where is this heat coming from? Your body! Heat flows from your body to the drops of water, making you feel cooler. When the water is gone, no more heat will flow from your body and you will get hot once again. However, this whole situation may be avoided by NOT GOING OUTSIDE WHEN IT IS C DEGREES!!! Heat is transferred by three unique mechanisms -conduction -convection -radiation In order for the water to evaporate, it must heat up. In order for it to heat up, it must take energy from something else, therefore cooling that "something else" down. So, since the water is in contact, with your skin. The process of the water evaporating cools the skin down Lecture 10, Pg 19 Lecture 10, Pg 20
6 Conduction -occurs when the atoms in one part of the substance vibrate or oscillate faster (higher temperature) than at another part of the substance (lower temperature) this vibrational/oscillation energy gets transferred along the chain of neighbours -thus, heat energy (really energy of motion) is transferred from one part of the object to another in an attempt to make all parts of the object equal energy-wise, a spreading out or sharing of the thermal energy through diffusion -some substances are better at this than others metals are generally good conductors of heat (and also good electrical conductors) or good thermal conductors while nonmetals (also called insulators) are generally poor thermal conductors mainly because of the atomic structure of the materials Lecture 10, Pg 21
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