Unit 6 Conservation of Energy & Heat

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Name: Unit 6 Conservation of Energy & Heat Hr: Grading: Show all work, keeping it neat and organized. Show equations used and include all units. Vocabulary Energy: The ability to do work. There are many different types of energy. We will focus on only mechanical energy, or the energy related to position (potential energy) and motion (kinetic energy). Vocabulary Potential Energy: Energy of position, or stored energy. An object gains gravitational potential energy when it is lifted from one level to a higher level. Therefore, we generally refer to the change in gravitational potential energy or ΔGPE, which is proportional to the change in height, Δh. Δ gravitational potential energy = (mass)(acceleration due to gravity)(δ height) or ΔGPE = mgδh It is important to remember that gravitational potential energy relies only upon the vertical change in height, Δh, and not upon the path taken. Vocabulary Kinetic Energy: Energy of motion. The kinetic energy of an object varies with the square of the speed. Kinetic energy = (/2)(mass)(speed) 2 or KE = 2 mv2 The SI unit for energy is the joule. Notice that this is the same unit used for work. Conservation of Energy Vocabulary According to the law of conservation of energy, energy cannot be created or destroyed. The total amount of mechanical energy in a system remains constant if no work is done by any force other than gravity. In an isolated system where there are no mechanical energy losses due to friction KE = GPE In other words, all the kinetic and gravitational potential energy before an interaction equals all the kinetic and potential energy after the interaction. KE i + PE i = KE f + PE f or 2 mv i 2 + mgh i = 2 mv f 2 + mgh f Example : Frank, a San Francisco hot dog vendor, has fallen asleep on the job. When an earthquake strikes, his 300-kg hot dog cart rolls down Nob Hill and reaches point A at a speed of 8.00 m/s. How fast is the hot dog cart going at point B when Frank finally wakes up and starts to run after it?

Solution: Because mass is contained in each of these equations, it cancels out and does not need to be included in the calculation. Also, the inclination of the hill makes no difference. All that matters is the change in height. Given: v i = 8.00 m/s Unknown: v f =? g = 0.0 m/s 2 Original Equation: KE = PE h i = 50.0 m h f = 30.0 m Solve: KE i + PE i = KE f + PE f 2 v f = v i 2 + gh i gh f 2 = or 2 mv i 2 + mgh i = 2 mv f 2 + mgh f 2 (8.00m s )2 + (0.0 m s 2)(50.0 m) (0.0m s2)(30.0 m) 2 = 2.5 m/s Vocabulary Heat: The transfer of energy between two objects that differ in temperature. Vocabulary Specific heat: A measure of the amount of heat needed to raise the temperature of kg of a substance by C. The common unit for specific heat is the joule per kilogram degree Celsius (J/kg C). The transfer of heat from an object depends upon the object s mass, the specific heat, and the difference in temperature between the object and its surroundings. change in heat = (mass)(specific heat)(change in temperature) OR Q = mc T The SI unit for heat is the joule (J), which is the same unit used for mechanical energy and work. The heat lost by one object equals the heat gained by another object. Heat lost = Heat gained or (mc T) lost = (mc T) gained For each object in the system, an mc T term is needed. Water has a very high specific heat. It makes a good cooling agent because it takes a long time for water to absorb enough heat to greatly increase its temperature.

Example 2: Remy is cooking his favorite peasant dish and needs to get 3.0 kg of water up to 98 C before adding the noodles. If he uses a 00-W hot plate to heat the water from 23 C, how much time will he have to prepare the vegetables before he can put the noodles in the hot water? Solution: First find the quantity of heat needed to heat the water. Given: m = 3.0 kg Unknown: ΔQ =? T i = 23 C Original Equation: Q = mc T Tf = 98 C c water = 4200 J/kg C Solve: Q = mc T Q = (3.0 kg)(4200 J/kg C)(98 C 22 C) = 960,000 J Then use this heat with the power of the hot plate to determine the time required for heating. Since all the energy expended by the hot plate is turned into heat, W = Q. We ignore the heat added to the pot because water has a much higher specific heat than any metal used in the pot, and absorbs much more of the available energy. Given: Q = 960,000 J Unknown: t =? P = 00 W Original Equation: P = W/t Solve: t = W/P t = (960,000 J)/(00 W) = 870 s = 4.5 minutes Exercises (For all of the exercise problems, assume there is no friction.) ) Sara is changing the tire of her jeep on the top of a 25.0 m high hill. She accidentally lets go of the 2.0 kg tire while reaching for her tire iron, which then rolls down the hill starting from rest. a) How fast is the tire rolling at the top of the next hill, which is 6.0 m high? b) How high can the tire climb before it stops (or rolls the other direction)? Explain.

2) Small worms that live inside Mexican jumping beans are the cause for their apparent bouncing. a) If a 3.0 g bean jumps up from your hand, how much gravitational potential energy has it gained while reaching its maximum height of 2.0 cm? b) What is the speed of the jumping bean right before it lands back on your hand? 3) Billy sets up a slip and slide down a 2.0 m tall hill. a) If Billy dives onto the slip and slide with an initial speed of 3.5 m/s, what is his speed as he falls off the slip and slide at the bottom of the hill? b) What are two ways that Billy could reach greater maximum speeds on his slip and slide?

4) Astronauts on the moon were known to hop around instead of walking, as an easier means of travel under the influence of reduced gravity. If Buzz Aldrin jumped off the surface of the moon with an initial speed of.5 m/s and reached a maximum height of 0.70 m, what is the acceleration due to gravity on the moon? (Calculate this and show your work!) 5) Julie, who has a mass of 58 kg, stays out snow shoeing for too long and her body temperature is only 3.0 o C when she gets back inside. Assuming that Julie normally has a body temperature of 37 o C, how much heat did Julie lose while outside? (c human body=3470 J/kg o C)

6) The water from Victoria Falls plummets 08 m before reaching the river below. Every second,.088 x 0 6 kg of water falls over the edge of the waterfall. a) How much work is done on the water by gravity as it falls? b) Assuming that all of the work done on the falling water generates heat as it crashes into the river below, how much will the temperature of the 3.4 x 0 6 kg of river water rise? c) If this amount of work is done on the river every second, then why doesn t the river s temperature increase steadily throughout time?