CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 1 IN OUR LAST LECTURE WE WERE TALKING ABOUT THE ENERGY INVOLVED IN PHASE CHANGE.

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1 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 1 CHM 105/106 Program 40: Unit 4 Lecture 11 IN OUR LAST LECTURE WE WERE TALKING ABOUT THE ENERGY INVOLVED IN PHASE CHANGE. SPECIFICALLY WE LOOKED AT THE ENERGY INVOLVED IN CHANGING FROM A LIQUID TO A GAS WHICH CALLED MOLAR HEAT OF VAPORIZATION AND THEN THE LAST THING WE LOOKED AT WAS AN EXAMPLE OF THE ENERGY NEEDED TO CHANGE FROM A SOLID TO A LIQUID, WHICH WE CALL THE MOLAR HEAT OF FUSION. NOW IS SHOULD STRESS IN BOTH OF THOSE CASES THAT IF WE MAKE THE REVERSE CHANGE, THAT IS A GAS BACK TO A LIQUID THE ENERGY INVOLVED IN GOING FROM A GAS TO A LIQUID STATE IS EXACTLY THE SAME AS THE AMOUNT OF ENERGY NEEDED TO GO FROM THE LIQUID TO THE GASEOUS STATE. OR, IN THE CASE OF THE SOLID, TO FREEZE A SOLID, TO GO FROM ITS NORMAL FREEZING POINT AS A LIQUID TO A SOLI D WE RE GOING TO GIVE OFF EXACTLY THE SAME AMOUNT OF ENERGY, THE MOLAR HEAT OF FUSION AS WE HAVE PUT IN TO CHANGE IT FROM A SOLID TO THE LIQUID STATE. OKAY, SO THE DIRECTION OF ENERGY MOVEMENT IS DIFFERENT, BUT THE MAGNITUDE OF ENERGY IS THE SAME REGARDLESS OF WHICH WAY WE RE GOING IN A PHASE CHANGE. NOW, ON THE LAST CHART THERE AND PREVIOUS ONE WHERE WE TALKED ABOUT THE HEATING A SOLID AND HEATING A LIQUID, WE DIDN T SAY ANYTHING NOW ABOUT THE HEATING PART, THE ENERGY INVOLVED IN THE HEATING PART. WE TA LKED ABOUT THE ENERGY INVOLVED IN THE PHASE CHANGE BUT DID NOT TALK SPECIFICALLY ABOUT THE ENERGY IN THE HEATING. IT TAKES ENERGY TO HEAT PARTICLES UP TO INCREASE THEIR KINETIC ENERGY. SO WHAT WE RE GOING TO FOCUS ON IN THIS LECTURE TODAY THEN IS HOW WE GO ABOUT CALCULATING THE ENERGY THEN OF NOT ONLY PHASE CHANGES BUT ALSO THE HEATING AND COOLING PROCESS. NOW ONE OF THE UNITS THAT WE ENCOUNTER WHEN WE RE TALKING ABOUT HEATING AND COOLING IS A UNIT CALLED SPECIFIC HEAT, AND SPECIFIC HEAT IS DEFINED AS THE AMOUNT OF ENERGY IN JOULES NEEDED TO HEAT ONE GRAM OF SUBSTANCE ONE CELSIUS DEGREE. THAT S REFERRED TO AS THE SPECIFIC HEAT OF A SUBSTANCE. THAT SPECIFIC HEAT OF COURSE IS AGAIN GOING TO DEPEND UPON THOSE TYPES OF FORCES THAT WE HAVE TO OVERCOME TO INCREASE THIS KINETIC ENERGY. SO SOMETHING THAT HAS A LOT OF BONDING FORCES IN IT PROBABLY HAS A

2 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 2 RELATIVELY HIGH SPECIFIC HEAT. THINGS THAT HAVE WEAKER BONDING FORCES ARE GOING TO HAVE LOWER SPECIFIC HEATS. WELL LET S LOOK AT A TABLE HERE OF SOME SPECIFIC HEATS JUST TO SEE WHAT THE RANGE LOOKS LIKE. THE FIRST ONE HERE WE SEE WATER. TO HEAT WATER AS A LIQUID REQUIRES JOULES FOR EACH GRAM WE WANT TO HEAT UP ONE CELSIUS DEGREE. AND NOTICE THAT S QUITE LARGE COMPARED TO THE REST OF THEM. WE GO DOWN HERE TO LEAD, LOOK AT THE METALS DOWN HERE AND WE SEE THAT THEY RE ABOUT 20 TIMES LESS ENERGY NEEDED TO HEAT THEM UP THAN WATER. ONE OF THE IMPORTANT PROPERTIES OF WATER, ONE OF THE IMPORTANT PROPERTIES OF WATER FOR MAINTAINING LIFE ON THE PLANET. WE EXPLORED MARS AND LOOKED FOR WATER AND ONE OF THE REASONS WE LOOKED FOR WATER ON MARS TO SEE IS IT FEASIBLE THAT SOME LIFE FORM EXISTED THERE. WELL ONE OF THE IMPORTANT PARTS IS THAT WATER ABSORBS A LOT OF HEAT ENERGY. THE REASON THE EARTH IS A FAIRLY CONSTANT, RATHER MODERATE TEMPERATURE IS BECAUSE OF ALL THE WATER WE HAVE. WHEN THE SUN IS HEATING THE EARTH S SURFACE DURING THE DAYTIME THE WATER IS ABSORBING JOULES FOR EVERY GRAM OF WATER FOR EVERY DEGREE IT WARMS UP. WHEN THE SUN SETS, THE FLOW OF HEAT IS THE OTHER WAY, THE AIR BECOMES COOLER, THE WATER BEGINS GIVING OFF ENERGY AND SO NOW THE WATER BEGINS TO HEAT THE ATMOSPHERE. AND SO IT IS THIS THAT MODERATES THE TEMPERATURE THEN, THAT CONTROLS THE TEMPERATURE ON OUR PLANET SO THAT IT IS FIT FOR HUMANS AND PLANTS AND ANIMALS TO LIVE ON IT. IN CONTRAST WHERE YOU HAVE A ALMOST NO ATMOSPHERE AND ALSO HAVE NO WATER, ON THE MOON S SURFACE FOR INSTANCE, YOU CAN HAVE A COUPLE OF HUNDRED DEGREES DIFFERENCE IN TEMPERATURE IF YOU RE STANDING IN THE SHADE VERSUS STANDING IN THE SUNSHINE. I MEAN IF I WERE TO WALK UNDERNEATH A PLATFORM THAT PROTECTED ME FROM THE SUN THERE WOULD BE A COUPLE OF HUNDRED DEGREES DIFFERENCE. ON THE ONE SIDE OF THE MOON THAT ISN T RECEIVING SUNLIGHT IT S WELL BELOW ZERO. THE OTHER SIDE IT S UP IN THE HUNDREDS OF CELSIUS DEGREES. SO WATER HELPS MODERATE THE EARTH S TEMPERATURE AND IT S BECAUSE OF THIS HIGH SPECIFIC HEAT THAT IT S SO EFFECTIVE IN DOING SO. WATER AS A GAS, WATER AS A SOLID. WE SEE THAT EVEN THOSE TWO THEY HAVE A LITTLE LESS ENERGY THAN LIQUIDS. SO LIQUID S

3 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 3 THE BEST FORM AS FAR AS FOR ABSORBING ALL OF THIS HEAT ENERGY ON THE EARTH S SURFACE. BUT AGAIN AS I SAY, NOTICE THE METALS ARE RELATIVELY LOW IN SPECIFIC HEAT AS COMPARED TO SOME OF THESE COMPOUNDS UP HERE. AN INTERESTING METAL THERE, ALUMINUM, ONE OF THE THINGS THAT WE DO IS WE USE ALUMINUM TO MAKE COOKWARE, AND ONE OF THE REASONS ALUMINUM IS PRETTY GOOD IN COOKWARE IS BECAUSE IT TENDS TO HOLD HEAT. TAKES A BIT TO HEAT IT UP BUT THEN IT HOLDS THE HEAT SO TEMPERATURE DOESN T FLUCTUATE AS ELECTRICITY AND THE COOLING COIL OR THE FLAME UNDERNEATH IT JUMPS BACK AND FORTH. THE TEMPERATURE IS MAINTAINED FAIRLY CONSTANT. BUT IF YOU WERE TO MAKE ONE OUT OF LEAD FOR INSTANCE, WHICH YOU WOULDN T DO BECAUSE OF THE FEAR OF LEAD POISONING, BUT IF YOU DID, TEMPERATURE THAT TYPE OF PAN WOULD JUMP AROUND A LOT MORE THAN THE TEMPERATURE OF THE PAN UP THERE MADE FROM ALUMINUM. OKAY. WELL HOW DO WE GO ABOUT CALCULATING THESE SPECIFIC HEATS? AND LET S TAKE A LOOK AT AN EXAMPLE PROBLEM HERE THEN. WE HAVE A SOLUTION AND WE FOUND THAT 1.75 GRAMS OF THIS SOLUTION REQUIRED 13.5 JOULES OF ENERGY TO HEAT IT FROM 13.9 TO 26.4 CELSIUS DEGREES, AND WE WANT TO CALCULATE THE SPECIFIC HEAT. SO WE D SAY SPECIFIC HEAT, NOW THE CLUE WILL BE OUR UNITS. NOTICE OUR UNITS UP THERE, JOULES OVER GRAMS CELSIUS DEGREE. OH, IN THIS PARTICULAR CASE WE ALREADY SEE THE JOULES, 13.5 JOULES. THE NUMBER OF GRAMS, 1.75 GRAMS, AND THE TEMPERATURE THERE IS ACTUALLY THE CHANGE IN TEMPERATURE. NOW SPECIFIC HEAT IS PER ONE DEGREE BUT TO CALCULATE IT WE PUT IN WHATEVER DEGREE CHANGE WE HAVE TALKED ABOUT RELATIVE TO THE JOULES THAT HAVE GONE IN. SO LOOKING AT THIS THAT LOOKS LIKE WE RE TALKING ABOUT 12.5 CELSIUS DEGREES. SO, IF WE NOW CALCULATE THIS OUT WE HA VE 13.5 DIVIDED BY 1.75 AND DIVIDED BY WE HAVE A SPECIFIC HEAT FOR THAT PARTICULAR SOLUTION THEN OF JOULES OVER GRAM CELSIUS DEGREES. ANY QUESTION ON HOW WE GO ABOUT CALCULATING SPECIFIC HEAT? THESE CAN BE DOME EXPERIMENTALLY QUITE EASY AND NO WE DON T DO ONE, I WAS THINKING THAT WE DO A SPECIFIC HEAT MEASUREMENT BUT WE DO A CALORIMETRY FOR A PHASE CHANGE IN ONE OF OUR EXPERIMENTS, BUT NOT THE SPECIFIC HEAT. WELL LET S LOOK AT ANOTHER PROBLEM HERE THEN. HOW MANY JOULES OF ENERGY

4 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 4 ARE RELEASED FORMA 40 GALLON HOT WATER HEATER IF THE WATER WAS TO COOL FROM 98 CELSIUS DEGREES TO 18 CELSIUS DEGREES? HOW MANY JOULES OF ENERGY ARE GOING TO BE RELEASED? WELL AGAIN THEN WE CAN LOOK AT UNITS HERE. WE WANT JOULES. WE RE STARING WITH, WELL WE CAN GO TO GALLONS. 40 GALLONS. WE RE GOING TO MULTIPLY THAT BY 3.78 LITERS PER ONE GALLON. SO NOW WE RE AT LEAST IN METRIC UNITS OF VOLUME. NOW SPECIFIC HEAT HOWEVER OF COURSE IS IN UNITS OF JOULES GRAM CELSIUS DEGREE. SO QUITE OBVIOUS WE CAN T BE CHANGING HERE WITH A SPECIFIC HEAT CONVERSION THERE WE RE NOT TRYING TO GET RID OF LITERS. SO WE RE GOING TO HAPPEN TO HAVE SOMETHING ELSE IN HERE. WELL IT JUST SO HAPPENS I THINK I VE MENTIONED THIS BEFORE. THE DENSITY OF WATER IS CONSIDERED ONE GRAM PER MILLILITER. AND THAT IS ITS DENSITY AT 4 DEGREES CELSIUS. IT IS ACTUALLY ONE GRAM PER MILLILITER. SO WE CAN GO AHEAD AND CONVERT THEN 1000 MILLILITERS PER LITER, MULTIPLIED NOT BY ONE GRAM PER ONE MILLILITER. NOW WE RE AT THE NUMBER OF GRAMS OF WATER IN OUR WATER HEATER. IF WE WERE TO MULTIPLY NOW BY THE SPECIFIC HEAT OF WATER, AND WE RE USING LIQUID WATER HERE. SO THAT S JOULE OVER GRAM CELSIUS DEGREE. GRAMS WILL CANCEL AND WE CAN SEE WE HAVE ONE THING LEFT TO PUT IN THERE AND THAT IS HOW MUCH ARE WE HEATING OR COOLING THE WATER? THE TEMPERATURE CHANGE THAT S INVOLVED? OKAY, AND WE SEE THAT WE RE CHANGING FROM 98 TO 18 SO WE HAVE AN 80 CELSIUS DEGREE CHANGE SO TIMES 80 CELSIUS DEGREES. SO NOW OUR DEGREES CANCEL AND WE SEE THE ONLY THING WE HAVE LEFT ARE JOULES. OKAY, SO WE HAVE 40 TIMES TIMES 1000, UH TIMES AND TIMES 80. AND WE HAVE AN ANSWER OF 5.07 TIMES 10 7 JOULES. OR, KEEPING IN MIND, I LL GET YOUR QUESTION IN A SECOND, KEEPING IN MIND NOW THAT OF COURSE WE COULD EXPRESS IT IN KILOJOULES THERE ARE 1000 JOULES PER KILOJOULE, SO THAT WOULD BE 5.07 X 10 4 KILOJOULES OF ENERGY. YES, WE HAD A QUESTION BACK HERE? (STUDENT NOT AUDIBLE) UH, THIS TIME WE RE NOT ASKED, WE, WE RE NOT CALCULATING SPECIFIC HEAT. WE RE CALCULATING THE JOULES, AND WE RE USING THE SPECIFIC HEAT IN THIS PARTICULAR CASE TO DETERMINE THE JOULES. SO IF WE RE GOING TO GO FROM SPECIFIC HEAT TO JOULES WE HAVE TO GET RID OF GRAMS WHICH WE DID ALL THIS FIRST PART TO DO, AND WE HAVE TO GET RID OF UNITS OF

5 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 5 DEGREES WHICH W E DID BY DELTA T ( Τ), THE TEMPERATURE CHANGE. OKAY, ALRIGHT, ANY OTHER QUESTIONS ON THAT PARTICULAR PROBLEM THEN? OKAY, WELL LET S SEE. HERE S ANOTHER ONE. SAYS WE RE GOING TO HEAT 115 GRAMS OF LEAD FROM 150 CELSIUS DEGREES TO 328 CELSIUS DEGREES, AND THE QUESTION IS THEN HOW MUCH ENERGY IS THIS GOING TO REQUIRE? WELL NOW WE D HAVE TO GO BACK FIRST OF ALL TO OUR MELTING POINT FOR LEAD AND SO IF WE CAN GO BACK, IF I CAN FIND THAT QUICKLY OR MAYBE SOMEBODY CAN FIND IT IN THE BOOK QUICKER THAN I CAN. WE NEED THE MELTING POINT OF LEAD, UH HERE WE GO. MELTING POINT OF LEAD IS 328 DEGREES. OKAY, SO LET S PICTURE HERE WHAT THE QUESTION IS ASKING. WE RE GOING TO HEAT, SO WE RE GOING TO START WITH NOW SOLID LEAD, Pb SOLID WHICH IS AT A TEMPERATURE OF 150 DEGREES, AND WE RE GOING TO HEAT IT TO LEAD SOLID AT ITS MELTING POINT OF 328 CELSIUS DEGREES. NOW WE HAVEN T PUT IN ANY ENERGY FOR THE PHASE CHANGE, ALL WE RE PUT IN SO FAR IS THE HEATING PART. NOW ONCE WE REACH 328 DEGREES WE CAN PUT IN THE ADDITIONAL HEAT. OKAY, SO WE RE GOING TO PUT IN SOME MORE HEAT NOW AND WE RE GOING TO GO TO Pb LIQUID NOW AT 328 CELSIUS DEGREES. SO WE RE HEATING UP THE SOLID, WE RE BRINGING ABOUT THE PHASE CHANGE AND THEN WE RE GOING TO ADD THE TWO TOGETHER TO DETERMINE TOTAL AMOUNT OF ENERGY NEEDED IN THE PROCESS. OKAY SO LET S CALCULATE HERE THEN HOW MANY GRAMS, OR HOW, EXCUSE ME, HOW MANY JOULES OF ENERGY OR KILOJOULES DO WE NEED TO GO FROM HERE TO HERE? SO JOULES, WELL LET S DO IT IN KILOJOULES. KILOJOULES WILL BE EQUAL TO 115 GRAMS OF LEAD. WE RE GOING TO HEAT IT 178 DEGREES. FROM WHERE WE STARTED AT 150 UP TO WHEN IT S GOING TO GET READY TO MELT, BUT NOT MELTING YET, 178 DEGREE CHANGE, TIMES THEN THE SPECIFIC HEAT. HOW MUCH ENERGY DOES IT TAKE TO HEAT LEAD? ALRIGHT, SO AGAIN IF WE LOOK AT OUR TABLE OF SPECIFIC HEATS WE SEE LEAD WAS JOULE PER GRAM CELSIUS DEGREE. SO GRAMS HAVE CANCELLED, DEGREES HAVE CANCELLED. WE RE LEFT WITH JOULES. WE LL MULTIPLY THEN BY ONE KILOJOULE TIMES 10 3 JOULES SO THAT WE LL END UP WITH KILOJOULES FOR A UNIT. SO, SETTING UP OUR PROBLEM THEN WE RE GOING TO HAVE 115 MULTIPLIED BY 178, MULTIPLIED BY.209 AND DIVIDED BY 1000 AND WE LL HAVE AN ANSWER OF 4.28 KILOJOULES OF ENERGY. THAT S THE HEATING. WHAT ABOUT THE PHASE CHANGE? HOW

6 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 6 MANY KILOJOULES OF ENERGY ARE GOING TO BE INVOLVED IN THE PHASE CHANGE? WELL, WE NOTICE THAT THE UNIT FOR PHASE CHANGE, MOLAR HEAT OF FUSION, IS KILOJOULES PER MOLE. SO WE RE GOING TO HAVE TO KNOW HOW MANY MOLES OF LEAD WE HAVE IN ORDER TO USE THAT. WELL WE LL START 115 GRAMS OF LEAD MULTIPLIED BY ONE MOLE OF LEAD AND WE, AND WE WOULD LOOK ON THE PERIODIC CHART THEN TO DETERMINE WHAT THE ATOMIC WEIGHT OF LEAD IS AND WE SEE LEAD IS GRAMS. SO NOW WE HAVE MOLES OF LEAD. IF WE NOW MULTIPLY BY 5.12 KILOJOULES PER MOLE WE SEE THAT MOLES WILL CANCEL AND WE RE LEFT WILL KILOJOULES. THIS IS FOR THE PHASE CHANGE PART OF THE PROCESS. SO, WE HAVE 115 DIVIDED BY AND MULTIPLIED BY 5.12, AND WE SEE THAT IT TAKES 2.92, 2.92 UH, KILOJOULES. OKAY, CHECK THAT AGAIN, 115, 207.2, MULTIPLIED BY WHOOPS, SOMETHING WRONG, I DIDN T THINK THAT WAS RIGHT ALRIGHT. SO OUR HEATING PROCESS REQUIRED 4.28 KILOJOULES OUR PHASE CHANGE PROCESS REQUIRED 2.84 KILOJOULES AND SO THE QUESTION IS WHAT WAS THE TOTAL AMOUNT OF ENERGY NEEDED AND THEN OF COURSE WE CAN GO AHEAD DOWN HERE AND Q, WE USE THE LETTER Q TO TALK ABOUT HEAT ENERGY. Q TOTAL THEN WOULD BE THE 4.28 KILOJOULES FROM THE HEATING PLUS 2.84 FROM THE PHASE CHANGE TO GIVE 7.12 KILOJOULES TOTAL ENERGY NEEDED TO HEAT THE LEAD TO ITS MELTING POINT AND CONVERT IT TO A LIQUID STATE. SO FOR INSTANCE YOU WERE MAKING HOMEMADE SHELLS FOR INSTANCE YOU MIGHT BE INTERESTED IN THAT BECAUSE THAT WOULD GIVE YOU SOME IDEA OF THE AMOUNT OF ENERGY YOU HAVE TO PUT IN TO MELT THE LEAD TO PUT IT IN SO THAT YOU CAN POUR IT INTO FORMS TO PRODUCE THE BULLETS. OKAY, ANY QUESTIONS ON ANY OF THAT ONE? HOW MANY KILOJOULES OF ENERGY ARE NEEDED TO HEAT 50 GRAMS OF WATER FROM 50 CELSIUS DEGREES TO 250 CELSIUS DEGREES? AGAIN LET S KIND OF BUILD A ROADMAP HERE WHERE WE RE GOING TO BE GOING BECAUSE WE RE GOING TO START WITH WATER AT 50 DEGREES, WATER S A LIQUID. HOPEFULLY WE WOULD KNOW THAT. WE RE GOING TO HEAT IT NOW TO WATER, STILL AS A LIQUID AT 100. SO WE RE GOING TO HEAT FROM 50 TO 100. THE REASON WE RE GOING TO 100 IS BECAUSE THE BOILING POINT OF WATER, WHERE IT UNDERGOES ITS PHASE CHANGE, OCCURS AT 100 DEGREES. THEN WE RE GOING TO CONVERT THE LIQUID WATER TO GASEOUS WATER. BUT IT LL STILL BE AT 100 DEGREES. PHASE CHANGE

7 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 7 DOES NOT PRODUCE A TEMPERATURE CHANGE, AND THEN FINALLY WE RE GOING TO TAKE THE GASEOUS WATER NOW AND HEAT IT UP TO 250 CELSIUS DEGREES. SO WHAT WE SEE HERE IS THAT WE HAVE A HEATING AND THEN WE HAVE A PHASE CHANGE, AND THEN WE HAVE A HEATING. SO WE RE GOING TO NEED TO MAKE 3 CALCULATIONS. NOW ONE THING TO NOTE WHEN YOU RE DOING HEATING DEGREES COMES IN TO THE CALCULATION. HEATING OR COOLING, YOU RE GONNA HAVE DEGREES IN THERE. WHEN YOU RE DOING A PHASE CHANGE THERE S NOTHING IN MOLAR HEAT OF VAPORIZATION OR MOLAR HEAT OF FUSION THAT SAYS ANYTHING ABOUT DEGREES BECAUSE IT S A CONSTANT TEMPERATURE. PHASE CHANGES DO NOT CHANGE TEMPERATURE. ALRIGHT. SO MAKING OUR CALCULATION HERE FOR THE FIRST PART. Q OF HEATING. THIS WOULD BE HEATING NOW THE LIQUID. WE RE GOING TO TAKE THE AMOUNT OF WATER, 50 GRAMS AND WE RE GOING TO MULTIPLY IT BY JOULES PER GRAM CELSIUS DEGREES, THAT S THE SPECIFIC HEAT OF LIQUID WATER, AND WE RE HEATING IT 50 DEGREES, SO WE LL MULTIPLY BY 50 DEGREES. SO WE DEGREES HAVE CANCELLED, GRAMS HAVE CANCELLED AND WE HAVE THE AMOUNT OF ENERGY INVOLVED IN THE HEATING PROCESS. AND LET S SEE, GETTING A NUMBER THERE, WE D HAVE 50 TIMES 50 DIVIDED BY, WHOOPS, MULTIPLIED BY AND WE WOULD HAVE THE VALUE THEN OF, AND WE LL PUT IT IN KILOJOULES INSTEAD, SO 10.5 KILOJOULES WOULD BE FOR OUR FIRST STEP. NOW Q OF THE PHASE CHANGE IS GOING TO BE EQUAL TO 50 GRAMS OF WATER AGAIN, BUT REMEMBER MOLAR HEATS OF VAPORIZATION AND MOLAR HEATS OF FUSION ARE GIVEN IN KILOJOULES PER MOLE. SO WE HAVE TO CONVERT TO MOLES FIRST. SO MULTIPLY BY ONE MOLE OF WATER OVER GRAMS. SO THAT GIVES US MOLES. NOW WE NEED TO KNOW THE HEAT OF VAPORIZATION OF WATER. WELL THE HEAT OF VAPORIZATION WOULD BE FOUND BACK IN THE EARLIER PART OF THE CHAPTER WHEN WE WERE TALKING ABOUT LIQUIDS, AND SEE IF I CAN FIND THAT TABLE HERE QUICKLY. UH THE, WHOOPS, YES. THE MOLAR HEAT OF VAPORIZATION OF WATER IS 40.7 TABLE 10.2, 40.7 KILOJOULES PER MOLE. SO NOW WE SEE THAT WE RE LEFT THEN WITH KILOJOULES. SO WE HAVE UM, 50 DIVIDED BY AND MULTIPLIED BY 40.7 WE HAVE 113 KILOJOULES OF ENERGY FOR THE PHASE CHANGE. BY THE WAY, THIS IS ONE OF THE REASONS THAT STEAM BURNS ARE SO SEVERE. IF WE GO FROM GASEOUS WATER TO LIQUID WATER WITH

8 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 8 NO TEMPERATURE CHANGE, EVERY MOLE OF WATER, EVERY 18 GRAMS OF WATER, THAT CONDENSE FROM GASEOUS STEAM TO LIQUID STEAM ARE GOING TO RELEASE 40.7 KILOJOULES OF HEAT ENERGY. AND THAT S QUITE A LOT OF HEAT ENERGY. IN CONTRAST, IF ONE WE TO SPILL SOME 100 DEGREE LIQUID WATER ON YOUR SKIN AND SAY IT DROPPED TO 99 DEGREES YOU WOULD ONLY END UP WITH JOULES OF ENERGY RELEASED TO THE SKIN VERSUS 40.7 KILOJOULES OF ENERGY RELEASED TO THE SKIN. AND OF COURSE IT S ALSO THIS REASON, THIS HIGH MOLAR HEAT OF VAPORIZATION THAT MAKES WATER SUCH A GOOD ENERGY TRANSPORT MECHANISM FOR HEATING BUILDINGS. FOR INSTANCE THE BUILDINGS ON THIS CAMPUS ARE HEATED BY STEAM. WE HAVE A POWER PLANT WHICH IS CONVERTING GAS, LIQUID WATER TO GASEOUS WATER. THE GASEOUS WATER COMES THROUGH PIPES TO THE BUILDINGS. IN THE BUILDINGS IT IS CONVERTED BACK TO LIQUID WATER AND FOR EVERY 18 GRAMS OF LIQUID WATER THAT CONDENSES 40.7 KILOJOULES OF HEAT ENERGY ARE RELEASED, AND SO WE CAN HEAT BUILDINGS THEN. SO IT S A VERY GOOD ENERGY TRANSPORT MEDIUM. MUCH BETTER THAN MOST ANYTHING ELSE BECAUSE OF THIS HIGH MOLAR HEAT OF VAPORIZATION. OKAY, WE STILL HAVE ONE MORE STEP TO GO AND THAT IS WE NEED TO HEAT UP THE GAS. SO Q FOR HEATING AGAIN, STILL GOING TO BE 50 GRAMS. THIS TIME HOWEVER WE RE GOING TO HEAT IT FROM 100 TO 250 SO THAT S 150 CELSIUS DEGREES. AND W E NEED THE SPECIFIC HEAT, BUT WE NEED THE SPECIFIC HEAT OF GASEOUS WATER, NOT LIQUID WATER, BECAUSE THEY RE NOT THE SAME. WELL GASEOUS WATER IS UH JOULES OVER GRAM CELSIUS DEGREES. SO YOU SEE OUR GRAMS HAVE CANCELLED, OUR DEGREES HAVE CANCELLED, AND WE RE LEFT WITH JOULES. AGAIN WE LL WRITE IT AS KILOJOULES TO BE CONSISTENT. SO, AH THERE IT IS RIGHT THERE, WE HAVE THEN 50 MULTIPLIED BY 150, MULTIPLIED BY 150, MULTIPLIED BY 1.933, AND WE WOULD HAVE THEN 14.5 KILOJOULES OF ENERGY. NOW THE QUESTION WAS HOW MUCH ENERGY DO WE NEED TO PUT IN TOTAL TO CARRY OUT THIS PROCESS, AND SO Q TOTAL THEN WOULD BE EQUAL TO 10.5 KILOJOULES PLUS 113 KILOJOULES PLUS 14.5 KILOJOULES AND LET S SEE WE WOULD END UP THEN WITH 24, 2, 138 KILOJOULES OF ENERGY WOULD BE NEEDED TO CARRY OUT THE ENTIRE PROCESS OF HEATING LIQUID WATER UP, DO THE PHASE CHANGE, AND HEAT THE GASEOUS WATER WHICH WAS PRODUCED FROM THE PHASE CHANGE TO A HIGHER

9 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 9 TEMPERATURE. ANY QUESTION ON ANY ONE OF THOSE? OKAY. WELL A PROBLEM WITH JUST A LITTLE BIT OF A TWIST TO IT, STILL INVOLVING THE ENERGY. I HAVE 50 GRAMS OF WATER IN A CONTAINER AND IT S AT 25 CELSIUS DEGREES AND I ADD 2250 JOULES OF ENERGY TO THE CONTAINER. QUESTION IS WHAT WILL THE FINAL TEMPERATURE OF THE WATER BE? WHAT WILL THE FINAL TEMPERATURE OF THE WATER BE? WELL NOW WE KNOW THAT SPECIFIC HEAT, AND THIS IS JUST GOING TO BE A HEATING PROCESS THEN, SPECIFIC HEAT IS AGAIN JOULES OVER GRAMS CELSIUS DEGREES, AND IF WE KNOW THE SPECIFIC HEAT OF LIQUID WATER, WHICH WE DO, IF WE KNEW HOW MANY JULES OF ENERGY WE WERE PUTTING IN, WHICH WE DO, IF WE KNEW HOW MANY GRAMS OF WATER WE WERE GOING TO HEAT, WHICH WE DO, WE COULD CALCULATE THIS IS REALLY DELTA T AT THIS POINT WE COULD CALCULATE THE CHANGE IN TEMPERATURE. AND IF WE CALCULATE THE CHANGE IN THE TEMPERATURE WE KNOW WHAT WE STARTED OUT, WE COULD CALCULATE WHAT THE FINAL TEMPERATURE WOULD BE. SO WE RE GOING TO REARRANGE THIS EQUATION A LITTLE BIT THEN. WE RE GOING TO ASK FOR T. SO T IS GOING TO BE WHICH WOULD BE CELSIUS DEGREES, OR DELTA T. DELTA T IS GOING TO BE EQUAL TO, UH LET S SEE HOW DO I WANNA WRITE THIS? 2250 JOULES MULTIPLIED BY ONE GRAM OVER DELTA T, OR CELSIUS DEGREES, OVER JOULES. ALL I M DOING THERE IS I M REORDERING THOSE AND ACTUALLY THIS WOULD BE JOULES. THE NUM BER OF JOULES OF ENERGY PER GRAM CELSIUS DEGREE FOR LIQUID WATER, OR SPECIFIC HEAT, IS WE NOTICE JOULES WILL HAVE CANCELLED AND WE NOW HAVE GRAMS CELSIUS DEGREE AND WE COULD SAY MULTIPLIED BY 1 OVER 50 GRAMS. NOW GRAMS WILL CANCEL AND THE ONLY UNIT WE HAVE LEFT ARE CELSIUS DEGREES. OKAY, SO IF WE WORK THIS OUT LET S SEE WHAT WE WOULD GET HERE WE WOULD GET 2250 DIVIDED BY DIVIDED BY 50 AND IT TELLS ME THAT DELTA T WOULD BE 10.8 CELSIUS DEGREES. NOW IF I M ADDING JOULES OF ENERGY TO THE WATER OF COURSE I WOULD PREDICT THAT THE TEMPERATURE WOULD DO WHAT? GET SMALLER OR LARGER? IF I M ADDING HEAT ENERGY TO THIS WATER IS THE FINAL TEMPERATURE GONNA BE HIGHER THAN 25 OR LOWER THAN 25? HIGHER THAN 25! SO. THE FINAL TEMPERATURE, WHICH IS WHAT W E WERE TRYING TO ACHIEVE WOULD BE EQUAL TO THE TEMPERATURE WE STARTED WITH PLUS THE CHANGE THAT WE VE MAE AND WE LL CALL IT 211

10 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 10 BECAUSE WE CAN ONLY HAVE 2 PLACES ANYWAY AND SO OUR FINAL TEMPERATURE WOULD BE 36 CELSIUS DEGREES. SO WE CAN USE THE ENERGY IDEA AND SPECIFIC HEATS ET CETERA TO DO A LOT OF DIFFERENT THINGS, SORT OF LIKE WE DID WITH THE IDEAL GAS EQUATION IN THE CHAPTER ON GASES, BUT AGAIN WE CAN DO MORE THAN ONE PARTICULAR TYPE OF CALCULATION FROM THIS. ANYONE HAVE A QUESTION OVER ANY OF THESE PARTICULAR IDEAS? OKAY. NOW, THE LAST PART OF THE CHAPTER TALKS ABOUT PROBABLY ONE OF OUR MOST COMMON CHEMICAL SUBSTANCES, AND ONE MIGHT WONDER WHY YOU WOULD SPEND TIME TALKING ABOUT SOMETHING THAT IS SO COMMON AS WATER, BUT WATER HAS SOME EXTREMELY UNIQUE PROPERTIES TO IT. WATER S PRETTY SMALL FIRST OF ALL. WHEN WE CONSIDER IT HAS A MSS OF ONLY 18, THAT S A PRETTY SMALL MOLECULE, AND ONE WOULD KIND OF PREDICT THAT A VERY SMALL MOLECULE LIKE THIS WOULD HAVE VERY LITTLE ATTRACTIVE FORCES. WE VE ALREADY TALKED ABOUT THE FACT THAT THAT S NOT TRUE. WATER AHS A LOT OF ATTRACTIVE FORCES BECAUSE IT HAS DIPOLE/DIPOLE, IT HAS HYDROGEN BONDING VERY STRONG HYDROGEN BONDING AS A MATTER OF FACT MIGHT SAY, AND ALSO OF COURSE SOME DISPERSION FORCE. BUT THE FIRST TWO ARE THE MOST EXTENSIVE. IF WE WERE TO LOOK AT A SERIES OF COMPOUNDS INCLUDING HYDROGEN AND THE FAMILY OF OXYGEN. IN OTHER WORDS IF WE GO OVER HERE WE HAVE H 2 O, THEN WE COULD HAVE H 2 S, H 2 SC, H 2 Te. AND AS WE DID THAT NOTICE THAT WE WOULD BE GOING FROM A MASS FROM THE FIRST ONE 18 TO 34 TO 81 TO 130 FOR MASS. AND EVERYTHING ELSE BEING THE SAME, ONE WOULD PROBABLY PREDICT THAT WATER SHOULD HAVE THE LEAST ATTRACTIVE FORCES/ WATER SHOULD HAVE THE LOWEST MELTING POINT OR BOILING POINT OF ANY OF THOSE MATERIALS, AND YET WE FIND THAT THAT S NOT THE CASE AT ALL. IF WE LOOK AT THE BOILING POINTS OF THIS FAMILY OF COMPOUNDS HERE WE SEE A SORT OF A GENERAL TREND AS WE WOULD PREDICT. THE HEAVIEST WOULD HAVE THE HIGHEST BOILING POINT, AND THEN THE BOILING POINT COMES DOWN BECAUSE THE MASS IS GETTING LESS AND THE MASS IS GETTING LESS. AS A MATTER OF FACT, IF WE WERE TO PREDICT JUST BASED ON THE OTHER FAMILIES, OTHER FAMILY MEMBERS, WE WOULD PREDICT THAT WATER WOULD HAVE A BOILING POINT OF ABOUT MINUS 60 CELSIUS DEGREES. BUT, IT DOESN T. IT HAS A BOILING

11 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 11 POINT OF 100 CELSIUS DEGREES. SO WE SEE THAT THERE MUST BE SOME VERY VERY STRONG FORCES INVOLVED IN WATER THEN THAT ACCOUNT FOR THIS DRASTIC DIFFERENCE THAT WE SEE IN THIS, AND OF COURSE THE FACT IS THAT AGAIN WATER HAS HYDROGEN BONDING. THESE WOULD ALL HAVE DIPOLE-DIPOLE FORCES, BUT ONLY HAS THE HYDROGEN BONDING, AND OF COURSE THIS WOULD HAVE GREATER DISPERSION FORCES THAN WATER BUT DISPERSION FORCES ARE THE WEAKEST OF THE FORCES INVOLVED IN LIQUIDS. SO THAT S ONE PARTICULAR PROPERTY OR BEHAVIOR OF WATER THAT IS UNUSUAL. ANOTHER PROPERTY OF WATER THAT MAKES IT UNUSUAL, UH-OH, THERE WE ARE. IF I LOOK AT DENSITY VERSUS TEMPERATURE OF WATER WE LL START DOWN HERE AT ZERO AND TO A HUNDRED. OKAY, SO FROM JUST AT ITS BOILING POINT DOWN JUST TO ITS FREEZING POINT. AS WE COOL LIQUID WATER WE SEE SOMETHING LIKE THIS HAPPENING. THE DENSITY COMES DOWN DOWN DOWN DOWN DOWN DOWN DOWN DOWN DOWN DOWN DOWN DOWN DOWN AS WE WOULD PREDICT. DOWNA LITTLE BIT FURTHER, BUT THEN ALL OF A SUDDEN THE DENSITY GOES THIS WAY AND THE DENSITY GOES THAT WAY AT 4 CELSIUS DEGREES. WATER REACHES ITS MAXIMUM DENSITY AT 4 CELSIUS DEGREES. WHEN WE COOL WATER BELOW THAT ITS DENSITY ACTUALLY DECREASES, MEANING THAT ITS VOLUME INCREASES. IT S THE SAME AMOUNT OF WATER, IT S JUST THE ONLY THING THAT CAN CHANGE DENSITY IS HOW MUCH VOLUME IT S TAKING UP. WELL THIS IS A VERY IMPORTANT PROPERTY OF WATER HERE. THE ONE THING IS THAT WHEN LAKES COOL OFF IN THE FALL, ESPECIALLY IN AREAS WHERE WE GET COLD ENOUGH TO HAVE TEMPERATURES THAT GET NEAR FREEZING, WHEN THE LAKE COOLS OFF IN THE FALL AND ITS DENSITY BECOMES MORE DENSE, THE WATER THAT WAS ON THE SURFACE OF THE LAKE BEGINS SINKING TO REPLACE THE WATER BELOW IT. THUS, THE WATER HAS A NATURAL TURNOVER PHENOMENON AS THAT DENSITY CHANGES. THAT MEANS THAT THE WATER ON TOP HAS A LOT OF OXYGEN DISSOLVED IN IT, SETTLES TO THE BOTTOM CARRYING THE OXYGEN AND THE WATER THAT S BEEN IN THE BOTTOM OF THE LAKE ALL SUMMER AND NOW PROBABLY OXYGEN DEFICIENT IS MOVED BACK TO THE SURFACE. SO WE GET THIS NATURAL TURNOVER PHENOMENON OCCURRING IN LAKES. THAT ACCOUNTS FOR A LOT OF THEIR ABILITY TO SUSTAIN AQUATIC LIFE AT GREAT DEPTHS. OKAY, THAT S UNIQUE, THAT WATER S DENSITY

12 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 12 DECREASES ONLY TO SOME POINT AND THEN BEGINS TO EXPAND BECAUSE ALMOST ALL OTHER COMPOUNDS, ELEMENTS, WHATEVER, THEIR DENSITY CONTINUES, OOPS, I PLOTTED THAT THE WRONG WAY, BECAUSE I M SHOWING THE DENSITY TO BE THE LOWEST AT 4 DEGREES INSTEAD OF THE DENSITY TO BE THE HIGHEST AT 4 DEGREES. WELL WE CAN TAKE CARE OF THAT. OKAY, THIS IS A NEW PLOT, DENSITY AS WE COOL THE WATER INCREASES TO A MAXIMUM AND THEN IT BEGINS TO DECREASE ONCE AGAIN. NOW, ANOTHER CHARACTERISTIC HOWEVER IS THAT AS WE GO TO THE SOLID STATE, AND THIS IS AGAIN VERY DIFFERENT, AS WE GO TO THE SOLID STATE THE DENSITY OF WATER EVEN FURTHER DECREASES. AGAIN, PUTTING THIS THING UPSIDE DOWN OKAY. SO WHAT THAT MEANS IS THAT SOLID WATER HAS A DENSITY LESS THAN LIQUID WATER. THAT S WHY AN ICE CUBE FLOATS ON TOP OF THE GLASS, NOT ON THE BOTTOM. THAT S WHY LAKES FREEZE FROM THE TOP DOWN INSTEAD OF THE BOTTOM UP. THAT S IMPORTANT AS FAR AS OUT PLANT IS CONCERNED BECAUSE IF IT DIDN T, IF THE HEAVIER WATER, THE WATER THAT WAS FREEZING SETTLED TO THE BOTTOM THEN THE LAKE WOULD FREEZE FROM THE BOTTOM UP. NOW IN THE SPRING WHEN THE SUN IS TRYING TO MELT IT, IN ORDER TO GET TO THE ICE NIT WOULD HAVE TO PENETRATE THROUGH ALL OF THE LIQUID WATER FIRST TO GET DOWN FAR ENOUGH TO HIT THE ICE TO PROVIDE ENERGY TO MELT IT, WHICH IT WOULDN T DO AND SO OUR PLANT WOULD BASICALLY BE A BLOCK OF ICE WITH A COUPLE OF INCHES OF THIN MELTED WATER IN THE PEAK OF SUMMER TIME AND THAT S NOT GOOD FOR WATER SKIING OR DIVING OR ANYTHING LIKE THAT TO HAVE THAT LITTLE BIT OF WATER, AND ALSO IT WOULDN T BE VERY COMFORTABLE BECAUSE IT WOULD BE RIGHT AT ABOUT ZERO DEGREES, AND SECONDLY IT SHOULD KILL ALL OF OUR AQUATIC LIFE IF IT FROZE FROM BOTTOM UP. BECAUSE THEY RE GOING TO BE THERE, THEY RE GOING TO BE TRAPPED, AND THIS IS A VERY UNIQUE PROPERTY OF WATER, AND THE REASON THIS OCCURS IS IF WE LOOK AT WATER MOLECULES WHEN THEY RE IN THE LIQUID STATE, AND THESE ARE THESE POLAR MOLECULES OF THE HYDROGENS AND THE OXYGENS. SEE WE CAN HAVE ONE LIKE THIS AND WE CAN HAVE A HYDROGEN AND AN OXYGEN AND A HYDROGEN LIKE THAT AND THEY RE IN CONSTANT MOTION SO THEY RE SORT OF SCATTERED AROUND, BUT WHEN THEY GO TO THE SOLID STATE THEY MUST TAKE VERY SPECIFIC CRYSTAL POINTS TO MAINTAIN THEIR

13 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 13 HYDROGEN BONDING BECAUSE THEY RE NOT MOBILE TO MAKE INTERACTIONS ANYMORE SO THEY HAVE TO LINE THEMSELVES UP TO GET THE RIGHT HYDROGEN BONDING. SO IF WE HAVE ONE THERE WE RE GONNA HAVE THE NEXT WATER LIKE THIS AND THEN DOWN ON AROUND AND THEN WE GET THEN THIS NICE HEXAGONAL STRUCTURE. I MEANT TO BRING A TRANSPARENCY OF THAT BUT IT S IN THE BOOK, THIS NICE HEXAGONAL RING STRUCTURE INVOLVING THE MOLECULES. THAT S WHAT ACCOUNTS FOR THE FACT THAT THE WATER ACTUALLY HAS TO EXPAND AS IT BECOMES A SOLID INSTEAD OF CONTRACTING AS MOST OTHER LIQUIDS DO. OKAY, WELL OF COURSE GOOD IN SOME WAYS BUT BAD IN SOME WAYS. THE REASON THAT OUR ROADS ARE DESTROYED IN THE WINTERTIME IS BECAUSE OF THIS EXPANSION. THAT S THE TENDENCY TO BE ABLE TO FORM THESE HYDROGEN BONDS HAS A LOT OF FORCE INVOLVED IN IT SO THAT IF WE GET A LITTLE BIT OF WATER INTO A CRACK IN THE CONCRETE OR THE ASPHALT AND THEN IT GETS COLD ENOUGH TO FREEZE, THE WATER FREEZES AND EXPANDS, AND WHEN IT EXPANDS IT POPS THE CONCRETE APART AND STARTS KNOCKING SOME OF THE ROCK, THE SAND, WHATEVER IT IS OUT. AND IF THIS IS GOING ON FOR A LONG PERIOD OF TIME VER WINTER OF COURSE YOU FINALLY END UP WITH POTHOLES IN YOUR STREETS AND IT S BECAUSE OF THIS CONSTANT FREEZING, EXPANSION, THAWING, PENETRATING, FREEZING, EXPANDING, THAWING, PENETRATING AND SO ON THAT CAUSES THIS GREAT AMOUNT OF DAMAGE. IF YOU VE EVER ACCIDENTALLY PUT SAY A BOTTLE OF SODA OR SOMETHING IN THE FREEZER BECAUSE YOU WANTED IT TO COOL QUICK AND YOU FORGOT ABOUT IT AND WENT BACK YOU CAN SEE THE FORCE. THE BOTTLE WILL EITHER BE SHATTERED OR IF IT S A CAN THE CAN WILL RUPTURE, OR THE STOPPER, THE CAP WILL BLOW OFF THE BOTTLE. IT' A TREMENDOUS AMOUNT OF FORCE WHEN WATER GOES FROM LIQUID TO SOLID DUE TO THAT HYDROGEN BONDING. MATTER OF FACT, IT WAS USED AT ONE TIME IN CIVIL WAR THAT S THE WAY THEY WOULD SPLIT CANNON BALLS TO REMELT THEM. THEY WOULD ACTUALLY PUT WATER IN AND THEN DRIVE A METAL PLUG IN AND THEN FREEZE IT AND YOU COULD GET THE STEEL TO ACTUALLY RUPTURE FROM THE PRESSURE THAT THE WATER GAVE. SO IT S QUITE A FORCE. OKAY, LAST THING THEN, UH AT THE END OF THE CHAPTER IS A LITTLE ARTICLE THAT TALKS ABOUT WHAT IF WE LIVED ON A PLANET THAT DIDN T HAVE ANY

14 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 14 WATER? AND SOMEBODY HAD DISCOVERED IT, AND IT S SORT OF A LITTLE BIT FICTITIOUS BUT BY THE SAME TOKEN WE CAN APPLY THE SAME THING TO THE WAY THAT WE LOOK AT NEW SCIENTIFIC DISCOVERIES AND HOW BROAD-MINDED WE ARE TO NEW IDEAS AS THEY RE PRESENTED. HOW WELL DO WE EVALUATE AND REALLY MAKE A GOOD LEARNED CITIZEN DECISION WHETHER OR NOT SOMETHING SHOULD OR SHOULDN T BE USED? AND I SAID BEFORE, ONE OF THE THINGS THAT WE HAVE TO WORRY ABOUT IS THEN RISK ANALYSIS, AND SO WE LOOK AT THINGS AND EVALUATE THE RISK. BUT AGAIN, IF ONE IS TOO NARROW MINDED AND DON T LOOK BEYOND THE IMMEDIATE OBVIOUS SOMETIMES YOU MIGHT DRAW A WRONG CONCLUSIONS. BUT ON PAGE 356 THERE IT TALKS ABOUT THE EMPIRICAL CHEMICAL INDUSTRIES THAT DISCOVERED THIS NEW FIRE-FIGHTING AGENT KNOWN AS WATER, WONDERFUL AND TOTAL EXTINGUISHING RESOURCE, AND THEN THEY GO ON AND THEY TALK ABOUT AND THEN THEY TALK ABOUT SOME OF THE PEOPLE THAT ARE PROTESTING THIS WATER. SUCH THINGS AS PROFESSOR CONNIE BERRINER POINTED OUT THAT IF ANYONE IMMERSED THEIR HEAD IN A BUCKET OF WATER IT WOULD PROVE FATAL IN AS LITTLE AS 3 MINUTES. WELL THAT S TRUE, BUT HOW MANY PEOPLE IMMERSE THEIR HEAD IN THE BUCKET OF WATER FOR 3 MINUTES. IT IS TRUE THAT DIOXIN MAY BE VERY TOXIC TO ANIMALS, BUT IF WE KNOW THAT WE AREN T GOING TO INTENTIONALLY FEED IT TO THE HORSES AND THE CATTLE. OKAY, SO WE EVALUATE IT BY THE RISK AND THE BENEFIT ONCE AGAIN. UH, IT HAS BEEN REPORTED THAT WATER IS A CONSTITUENT OF BEER. SO THEY RE WORRIED ABOUT THAT THE FIREMEN WOULD BECOME INTOXICATED IF THEY WERE USING THIS NEW WATER TO DO THE FIRE EXTINGUISHING. AND SO ON, SO I JUST POINT THAT OUT THAT WATER IS UNIQUE. WATER IS UNIQUE TO OUR PLANET, AT LEAST WITHIN OUR SOLAR SYSTEM TO ANY GREAT EXTENT AND IT IS ITS PROPERTIES THAT ACCOUNT FOR A LOT OF THE THINGS THAT HELP SUSTAIN LIFE ON EARTH AND HELP MAINTAIN A PLANET WITH A FAIRLY MODERATE CLIMATE AND PROVIDE OF COURSE THE NECESSARY CHEMICALS FOR US TO ALLOW PLANT GROWTH AND ALL OF THE REST OF THE THINGS THAT WE ASSOCIATE WITH OUR PLANET. SO READ OVER THAT LITTLE ARTICLE THAT THE END, THIS WAS IN A CHEMICAL AND ENGINEERING NEWS MAGAZINE THAT I RAN ACROSS ONE DAY. ANYONE HAVE ANY QUESTIONS BEFORE WE FINISH FOR TODAY THEN?

15 CHM 105 & 106 MO1 UNIT FOUR. LECTURE ELEVEN 15 OKAY, ALRIGHT, IN OR NEXT LECTURE THEN WE WILL BEGIN LOOKING AT CHAPTER ELEVEN ON REACTION RATES.