(THE STUDY OF MEASURING HEAT) CALORIMETRY (THE STUDY OF MEASURING HEAT)

WE HAVE LEARNT ABOUT TEMPERATURE LAST YEAR. TEMPERATURE IS THE DEGREE OF HOTNESS OR COLDNESS OF ANY SUBSTANCE. THIS IS MEASURED IN CENTIGRADE, FAHRENHEIT OR IN KELVIN. WHAT DO WE KNOW ABOUT HEAT? WE KNOW HEAT FLOWS FROM A BODY AT HIGHER TEMPERATURE TO A BODY AT LOWER TEMPERATURE. HEAT IS ALSO A MEASURABLE QUANTITY. THE BRANCH OF PHYSICS WHICH DEALS WITH THE MEASUREMENT OF HEAT IS CALLED CALORIMETRY. THE S.I UNIT OF HEAT IS JOULE (J). 1 CALORIE = 4.186 J WHAT IS ONE CALORIE? IT IS THE HEAT ENERGY REQUIRED TO RAISE TEMPERATURE OF 1gm OF WATER THROUGH 1 DEGREE CELSIUS

SPECIFIC HEAT CAPACITY: THIS IS MEASURED BY A CALORIMETER SPECIFIC HEAT CAPACITY: THIS IS MEASURED BY A CALORIMETER. THIS IS MADE OF COPPER BECAUSE IT IS A GOOD CONDUCTOR OF HEAT SO THAT THE VESSEL GETS THE SAME TEMPERATURE FAST IT HAS LOW SPECIFIC HEAT CAPACITY THE OUTER AND INNER SURFACES OF THE VESSEL IS POLISHED : TO REDUCE THE LOSS OF HEAT DUE TO RADIATION THE SPACE BETWEEN THE CALORIMETER AND THE WOODEN JACKET IS FILLED WITH WOOL, COTTON : TO REDUCE HEAT LOSS BY CONDUCTION IT IS COVERED WITH A WOODEN LID : TO AVOID HEAT LOSS BY CONVECTION

TO RAISE THE TEMPERATURE OF ANY SUBSTANCE, WE NEED TO SUPPLY HEAT SO WHAT DOES IT DEPEND ON? HEAT (H) SUPPLIED TO A BODY IS DIRECTLY PROPORTIONAL TO THE MASS OF THE SUBSTANCE H α m HEAT (Q) SUPPLIED TO A BODY IS DIRECTLY PROPORTIONAL TO THE RISE IN TEMPERATURE (ΔT) H α ΔT HEAT (H) SUPPLIED TO A BODY DEPENDS ON THE NATURE OF THE SUBSTANCE COMBINING THE EQUATIONS WE GET H α m Δ T (OR) H = mC Δ T

WE KNOW THAT H = mC Δ T c = H = Joules or Joules WHAT IS THE UNIT OF C ? c = H = Joules or Joules m Δ T g C kgC ARE ALL OF YOU CLEAR ABOUT THE UNIT ?

HOW TO ATTEMPT ANY PROBLEM? THERE ARE BASICALLY THREE TYPES OF PROBLEMS TYPE OF PROBLEMS WHERE THERE IS DIRECT APPLICATION OF THE FORMULA 2. TYPE OF PROBLEMS WHERE THE LAW OF CALORIMETRY IS APPLIED (i.e) HEAT LOST BY THE HOT BODY = HEAT GAINED BY THE COLD BODY 3. TYPE OF PROBLEMS WHERE A CALORIMETER IS ALSO INVOLVED TYPE 1 PROBLEM: FOLLOW THE STEPS READ THE PROBLEM CAREFULLY AND TRY TO WRITE THE FIGURES GIVEN WITH THE CORRECT SYMBOLS. IF YOU DO THIS CORRECTLY, YOU ARE ALREADY 50% CORRECT. DO NOT GO BEYOND THIS. PRACTICE THIS THOROUGHLY. ATTEMPT IN THE SAME WAY FOR ATLEAST 10 PROBLEMS. IF YOU DO THIS CORRECTLY FOR MAJORITY OF THE PROBLEMS , THEN YOU ARE GREAT

IF YOU HAVE UNDERSTOOD THIS “SUPERB” TYPE 2 PROBLEM WHERE THE PRINCIPLE OF CALORIMETRY OR PRINCIPLE OF MIXTURES NEEDS TO BE APPLIED. HEAT LOST BY A HOT BODY = HEAT GAINED BY A COLD BODY (e.g) if there are two objects say HOT WATER TEMPERATURE = 40 DEGREES COLD WATER TEMPERATURE = 10 DEGREES REMEMBER ΔT = HIGH TEMP. – LOW TEMP. THE FINAL TEMPERATURE WILL BE BETWEEN 10 DEGREES AND 40 DEGREES. ASSUME ‘X’ IS THE FINAL TEMPERATURE FOR HEAT LOST BY A HOT BODY = mc (40 – x) FOR HEAT GAINED BY A COLD BODY = mc (x – 10) IF YOU HAVE UNDERSTOOD THIS “SUPERB”

CONGRATS, NOW YOU HAVE MASTERED CALORIMETRY!!!! TYPE 3 PROBLEMS: WHERE CALORIMETER IS INVOLVED. YOU HAVE TO SEGREGATE THE PROBLEM AS HEAT LOST BY HOT BODY HEAT GAINED BY COLD BODY + HEAT GAINED BY CALORIMETER IT IS IMPORTANT FOR YOU TO SEGREGATE AND WRITE SEPARATELY. IF YOU ARE ABLE TO DO BOTH THE STEPS, IT MEANS YOU HAVE UNDERSTOOD CALORIMETRY VERY WELL. MAKE SURE YOU ARE ABLE TO SUBSTITUTE CORRECTLY. YOU CAN PAT N YOUR BACK IF YOU ARE ABLE TO ACHIEVE THIS. CONGRATS, NOW YOU HAVE MASTERED CALORIMETRY!!!!

QUESTIONS BASED ON APPLICATIONS OF CALORIMETRY ALCOHOL HAS A SPECIFIC HEAT CAPACITY OF 2180 J/kgK WHAT DOES IT MEAN? IT MEANS THAT IF THE TEMPERATURE OF 1 KG OF ALCOHOL IS SAY 10 DEGREES C, IT REQUIRES 2180J OF HEAT ENERGY TO RAISE IT TO 11 DEGREES C. QUESSTION: WATER TUBS ARE KEPT IN WAREHOUSES STORING FRUITS AND VEGETABLES IN COLD COUNTRIES DURING WINTER. ANSWER: IN COLD COUNTRIES IF THE VEGETABLES AND FRUITS ARE EXPOSED, WHAT WILL HAPPEN??? WHAT HAPPENS IF YOU STORE THEM UNDER WATER? SINCE THE OUTSIDE TEMPERATURE IS VERY LOW, IT WILL SLOWLY TAKE AWAY THE HEAT FROM WATER. FOR EVERY DEGREE LESS TEMPERATURE , EVERY kg OF WATER WILL RELEASE 4200J OF HEAT ENERGY OUTSIDE. THIS WILL ACT AS A BLANKET OF SAFETY FOR THE VEGETABLES AND FRUITS. DONT THEY?

IT IS TIME WE ALSO KNOW ABOUT WHAT IS HEAT CAPACITY OR THERMAL CAPACITY. HEAT CAPACITY IS THE AMOUNT OF HEAT ENERGY REQUIRED TO RAISE THE TEMPERATURE OF THE ENTIRE BODY BY 1 DEGREE CELSIUS. HEAT CAPACITY = MASS X SPECIFIC HEAT CAPACITY WHERE IS THIS APPLIED? ON MATERIALS WHERE THE MASS IS DELIBERATELY INCREASED: (e.g) the base of a cooking pan is made thick: why? Because it is thick, the thermal capacity becomes more. So it gets heated slowly and contains a large amount of heat. (e.g) the base of an electric iron is made heavy: why? The thermal capacity becomes more. So it remains hot for longer time even after the current is switched off.

LATENT HEAT YOU SHOULD BE CONVERSANT WITH CHANGE OF STATE MELTING FREEZING SOLIDIFICATION CHANGE IN VOLUME ON MELTING: LEAD, WAX EXPAND ON MELTING WHILE ICE CONTRACT ON MELTING EFFECT ON PRESSURE ON MELTING POINT: MELTING POINT OF ICE DECREASES WITH THE INCREASE IN PRESSURE, WHEREAS FOR SUBSTANCES LIKE LEAD, WAX, THE MELTING POINT INCREASES WITH INCREASE IN PRESSURE. EFFECT OF IMPURITIES ON MELTING POINT : MELTING POINT OF A SUBSTANCE DECREASES BY THE PRESENCE OF IMPURITIES. (e.g) by adding salt to ice.

LATENT HEAT SPECIFIC LATENT HEAT OF MELTING: HEAT ENERGY REQUIRED TO CONVERT A UNIT OF A SUBSTANCE FROM SOLID STATE TO LIQUID STATE WITHOUT CHANGE IN TEMPERATURE SPECIFIC LATENT HEAT OF FUSION: HEAT ENERGY RELEASED WHEN A UNIT MASS OF A SUBSTANCE CONVERTS FROM LIQUID TO SLID PHASE WITHOUT CHANGE IN TEMPERATURE. FOR A PURE SUBSTANCE BOTH ARE ONE AND THE SAME UNITS: J/kg OR cal/g The latent heat of fusion of ice = 336000 J/kg: WHAT DOES IT MEAN??? IT MEANS THAT 1 Kg OF ICE AT 0 DEGREES CELSIUS ABSORBS 336000 J OF HEAT ENERGY TO CONVERT INTO WATER AT 0 DEGREES CELSIUS.

APPLICATIONS OF LATENT HEAT SNOW ON MOUNTAINS DO NOT MELT IMMEDIATELY: ICE HAS A HIGH SPECIFIC LATENT HEAT OF FUSION. SO IT MELTS SLOWLY WHEN ICE IN A FROZEN LAKE STARTS MELTING, THE SURROUNDING BECOMES COLD: ICE ABSORBS A LARGE AMOUNT OF HEAT ENERGY FROM ATMOSPHERE FOR MELTING. SO THE SURROUNDING BECOMES COLD. DRINKS GET COOLED QUICKLY BY ADDING ICE THAN ADDING ICE COLD WATER ICE HAS A HIGH SPECIFIC LATENT HEAT OF FUSION 1G OF ICE ABSORBS 336J OF HEAT ENERGY FROM THE DRINK. WHERAS IF ICE COOLED WATER IS ADDED, ONLY 4.2J OF HEAT ENERGY IS ABSORBED.

NUMERICAL PROBLEMS: CONVERTING ICE AT MINUS 10 DEGREES TO WATER AT 80 DEGREES CELSIUS REMEMBER YOU ARE FINDING THE TOTAL HEAT ENERGY: THEREFORE UNIT IS IN JOULES CONVERT MINUS 10 DEGREE ICE TO 0 DEGREE ------ mCΔT --- (same state) CONVERT 0 DEGREE ICE TO 0 DEGREE WATER ------- mL ---- ( change of state) CONVERT 0 DEGREE WATER TO 80 DEGREE WATER ---- mCΔT --- (same state) ADD ALL THE VALUES. THAT’S ALL!!!!!! IF A 40W HEATER IS USED TO MELT THE ICE, HOW MUCH TIME DOES IT TAKE? ANSWER: YOU HAVE ALREADY FOUND THE HEAT ENERGY REQUIRED TO MELT. THEN THE NEXT STEP IS POWER X TIME = ENERGY (HEAT) ; REMEMBER: POWER = RATE OF CHANGE OF WORK. POWER = WORK / TIME: IN OTHER WORDS, WORK = POWER X TIME THEREFORE TIME REQUIRED TO MELT = {ENERGY OR WORK} / POWER 40 W = 40 J/s ---- remember

HAVE CONFIDENCE IN YOURSELF!!!! HOPE THESE SLIDES HAVE HELPED YOU. IF YOU FEEL I HAVE NOT COVERED SOME OTHER TOPICS, IT IS BECAUSE, I HAVE DELIBERATELY LEFT THEM OUT SO THAT WITH THESE BASICS, YOU WILL UNDERSTAND THE REST. REMEMBER WHAT IS IMPORTANT IS YOU CAN DO IT!!!! HAVE CONFIDENCE IN YOURSELF!!!!