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Calorimetry and Specific Heat Heat and Temperature Basics • Temperature does not depend on the amount • If two samples of identical material are at the same temperature, the sample with more mass has more thermal energy (internal energy) • Heat is thermal energy transferred • Internal energy is thermal energy in something Which Contains More Thermal Energy? A cup of boiling water or a swimming pool frozen solid? Answer: the swimming pool. What it loses in temperature it more than makes up in mass This will become clearer as we learn more… Mixtures I • 100 g of water at 50oC is added to 100g water at 70oC. What will be the final temperature? • You guessed it: 60oC • Mix a liter of 20oC water with two liters of 30oC water and you’ll get… • Water at 2/3rd of the way up from 20 degrees to 30 degrees or 26 2/3 oC Calories • 1 calorie raises the temperature of 1 gram of water by 1 Celsius degree. • 1 kilocalorie (kcal or Calorie) raises temperature of 1 kg of water by 1 degree Celsius • 1 British Thermal Unit (BTU) raises temperature of 1 pound of water by one degree Fahrenheit Application • If 10 calories of heat go into a gram of water, how much will the temperature increase? Answer : 10 degrees C • How much heat is needed to raise the temperature of 10 grams of water by one degree C? Answer: 10 calories • How much heat is needed to raise the temperature of 10 grams of water by 10 degrees C? • Answer: 100 calories Specific Heat • How much does temperature rise when heat is put into something? • It depends on the material as well as the mass and the quantity of heat: Q = m c Dt c is specific heat in calories/g oC • Water has the highest specific heat of any common material, 1 calorie/g oC • Metals generally have low specific heats, which makes them easy to cool or heat. Example • How much heat is required to raise the temperature of 1000g water from room temperature (20oC) to boiling (100oC)? • Q = m c Dt = 1000g x 1 cal/g oC x 80 oC = 80,000 calories (or 80 kilocalories) Fact: It would take about a tenth as much heat to raise the temperature of an equal amount of iron this much Hot Stuff • What would happen if 1 kg iron (specific heat 0.11 calories/ g oC) at 300 oC were placed in 200g water at 20 oC? • Heat lost by iron = heat gained by water • Let TW be initial temp. of water; TI that of iron; TF final temp of both • mIcI(TI-TF) = mW cW (TF – TW) • mIcI TI – mIcITF = mW cW TF - mW cW TW • mIcI TI – mIcITF = mW cW TF - mW cW TW • mW cW TW + mIcI TI = (mW cW + mIcI)TF • TF = {mW cW TW + mIcI TI }/ (mW cW + mIcI) • TF ={200x1x20 + 1000x0.11x300}/(200x1+1000x0.11) • TF ={4000 + 33000}/310 = 119 oC • Boils first Method of Mixtures • How can the specific heat of an unknown liquid such as antifreeze be determined? Design an experiment to do this. Why do we put a mixture of water and antifreeze in our car and just pure antifreeze? Use Calorimeter • Cup within a cup, air insulated • Obtain hot water in known amount with known temperature. • Put into known amount of antifreeze in cup of known mass and specific heat and stir • Heat lost by water = heat gained by AF and cup • mWcW(TW-TF) = mAcA(TF-TA) + mCcC(TF-TA) • cA = {mWcW(TW-TF) - mCcC(TF-TA)}/ mA(TF-TA) High Specific Heat of Water • Makes it a good coolant (water also has high conductivity although this is not the same) • Large bodies of water such as oceans moderate climate – Gives coastal communities relatively mild summers and winters • Another peculiar fact about water. It’s highest density (and smallest volume) is at 4oC. – Water at bottom of frozen lake is always 4oC Change of Phase Required to… • Melt one gram of ice at 0oC, adding about 80 calories – Called latent heat of fusion – Equal amount is given off when 1g water freezes • Boil or vaporize one gram of water at 100 oC, about 540 calories – Called latent heat of vaporization – Equal amount given off when one g steam condenses Examples • How many calories are required to melt 100 g ice? • Q = ml = 100g x 80 cal/g = 8000 calories • How many calories are needed to boil 100 water at 100 degrees Celsius? • Q = ml = 100g x 540 cal/g = 54,000 calories