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 50 o C is added to 100g water at 70 o C. What will be the final temperature? You guessed it: 60 o C Mix a liter of 20 o C water with two liters of 30 o C water and you’ll get… Water at 2/3 rd of the way up from 20 degrees to 30 degrees or 26 2/3 o C

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  t c is specific heat in calories/g o C Water has the highest specific heat of any common material, 1 calorie/g o C 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 (20 o C) to boiling (100 o C)? Q = m c  t = 1000g x 1 cal/g o C x 80 o C = 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 o C) at 300 o C were placed in 200g water at 20 o C? Heat lost by iron = heat gained by water Let T W be initial temp. of water; T I that of iron; T F final temp of both m I c I (T I -T F ) = m W c W (T F – T W ) m I c I T I – m I c I T F = m W c W T F - m W c W T W

m W c W T W + m I c I T I = (m W c W + m I c I )T F T F = {m W c W T W + m I c I T I }/ (m W c W + m I c I ) T F ={200 x 1 x x 0.11 x 300}/(200 x x0.11) T F ={ }/310 = 119 o C 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 m W c W  T W -T F ) = m A c A (T F -T A ) + m C c C (T F -T A ) c A = {m W c W  T W -T F ) - m C c C (T F -T A )}/ m A (T F -T A )

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 4 o C. –Water at bottom of frozen lake is always 4 o C

Change of Phase

Required to… Melt one gram of ice at 0 o C, 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 o C, 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