Download Calorimetry and Specific Heat

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