Unit 2 – Matter and Energy
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Transcript of Unit 2 – Matter and Energy
Unit 2 – Matter and EnergyMrs. Callender
Lesson Essential Question
How do I calculate energy in a system?
Hydrocarbons and Heat
For example, the amount of energy a food releases when burned, can tell us about it’s caloric content (fats release lots of energy).
Most hydrocarbons are used as fuels. Knowing how much energy a fuel provides can tell us if it is useful for a certain application.
Bomb Calorimeter
Measures heat released during combustion.
C57H104O6 + 80 O2 ---> 57 CO2 + 52 H2O + energy
Combustion Equation for a peanut:
Calories and Joules
1 Calorie (food) = 4200 joules
1000 calories = 1 dietary Calorie
Food Energy ProblemThe fuel value of peanuts is 25 KJ/g. If an average adult needs 2800 kilocalories of energy a day, what mass of peanuts would meet an average adult’s energy needs for the day? Assume all of the fuel value of the peanuts can be converted to useful energy.
X 1 Calorie1 Kcalories
2800 Kcaloriesda
y
X 4200 J1 Calorie
X1000 J
1 KJX
25 KJ
1 g
= 470 g of peanuts
Law of Conservation of Energy
Reminder:
Energy Gained = Energy Lost in a system
q gained = q lost
Sample Problem #1If a piece of aluminum with mass 3.90 g and a temperature of 99.3 oC is dropped into 10.0 cm3 of water at 2.6 oC, what will be the final temperature of the system? (Recall the density of water is 1.00 g/cm3)
System Example #1LOSS GAINED
m = 3.90 g
m = 10.0 g
Cp = 0.9025 J/goC
Cp = 4.184 J/goC
Ti = 99.3 oC Ti = 22.6 oC
Tf = ? oC Tf = ? oC
(3.90 g)
mCpΔT = mCpΔT
(99.3 - Tf)
(0.9025 J/goC)
qloss = qgained
= (10.0 g)
(4.184 J/goC)
(Tf - 22.6)
(3.51975)
(99.3 - Tf)
41.84 Tf – 945.584
= (41.84 )
(Tf - 22.6) =349.51117
5 - 3.51975 Tf +
945.584+ 945.584 1295.09517
5 - 3.51975 Tf
= 41.84 Tf +
3.51975 Tf
+ 3.51975 Tf
= 45.35975 Tf 45.35975 45.35975
Tf=28.6 oC
1295.095175