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