Ü Energy Involved in Making Or Breaking Attractions Between Particles of a Substance

Quantitative Energy Calculations 2013

1. Energy of physical change (DH)

ü  energy involved in making or breaking attractions between particles of a substance.

ü  energy amount is a physical property of a substance

ü  DH to make the attractions is equal but opposite to the DH to break them.

E.g. for water (Joules (J) is the unit for energy)

make attractions DHSOL = (-) 334 J/g SOL = solidification (freezing)

energy released

break attractions DHFUS = (+) 334 J/g FUS = fusion (melting)

energy absorbed

make attractions DHCOND = (-) 2260 J/g COND = condensation

energy released

break attractions DHVAP = (+) 2260 J/g VAP = vaporization

energy absorbed (boiling or evaporation)

(-) means LEAVING the system (exothermic)

(+) means LEAVING the system (endothermic)

2. (Specific) Heat Capacity (c)

ü  how quickly a substance absorbs heat and how quickly it releases heat.

E.g. Metals get hot quickly and lose heat quickly so they have small heat capacities (c).

Water takes longer to heat up and longer to release heat so it has a larger heat

capacity (c).

Specific heat for metals:

Aluminum 0.90 J/g oC Iron 0.46 J/g oC Silver 0.24 J/goC

Specific heat for water:

Liquid water 4.18 J/g oC Solid water 2.1 J/g oC

3. Heat terms and units:

Heat Q Joules (J)

Heat of physical change DH J/g

Heat capacity C J/g oC

Mass m g

4. Formulas for calculating energy of physical change:

Plateau (phase change) Q = m DH

Slant (temperature change) Q = mC DT

5. Solving quantitative energy calculations:

ü  draw heating or cooling curve and identify where “story” is happening on curve.

ü  fill in values for correct equation for each segment of curve and solve.

E.g. Gold has a specific heat capacity of 0.129 J/g. How much energy is required to raise

the temperature of 15 g of gold from 22 oC to 85oC?

DT = 85 – 22 = 63 oC

Q = mCDT = (15 g)(0.129 J/g)(63oC)

= 121.91 J = 120 J

E.g. A bag of ice (450 g) at 0.0oC sits on the counter and melts to liquid water. How much

energy is absorbed by the ice? DHFUS = 334 J/g for water.

Q = mDH = (450 g)(334 J/g) = 150300 J

= 150000 J

= 150 KJ