Thermodynamics

Ch 10 EnergySections 10.4-10.6

ThermodynamicsThe Law of Conservation of Energy is also known as The 1st Law of Thermodynamics.

Internal Energy (E) = kinetic energy + potential energy

It can be stated as “the energy of the universe is constant.”

ΔE = q + w

Δ (delta) means a changeq = heat absorbed by the system

w = work done on the system

Units for Measuring HeatThe Joule is the SI system unit for measuring heat:

The calorie is the heat required to raise the temperature of 1 gram of water by 1°Celsius

2

2111smkgmeternewtonJoule

1 cal = 4.184 J

Energy & Temperature of Matter

The amount the temperature of an object increases depends on the amount of heat added (q).

The amount the temperature of an object increases when heat is added depends on its mass

– If you double the added heat energy the temperature will increase twice as much.

– If you double the mass it will take twice as much heat energy to raise the temperature the same amount.

Converting Calories to JoulesHow many calories of energy correspond to 28.4 J?

Express 34.8 cal of energy in units of joules.

28.4 J x 1 cal = 6.79 cal 4.184 J

34.8 cal x 4.184 J= 146 J 1 cal

Express 47.3 J of energy in units of calories.47.3 J x 1 cal = 11.3 cal

4.184 J

Specific Heat

The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.

1

2

3

45 6

7

8

9

1 102

3

45 6

7

8

9

11

Calculations Involving Specific Heat

c = Specific Heat

Q = Heat lost or gained

T = Temperature change

OR

m = Mass

TmQc

cTmQ

Specific HeatThe amount of heat required to raise the temperature of one gram of substance by one degree Celsius.Substance Specific Heat (J/g·°C)

Water (liquid) 4.18Ethanol (liquid) 2.44Water (solid) ICE 2.06Water (vapor) STEAM 1.87Aluminum (solid) 0.897Carbon (graphite,solid) 0.709Iron (solid) 0.449Copper (solid) 0.385Mercury (liquid) 0.140Lead (solid) 0.129Gold (solid) 0.129

Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C.

cTmQ c = Specific HeatQ = Heat lost or gainedT = Temperature changem = Mass

Specific heat of water = 4.184 C g

J

°Mass = 7.40 g

Temperature change = 46.0°C – 29.0°C = 17.0°C

J 526 C17.07.40gC g

J 4.184 Q Heat

A 1.6 g sample of metal that appears to be gold requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold?.

mTcQ c = Specific HeatQ = Heat lost or gainedT = Temperature changem = MassSpecific heat of gold = 0.13

C g

J

°

Mass = 1.6 g

Temperature change = 41.0°C – 23.0°C = 18.0°C

T mQ c

Tmc Q

C gJ0.20

C18 x g 1.6J 5.8 c

The metal cannot be pure gold!

Calculate the joules of energy required to heat 454g of water from 5.4°C to 98.6°C ?

Remember, 4.184 J of energy is required to change the temperature of each gram of water by 1°C, so………

4.184 J x 454 g x 93.2°C = 1.77 X 105 J g °C

Determine the amount of energy as heat that is required to raise the temperature of 1.0 L of water from 25.0 C to boiling. (in Joules? in Calories?)

4.184 J x 1000 g x 75.0°C = 313,800 J g °C

if 1 mL of H2O = 1 g of H2Othen 1.O L of H2O = 1.0 kg of H2O

= 314,000 J = 314 kJ

Energy in Joules

Energy in Calories 314,000 J x 1 cal = 7.50 x 104 cal = 75.0

kcal 4.184 J

If 455 J of heat is transferred to 25.0 g of water at 45.0°C, what is the final temperature of the water?

4.184 J x 25.0 g x (Tfinal - 45.0°C) = 455 J g °C

Let’s set of the equation, then use algebra to solve for temperature

(Tfinal – 45.0°C) = 455 J 4.184 J x 25.0 g g °C

(Tfinal – 45.0°C) = 455 104.6

Tfinal = 45.0 + 4.34 = 49.3 °C

Heat of SolutionThe Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

Substance

Heat of Solution (kJ/mol)

NaOH -44.51NH4NO3 +25.69KNO3 +34.89KNO3 +34.89KNO3 +34.89HCl -74.84