Ch. 21Temperature, Heat, and Expansion

Question to the class:

• Discuss with your neighbor, in what terms have we discussed energy so far?– What equations were used? – What happened to Total Energy?

Energy cannot be created or destroyed, although in many processes energy is

transferred to the environment as heat.

Temperature: measure of how hot or cold something isTemperature ScalesCelcius (°C)Fahrenheit (°F)Kelvin (K)

Temperature and Kinetic Energy• Kinetic Energy: Energy based on motion• Temperature is proportional to the average

motion of particles. As Temperature ↑, K.E.↑

• At the atomic and molecular levels, all matter is continuously in motion!!!– For example: individual molecules of nitrogen,

oxygen, and other gases that make up the air inside a balloon move at varying speeds in random directions, vibrating and rotating.

Internal Energy

• Grand total of ALL energy in a substance.• Includes the energy of random motion of the

object’s atoms and molecules, often referred to as thermal energy.

** A substance does not contain heat, rather it contains internal energy***

HEAT

• Energy that is transferred from one object to another – because of differences in temperature – through work done by or on a system

In this sense, both heat and work have meaning only as they describe energy exchanges into and out

of the system, adding or subtracting from a system’s store of internal energy.

Heat Flow

• The transfer of internal energy from one system to another, because of a temperature difference.

Three basic kinds of heat flow: – Conduction– Convection– Radiation

• As heat is transferred to a system (object), the temperature of the system (object) may increase. – Substances vary in the amount of heat necessary

to raise their temperatures by a given amount. – More mass in the system requires more heat for a

given temperature change.

Heat flows from hot to cold

Example: A copper bar with one end hot and the other cold. Atoms

on the hot end have more kinetic energy than the atoms on the cold end.

Over time, kinetic energy will be transferred from the hot end of the bar to the cold end, and all the atoms will have nearly the same kinetic energy.

The change can be interpreted as heat flowing from hot to cold until the temperature of the bar is uniform.

Heat flows from hot to cold

Thermal Equilibrium

• When there is no heat flow between objects in contact.

• Heat Transfer is zero!!!

Also called Thermodynamic Equilibrium

• Students often confuse temperature and heat. –Temperature is a measure of the average

kinetic energy of a molecule–Heat is energy that is transferred from one

object to another

Measures of Heat

• Calorie: the amount of heat required to raise the temperature of 1 gram of water by one degree Celsius.

• Kilocalorie: 1000 calories

1 cal = 4.184 Joules

Joules (J) is the international unit for all measures of energy transfer.

Specific Heat Capacity, c• The quantity of heat required to raise the

temperature of a unit of mass of a substance by 1 degree.

Q = mcΔT

Q : the internal energy added by heat transfer to the system from the surroundings (joules or cal)

m : mass of the substance (g)c : specific heat (J/gram oC or cal/gramoC)ΔT : The difference in temperature between the

final and initial states of the system (oC)

Example:

Calculate the amount of heat needed to change the temperature of 70 grams of water by 25 oC

The specific heat of water is 1 cal/ g oC

Q = mcΔTQ =

Example:

Calculate the amount of heat needed to change the temperature of 70 grams of water by 25 oC

The specific heat of water is 1 cal/ g oC

Q = mcΔTQ = (70g) (1 cal/ g oC )(25oC)Q =

Example:

Calculate the amount of heat needed to change the temperature of 70 grams of water by 25 oC

The specific heat of water is 1 cal/ g oC

Q = mcΔTQ = (70g) (1 cal/ g oC )(25oC)Q = 1750 cal

Your turn!!

When a 50 gram piece of metal at 75 oC is placed in water, it loses 650 calories of heat while cooling to 28 oC . Calculate the specific heat capacity, c.

Your turn!!

When a 50 gram piece of metal at 75 oC is placed in water, it loses 650 calories of heat while cooling to 28 oC . Calculate the specific heat capacity, c.

Q = mcΔTc = Q / mΔTc = 650 cal (50 g)(75 oC - 28 oC)

c = 0.28 cal /g oC

Another one for you to try:A 45 g sample of Iron is dropped into a

container of water and gives off 175 calories in cooling. The specific heat of iron is 0.11 cal/goC. Calculate the temperature change of the iron.

Another one for you to try:A 45 g sample of Iron is dropped into a container of

water and gives off 175 calories in cooling. The specific heat of iron is 0.11 cal/goC. Calculate the temperature change of the iron.

Q = mcΔTΔT = Q /mc

ΔT = 175 cal (45 g) (0.11 cal/goC)

ΔT = 35.35 oC

Writing to Learn: Q = mcΔT

• Using complete sentences explain the following: – What does Q stand for, what are the units for Q?– What does c stand for, what are the units for c?– Explain how to use the equation to solve for

temperature change due to heat transferred.

Once I check your work and it is complete, you will be given your practice problems.