Topic 3 – Lessons 3 and 4

Thermal physics

Today’s lesson

• Define specific heat capacity and thermal capacity.

• Solve problems involving specific heat capacities and thermal capacities.

Imagine if……..

Two beakers containing different amounts of water were heated by identical heaters for an equal amount of time.

Imagine if……..

The beaker with less water gets hotter. WHY?

Imagine if……..

Two beakers, one containing water and one containing kerosene (equal masses) were heated by identical heaters for an equal amount of time.

The beaker containing the kerosene would be hotter! WHY?

Since the amount of heat energy supplied is the same to both substances, it seems that different substances require different amounts of heat energy to cause the same temperature rise.

Heat Capacity

The relationship between the amount of heat energy a substance requires to raise its temperature by a given amount is called its thermal capacity. It is measured in J.°C-1 or J.K-1.

Definition to learn

• Thermal capacity is the amount of energy needed to raise the temperature of a substance/object by 1K.

Why are thermal capacities different?

• When a substance is heated, its internal energy increases (potential and kinetic). The stronger the force between the particles in the substance, the more heat energy goes into potential energy (and less into kinetic), so the temperature rise is less than in substances with little force between particles. Obviously the more particles there are too, the more heat energy can be absorbed.

Calculations using Thermal capacity

Energy absorbed = Thermal capacity x Temp rise

 

Q = CΔT

J J.°C-1 °C

NOT in the data booklet!

Specific heat capacity

Specific heat capacity is the amount of energy needed to raise the temperature of unit mass of a substance by 1K

Specific heat capacity of water = 4186 J.kg-1.°C-1

Specific heat capacity of kerosene = 2010 J.kg-1.°C-1

Specific heat capacity of mercury = 140 J.kg-1.°C-1

Calculations using S.H.C.

Energy absorbed = Mass x Specific Heat capacity x Temp rise

 

Q = mcΔT

J kg

J.kg-1.K-1 K

For example

500 g of olive oil is heated until its temperature rises by 120 K. If the specific heat capacity of olive oil is 1970 J.kg-1.K-1, how much heat energy was used?

Energy absorbed = Mass x Specific Heat capacity x Temp rise

Energy absorbed = 0.5 x 1970 x 120

Energy absorbed = 118200 J

An analogy: Water and wetness

“This analogy is one of my ideas!”

Richard Feynmann – Nobel prize winning Physicist, lock-picker and bongo

player

Two towels – same size/mass

• You can add the same amount of water (heat), but the cheaper towel will be “wetter” (temperature). They have different capacities for absorbing water

Let’s try some questions!

Investigation time!

Let’s do an experiment to measure specific heat capacities

Specific heat capacities

• Cwater = 4181.3 +/- 0.1 J.kg.°C-1

• Caluminium = 897 +/- 1 J.kg.°C-1

Things to note:

• If specific heat capacity is constant, the temperature will rise at a uniform rate so long as the power input is constant and no energy is lost to the outside.

• There are large potential heat losses if the substance is not well insulated. These can be accounted for in some experiments (how?).

• You should be able to think of a number of reasons why your value does not match that in the data book.