THERMAL PROPERTIES OF MATTER

Heat is a form of energy that is transferred from a hotter to a colder region by one or more of the following methods conduction, convection and radiation.

When heat is given to a body one or more of the following may happen

(i) Increase in temperature

(ii) Increase in length ( area, volume)

(iii) Change in state of the body

Solid Liquid Liquid Gas

(iv) Change in chemical composition

(v) Change in electrical properties

(vi) Change in colour

TEMPERATURE.

Temperature is a measure of the Heat energy that a body contains or a measure of the Kinetic energy of the molecules of the material.

By heating a material the kinetic energy of the molecules of the material also increase.

The temperature also controls the direction of flow of Heat Energy.

Heat Energy always flows from a higher temperature to a lower temperature.

S.I. Unit of temperature is Kelvin.

The Kelvin scale and the kelvin are named after the Scottish physicist and engineer William Thomson, 1st Baron Kelvin (1824–1907). Unlike the degree Fahrenheit and degree Celsius, the kelvin is not referred to as a "degree" it is written K. The Kelvin and the degree Celsius are often used together, as they have the same interval, and 0 Kelvin is -273.15 degrees Celsius.

Thermometer.

A thermometer is an instrument for measuring temperature.To calibrate a thermometer two temperature called FIXED POINTS are chosen and the interval in between them is divided into a number of equal spaces called DEGREES. In this way we obtain a temperature scale.The lower fixed point called the ICE POINT is the temperature of pure melting ice a standard pressure i.e. 0 0C

The upper fixed point called the MELTING POINT is the temperature of steam from pure boiling water at standard pressure i.e. 100 oC

On the Celsius or Centigrade scale

Lower point = 0 oC And Upper point = 100 oCThe interval divided into 100 degrees.

We may also use the Kelvin scale Where Kelvin = degree celsius + 273K = oC + 273

We know that to convert any temperature in degrees Celsius to Kelvin we must add 273

In many applications it is a change in temperature that is more important than the actual initial and final temperatures.

Where the change in temperature is delta T and each temperature T1 and T0 are given in degrees CelsiusTherefore T = T1 – T0

In degrees Celsius

T (oC) = T1 (oC) – T0 (oC) = T1 – T0

In Kelvin

T(K) = T1(K) – T0(K) = (T1(oC)+273) – (T0(oC)+273)

Since the 273 values cancel each other out

T(K) = T1(oC) – T0(oC) = T (oC)

So for changes in temperature the value of the change in temperature in degrees Celsius is the same as the value of the change in temperature in Kelvin.

CHANGE IN TEMPERATURE

The change in temperature which a material experiences depends on three factors

(i) The type of material

(ii) The mass of the material And

(iii) The amount of heat energy given to the material

When there are more than two factors which determine a value to compare only two of them we must make the third factor a common standard value. To compare how different materials react to the same amount of heat i.e how two of the above factors depend on each other we must make the third factor common i.e. all materials must have equal mass and we take unit mass or 1 kg as the standard value.

SPECIFIC HEAT CAPACITY

The specific heat capacity, c, of a material is defined as the heat energy required to change the temperature of unit mass (1 kg) of the material by one degree.

S.I. Units

The symbol used for heat energy is Q Therefore using symbols the equation becomes

Rearranging gives Q = m c T Joules

Is the quantity of heat energy required to raise the temperature of m kg of a substance, of specific heat capacity c, by T degrees

Typical Values

Substance specific heat capacity J / kg K

Water 4,200

Ice 2,100

Copper 400

Aluminium 900

Iron 460

Brass 380

Note 1.water has a very large specific heat capacity value in comparison to most other materials

2. Ice and water do not have the same specific heat capacity value.

Heat Flow between bodies at different temperatures.

When a body at a higher temperature is in direct contact with a body at a lower temperature heat energy flows between the two bodies.The heat energy flows from the hotter body to the cooler body until they are both at the same temperature.

If we assume that there is no heat loss to the surroundings as the heat flow occurs then

Heat lost = Heat gainedby hotter body by cooler body

Therefore when two or more bodies are in contact they have or will achieve the same temperature.

When doing problems where substances at different temperatures are being put together(i.e mixing substances at different temperatures) always

1. Assume that there is no heat loss to the surroundings unless otherwise told And

2. always use the equation

Heat lost = Heat gainedby hotter body/bodies by cooler body/bodies

Experiment to measure Specific Heat Capacity of a Metal by Method of Mixtures

In this experiment, a piece of copper is heated to 100oC and then transferred to some water whose temperature change is noted.A simple heat balance equation enables the specific heat capacity for the metal to be calculated if that for the water is known.

1. Weigh a piece of metal using the analytical balance, attach it to a thread and leave it in boiling water for about 5 - 10 minutes, as shown

thread

boiling water

solid metal

2. Place some water into a polystyrene or plastic cup and weigh the mass of water using the analytical balance m w . Also place a thermometer into the water and note the temperature T i.

thermometer

polystyrenebeaker with lagging

cup

3. Rapidly transfer the metal to the plastic cup contained in a beaker lagged with cotton wool. Use the thermometer to stir the mixture and note the highest temperature reached.

4. Assuming that the polystyrene cup has zero specific heat capacity and given that the specific heat capacity of water = 4,200 J kg-1 K-1

Heat lost by metal = Heat gained by water

i.e.mm cm ( 100 - Tf ) = mw cw ( Tf - Ti )

where Tf is the final temperature of the mixture and Ti is the initial temperature of the water (the subscripts m and w stand for metal and water respectively).

Calculate the value for the specific heat capacity of the metal.

Problem Sheet Specific Heat Capacity

m

Question 1. If 101 J of heat is added to a 257 g piece of aluminium at 16.0. 0C calculate the final temperature of the aluminium?

Question 2. A 235 g lead ball at a temperature of 84.4oC is placed in a light calorimeter of negligible specific heat capacity containing 178 g of water at 20.5oC. Calculate the equilibrium temperature of the system.

Question 3. To determine the specific heat of an object, a student heats it to 100oC in boiling water. She then places the 82.8g object in a 150 g copper calorimeter containing 100 g of water. The copper and water are initially at a temperature of 19.4C, and are thermally insulated from their surroundings. If the final temperature is 30.3C calculate the specific heat of the object?

Question 4. How much heat energy is required to raise the temperature of a 125g iron cylinder by 58oC

Question 5. A copper block of mass 1 kg is being heated by an electric immersion heater. If the temperature of the block changes from 38oC to 68oC in 12 minutes calculate the power rating of the immersion heater.Question 6. A 1 kW electric heater takes 6 mins 6 seconds to heat 2.5 kg of water from 10oC to 40oC. Calculate (a) The % heat loss to the surroundings and

(b) if the % heat loss remains constant how long will it take to bring the temperature of the water up to 100oC.

Question 7. A ceramic coffee cup of mass 116 g and specific heat capacity 1090Jkg-1K-1, is initially at room temperature (22.00C). If 215g of 80.3oC coffee and 12.2 g of 6.00oC cream are added to the cup, what is the equilibrium temperature of the system? Assume that no heat is exchanged with the surroundings, and that the specific heat of coffee and cream are the same as the specific heat of water.

EXPANSION.

Materials expand when heated and contract when cooled. This is true for all three states of matter however gases expand more than liquids which expand more than solids.

Expansion of Solids.

Different solids e.g metals expand by different amounts when heated through the same temperature range.

Solids expand in

(i) Length

(ii) Area And

(iii) Volume

For each we define a coefficient of expansion

Coefficient of expansion SymbolLinear ( linear) Area ( superficial ) Volume ( cubical )

Linear Expansion.

Experiments show that the increase in length or expansion of a solid depends of three factors

(i) The change in temperature

(ii) The material of which the solid is made And

(iii) the original length of the material

Bimetallic Strip.

A bi-metallic strip is used to convert a temperature change into mechanical displacement. The strip consists of two strips of different metals which expand at different rates as they are heated, usually steel and copper. The strips are joined together throughout their length. The different expansions force the flat strip to bend one way if heated, and in the opposite direction if cooled below its normal temperature. The metal with the higher expansion is on the outer side of the curve when the strip is heated and on the inner side when cooled.

The coefficient of linear expansion is defined as the length by which 1 m of a material expands for every degree rise in temperature.

S.I. Units

The coefficient of linear expansion is a constant for a given material.

lo is the original length of the material

T0 is the original temperature of the material

l1 is the final length of the material

T1 is the final temperature of the material

Then

Cross multiply gives

It can also be shown that

Typical valuesMaterial coefficient of linear

expansion (x 10-5 K-1)Aluminium 2.2

Brass 1.9

Copper 1.8

Glass 0.9

Iron 1.2

Steel 1.1

Zinc 2.6

Superficial expansion

Cubical expansion

PROBLEM SHEETQuestion 1.(a) A bar of steel is 150 mm long at 20oC calculate its length at 175oC

(b)Repeat part (a) for and original length of 25 m (c) Repeat part (a) for a zinc bar

Question 2.A concrete highway is built of slabs 12 m long (150C). How wide should the expansion cracks between the slabs be (at 150C).to prevent buckling if the range of temperature is - 250C to +500C?

Question 3. An aluminum sphere is 18 cm in diameter. What will be its change in volume if it is heated from 100C to 1000C?

Question 4. A glass and an iron bar are the same length at 10oC. At 120oC they differ in length by 1.2 mm, calculate their original lengths.

Question 5 A brass plug is to be placed in a ring made of iron. At 200C, the diameter of the plug is 26.80 cm and that of the inside of the ring is 26.699 cm. (a)Calculate the common temperature in they must both be brought to in order to fit? (b) Repeat part (a)if the plug were iron and the ring brass?

Laboratory Experiment to Measure the coefficient of the linear expansivity of copper

In this experiment you will measure the linear expansion of a copper bar at various temperatures ranging from room temperature to 100oC. A graph of expansion versus temperature change will enable you to find a value for the coefficient of linear expansion of copper.

One slight difficulty is that you will not be able to measure the expansion directly, because the scale which registers changes in length is not calibrated directly. Thus, prior to commencing the experiment you will need to produce a calibration graph for this scale.

Experimental Method

Fig.1

scale

thermometer

from steam generator

Ratchet and pawl micrometer screw thread

steam out

(a) Calibration of the scale

1. Turn the micrometer screw thread (see Fig.1) until the pointer just begins to move over the scale.The pitch of the screw thread is 0.5mm. Thus one complete turn of the screw will cause the bar to move against the ratchet and pawl device by 0.5 mm. Notice that the screw has a scale showing quarter turns

2. Continue turning the screw until the scale reads 50 divisions. Note the scale reading corresponding to every quarter turn of the screw.

3. Plot a calibration graph of Scale readings versus expansion, and refer to this graph when performing the main experiment.

(b) Measurement of the expansivity

1. Measure the length of the bar L at room temperature.Adjust the micrometer screw until the scale reading corresponds to zero expansion on your calibration curve.

2. Pass steam through the apparatus for a few minutes until the temperature is steady at about 100oC.

3. Turn off the steam and note the scale reading versus temperature as the bar cools.

4. Use your calibration graph to draw up a data table of expansion and temperature change.

5. Plot an graph of expansion versus temperature change

Given that

L = L0 (T)

obtain a value for the coefficient of linear expansion of copper from its slope.