GCSE Physics Notes: Definition of specific heat capacity and its measurement

SPECIFIC HEAT CAPACITY - explanation and its applications

How to determine the specific heat capacity of a material

Applications of data - thermal energy stores and calculations

Doc Brown's school physics revision notes: GCSE physics, IGCSE physics, O level physics,  ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old students of physics

Whenever any material is heated to a higher temperature you increase the thermal energy store of the material. A measure of how much energy is needed to raise the temperature of a given amount of material to a specific temperature is called the heat capacity of the material. The specific heat capacity of a substance can be defined as the amount of energy required to change the temperature of one kilogram of the substance by one degree Celsius. From the specific heat capacity of a material, the amount of material and the temperature change the material experiences, you can calculate the increase or decrease of that material's thermal energy store.

1. Explaining and defining the specific heat capacity of materials

2. Examples of questions involving specific heat

3a. How to measure the specific heat capacity of a substance - solid

3b. Measuring the specific heat capacity of a liquid like water

3c. Two ways of measuring the heat capacity of a solid using an indirect method

4. Applications of heat capacity data - examples of thermal energy storage systems - thermal energy transfers

1. Explaining and defining the specific heat capacity of materials

It's a good idea to read Examples of energy store conversions in systems first.

Specific latent heat is dealt with on a separate page.

Whenever you get an increase in temperature of a system, energy must be transferred from one energy store to another.

However, for the same quantity of heat energy transferred, the temperature rise will vary.

The temperature rise will depends on the amount of material heated and its structure.

Don't confuse heat and temperature!

When some object is heated, the thermal energy ('heat') transferred increases the thermal energy store of the object.

The temperature increases, but the temperature only indicates how hot or cold the object is.

When you heat a material, thermal energy is absorbed and its internal energy is increased due to an increase in its thermal energy and potential energy stores.

At a particle level this is due:

(i) An increase in the kinetic energy store caused by increased vibration of solid particles or increased kinetic energy of the free movement of liquid and gas particles from one place to another.

From kinetic particle theory, a temperature value is a measure of the average kinetic energy of the particles - much of the average internal energy of the material.

(ii) An increase in the potential energy caused by the increase in kinetic energy opposing the inter-particle forces of attraction - the particles on average a bit further apart with increase in temperature.

The internal energy store is the sum of the kinetic energy store plus the potential energy store - the latter can often be ignored in the situations described here concerning heat capacity.

The energy transferred to a given material acting as a thermal energy store to raise its temperature by a specific amount can vary quite widely.

e.g. you need over four times more heat energy to raise a given mass of water to specified temperature than that for the same mass of central heating oil or aluminium (they have different specific heat capacities - but more on this later).

Application: Solar panels may contain water that is heated by radiation from the Sun.

Water has a high heat capacity and can store a lot of thermal energy.

This water may then be used to heat buildings or provide domestic hot water.

Water is the usual conveyer of thermal energy in central heating systems.

Water is a very good thermal energy store in a hot bottle for cold winter nights in bed.

Different substances store different amounts of energy per kilogram for each °C temperature rise.

To put it another way, different materials require different amounts of heat energy to raise a given amount of material by the same increase in temperature.

This is called the specific heat capacity and varies from material to material, whether it be a gas, liquid or a solid - its all to do with the nature and arrangement of the particles - atoms, ions or molecules.

Materials with a high heat capacity will release lots of heat energy when cooling down from a higher to a lower temperature.

The specific heat capacity (SHC or just c) of a substance is the amount of energy required to change the temperature of one kilogram of the substance by one degree Celsius.

This is a way of quantifying an increase or decrease in a material's thermal energy store.

The formula for expressing the amount of heat transferred between energy stores is given by the equation.

change in thermal energy store (J) = mass (kg) x specific heat capacity (J/kgoC) x change in temperature (oC)

∆E = m x c x ∆θ

E = energy transferred in Joules (change in thermal energy)

m = mass of material in kilograms kg

c = SHC = specific heat capacity J/kgoC,

θ = ∆T = temperature change in Celsius oC

The specific heat capacity of water is 4180 J/kgoC (Joules per kilogram per degree),

this means it takes 4180 J of heat energy to raise the temperature of 1 kg of water by 1oC.

The amount of energy stored in (transferred to) or released from a system as its temperature changes can be calculated using the above equation.

Other specific heat capacity values (J/kgoC):

ice 2100, aluminium 902, concrete 800, glass 670,  steel 450, brass 380, copper 385, lead 130

Because each material has a different heat capacity, although you can heat the same mass of substance from one temperature to another, you cannot assume they store the same amount of thermal heat energy per kilogram.

The materials with the highest heat capacity will store the most thermal energy per kilogram for the same increase in temperature - they are effectively a more concentrated thermal energy store.

Conversely, when allowing materials to cool, the materials with the highest specific heat capacity will release more thermal energy per kilogram for the same decrease in temperature.

2. Example questions on specific heat

Q1 A 0.5 kg block of copper absorbed 1520 J of energy and its temperature rose by 8.0oC.

Calculate the specific heat capacity of copper.

∆E = m x c x ∆T

SHCcopper = ∆E / (m x ∆T)

SHCcopper = ∆E / (m x ∆T) = 1520 / (0.5 x 8.0) = 380 J/kgoC

Q2 A block bronze has a specific heat capacity of 400 J/kgoC.

If a 1500 g block of bronze absorbs 3000 J of energy, what would be the rise in its temperature?

∆E = m x c x ∆T,  1500 g = 1.500 kg

so rearranging: ∆T = ∆E / (m x SHCbronze) = 3000 / (1.5 x 400) = 5.0oC

Q3 Calculate the thermal energy in kJ required to heat 2 kg of water (SHC = 4180 J/kgoC) from 20oC to 100oC.

∆E = m x c x ∆T = 2 x 4180 x (100-20) = 6.68 x 105 J

Thermal energy required = 668 kJ

Q4 A storage tank of water contains 500 kg of hot water at 80oC. (SHC water = 4180 J/kgoC)

How many MJ of thermal energy can be transferred to a heating system if the temperature of the water falls to 25oC?

∆E = m x c x ∆T = 500 x 4180 x (80-25) = 1.15 x 108 J

Thermal energy transferred = 1.15 x 108 / 106 = 115 MJ (1.15 x 102 MJ, 3 sf)

Q5 A more complicated problem to solve - so 'double' think carefully!

Suppose 2 kg of steel at a temperature of 80oC is placed in 10 kg (~10 litres) of water at 15oC.

Calculate the final temperature of the materials.

SHC steel = 450 J/kgoC,  SHC water = 4180 J/kgoC

The loss in J from the steel energy store = gain in J of the thermal energy store of the water.

At 'thermal equilibrium' they will be both at the same temperature, lets call this T.

The temperature fall of the steel is (80 - T), the temperature rise of the water is (T - 15)

The loss from the steel thermal energy store = ∆E = m x c x ∆T = 2 x 450 x (80 - T) = 900 x (80 - T) J

The gain of the water thermal energy store  = ∆E = m x c x ∆T = 10 x 4180 x (T - 15) = 41800 x (T - 15) J

So, change in steel thermal energy store = change in water thermal energy store

900 x (80 - T) = 41800 x (T - 15)

72000 - 900T = 41800T - 627000

(changing sides, changing signs, we get)

72000 + 627000 = 900T + 41800T

699000 = 42700T

T = 699000 / 42700 = 16.4oC (3 sf)

Q6 A night storage heater contains 75 kg of concrete (SHC = 800 J/kgoC)

At night when electricity is cheaper and the house cooler, how many kJ of thermal energy is needed to raise the temperature of the concrete from 8oC to 35oC.

∆E = m x c x ∆T = 75 x 800 x (35 - 8) = 1.62 x 106 J

1.62 x 106 / 1000 = 1620 J Q7 This question is based on a physics experiment I did in school back in circa 1961.

Its a good little experiment to do.

Accurately weigh 100.0 g (~100 ml) of water into a beaker at room temperature.

A 50.0 g brass weight was held by a thin wire string or with a pair tongs in the centre of a roaring Bunsen flame.

After a few minutes the brass weight was plunged into a beaker of 100 g cold water whose initial temperature was 19.5oC.

After gently stirring carefully with a thermometer the temperature of the water rose to 55.5oC.

SHC of brass = 380 J/kgoC, SHC water = 4180 J/kgoC

(a) If the flame temperature is T, calculate the two thermal energy store transfers.

Thermal energy transfer from brass weight = ∆E = m x c x ∆T = (50/1000) x 380 x (T - 19.5) = 19.0 (T - 19.5) J

Thermal energy transfer from brass weight = ∆E = m x c x ∆T = (100/1000) x 4180 x (55.5 - 19.5) = 15048 J

(b) Calculate the flame temperature T

Thermal energy transfer from brass weight (J) = thermal energy transfer to water (J)

19.0 (T - 19.5) = 15048

19.0T - 370.5 = 15048

19.0T = 15418.5

T = 15418.5 / 19.0 = 812oC (3 sf)

(b) Suggest some sources of error in the experiment

(i) the weight loses some thermal energy to air on transfer - possibly quite a bit, since the experimental answer you get is often well below the real answer for the flame temperature of a bunsen flame ~1000-1200oC.

(ii) the glass beaker absorbs some heat, therefore its heat capacity has been ignored

(iii) the lower part of the wire will transfer a little heat - but if you use brass wire, you could weigh that too and make sure everything is immersed in the water!

Q8 A car of mass 1000 kg is moving at 20.0 m/s is brought to a halt by sharp braking.

Assume all the kinetic energy is converted to thermal energy by four steel brake drums.

If the total mass of the steel drums of the brakes is 25.0 kg, assuming no heat losses, calculate the maximum rise in temperature the brake drums will attain.

(a) First calculate the kinetic energy (KE) of the car.

KE = ½ mv2

KE = ½ x 1000 x 202 = 2.0 x 105 J

(b) Calculate the temperature rise of the braking system (SHC of steel = 450 J/kgoC)

∆E = m x c x ∆T

∆T = ∆E / (m x c) = 2.0 x 105 / (25 x 450) = 17.8oC (3sf)

The next two questions are more tricky and involve using electricity formulae: Electricity power calculations, P = IV

The specific heat capacity of water is 4180 J/kgoC. Be careful with all the units.

Q9 A small electric kettle is connected to the 240 V AC mains supply and uses a current of 8.0 A.

It contains 1.5 kg of water (~1.5 litres, 1500 ml, 1500 cm3) at 20oC.

(a) Calculate the power of the kettle and the rate of energy transfer.

P (W) = I (a) x V (p.d. in volts) = 8 x 240 = 1920 W = 1920 J/s

(b) If the kettle is switched on for 2.0 minutes, how much energy is transferred to the thermal energy store of the water?

P = 1920 J/s, energy transferred = power x time = 1920 x 2 x 60 = 230400 J = 2.3 x 105 J

(c) What temperature will the hot water rise too?

∆E = mass of water x SHCH2O x temperature rise

∆E (J) = m (kg) x c (J/kgoC) x ∆T (oC)

Rearranging: ∆T = ∆E / (m x SHCH2O)

∆T = 230400 / (1.5 x 4180) = 36.7 oC

Therefore temperature after 2 minutes of heating is 20 + 36.7 = 56.7oC

(d) How much energy is needed to raise the water temperature from 20oC to 100oC?

∆E = mass of water x SHCH2O x temperature rise

∆E = 1.5 x 4180 x (100 - 20) = 5.016 x 105 J = 5.01 x 105 J (3 sf)

(e) Starting with the cold water at 20oC, how long would it take to boil the kettle' in minutes and seconds?

What assumptions have you made for this calculation?

The kettle's power s 1920 W, that is 1920 J/s, and energy needed =  5.016 x 105 J

P = E / t, so t = E / P = 5.016 x 105 / 1920 = 261.25 s.

Time to boil = 4 mins and 21 seconds (to the nearest second).

This calculation assumes all the electrical energy converted to thermal energy, actually increased the thermal energy store of the water.

However, you would always get a small amount of wasted thermal energy transferring to the thermal energy store of the kettle casing by conduction, and convection and radiation transfer to the thermal energy store of he surrounding air.

Q10 A hot water cylinder (immersion heater) is quoted as having a capacity of '120 litres of water' (120 kg of water).

The tank is fitted with a 3 kW heating element working off 240 V mains electricity.

(a) How much energy is needed to heat fresh cold water at 12oC to a temperature of 70oC?

∆E = mass of water x SHCH2O x temperature rise

Energy needed = 120 x 4180 x (70 - 12) = 29 092 800 J = 2.91 x 107 J (3 sf)

(b) How long will it take to heat the water up? (to the nearest minute)

P (J/s) = E (J) / t (s), 3000 = 29 092 800 / t

time = E / P = 29 092 800 / 3000 = 9697.6 seconds

9697.6 / 60 = 161.6.  time required = 162 minutes (sf)

(c) What current is flowing through the heating element?

P = I x V,  I = P / V = 3000 / 240 = 12.5 A

(d) What is the resistance of the heating element?

V = IR,  R = V / I = 240 / 12.5 = 19.2 Ω

(e) If the immersion was fitted with a 6 kW heating element, how would this affect the time to heat water up to a certain temperature?

This is twice the power of the 3 kW heating element.

P = E / t,  t = E / P, for the same quantity of water and the same temperature rise, it would take half the time compared to the 3 kW heater,

Q11

?

3a. How to measure the specific heat capacity of a substance - solid The experiment apparatus and set-up for a block of solid material

You need a block of material of known mass eg 0.5 to 1.5 kg.

So you need a mass balance.

The block must be surrounded by a good layer of insulation to minimise heat losses to the surroundings. Polystyrene would be a good insulator and because it is mainly pockets of CO2 gas of low density with a low heat capacity (low thermal energy store), but watch you don't 'overheat' and soften the polystyrene! Layers of cotton or newspaper might do.

The block must have two holes drilled in it - one for a thermometer and another for the heating element.

Do extra diagram with a joulemeter?

Its mass should be accurately measured.

The heating element is connected in series with an ammeter (to measure the current I in amperes) and a d.c. power supply e.g. 5-15 volts. The voltmeter must connected in parallel across the heating element connections.

You also need a stop clock or stopwatch.

In the experiment electrically energy is transferred and converted to heat energy which is absorbed by the block, increasing its temperature and increasing its thermal energy store.

The electrical current in the circuit does work on the heater and so transferring electrical energy from the power supply to the heaters thermal energy store which in turn is transferred to the metal block's energy store and therefore its temperature rises.

Procedure and measurements

Method (i) one set of measurements using a 0.50 kg block of aluminium

Switch on the heater setting the voltage at eg 12V (but use the accurate digital voltmeter reading for calculations).

When the block seems to be heating up steadily, start the clock/stopwatch and record the temperature.

Record the p.d. voltage and the current in amps with an accurate digital ammeter, both readings of which should be constant throughout the experiment.

After e.g. 15 minutes, record the final temperature and check the voltage and current readings and still the same and turn of the power.

When the block has cooled down, you can repeat experiment.

Method (ii) multiple measurements using a 1.1 kg block of copper

Another approach is to take the temperature reading every minute for eg 15 minutes once the copper block seems to be steadily heating up. The voltage and current readings should be constant.

This produces more data AND more reliable results than method (i)  and sorts out inconsistencies in the temperature readings.

The procedure is the same as method (i) BUT taking more temperature readings between the initial and final thermometer readings over a longer period of time.

I have assumed the same current and voltage, however, there is a lot more work in the calculations!

How to calculate the specific heat capacity of the solid

The calculations assume that all the electrical energy does end up increasing the thermal energy store of the metal block.

In reality, you can't avoid a small loss of heat through the insulation.

Results data and calculation for method (i)

Mass of an eg aluminium block 500g = 0.50 kg

Initial temperature 29.5oC, final temperature 38.5oC, temperature rise ∆T = 9.0oC

Current 0.39A, p.d. 11.5V, time 15 mins = 15 x 60 = 900 s

Power P = current x p.d. =  I x V = 0.39 x 11.5 = 4.485 W = 4.485 J/s

therefore total electrical energy = heat energy transferred = P x time = 4.485 x 900 = 4036.5 J

(Note: You can do the experiment with a joulemeter, initially set at zero, so no need for the above calculations!)

energy transferred = E (J) = m x c x ∆θ = mass of Al (kg) x SHCAl (J/kgoC) x ∆T

4036.5 = 0.5 x SHCAl x 9.0 = SHCAl x 4.5

therefore on rearranging SHCAl = 4036.5 / 4.5 = 897

so, the specific heat capacity of aluminium = 897 J/kgoC

Note that this method relies on only two temperature readings.

In SHC experiments you can include in the power supply circuit a joule meter to measure the energy transferred, which makes the calculation a lot easier. By using a joulemeter you don't need the voltmeter or ammeter.

energy transferred = mass of water x specific heat capacity of water x temperature rise

energy transferred = E (J) = m x c x ∆θ = mass of aluminium (kg) x SHCAl (J/kgoC) x ∆T

rearranging gives: SHCAl = ∆E / (mass of Al x ∆T)

Let the temperature rise a good 10 degrees and repeat the experiment at least twice to get an average - for the most accurate result.

Data and calculation for method (ii) a lot of work!

From the voltage (V) and current (I) readings you calculate the total energy transferred for all the 15 minutes of readings.

total energy transferred = P x t = I x V x t = current (A) x p.d. (V) x time in seconds

So you then have 15 total transferred energy numbers, steadily increasing from 1 to 15 minutes

Let us assume the current, voltage as method (i)

I'm assuming the thermometer can be read to the nearest 0.5oC like a typical 0-100oC school laboratory thermometer (a more accurate thermometer, mercury or digital reading to 0.1oC is most desirable!)

Therefore P = IV = 0.39A x 11.5V = 4.485J/s, energy transferred per second.

So after 1 minute energy transfer = 4.485 x 1 x 60 = ~269 J,

this finally rises to 4.485 x 15 x 60 = ~4037 J

 Time / mins 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Energy transferred / J 0 269 538 807 1076 1346 1614 1884 2153 2422 2691 2960 3229 3498 3767 4037 Temperature / oC 29 29.5 30 31 31.5 32 32.5 33 33.5 34.5 35 35.5 36.5 37 38 38.5 You then plot a graph of temperature versus energy transferred from eg 29.5oC to 38.5oC. By assuming the temperature reading is at best to the nearest 0.5oC, it makes the 'calculated' data more realistic AND justifying the multiple reading method (ii).

Graph note: The block may not heat up steadily at first and you may get a curve upwards at the start, but eventually the plot should become linear AND that is where you measure the gradient.

Calculation

Mass of copper = 1.10 kg, let c = SHCCu

The specific heat capacity equation is: E = m x c x ∆θ

energy transferred = mass of Cu x SHCCu x temperature change

Rearranging ∆E = m x c x ∆θ gives ...

∆θ = E / (m x c) and ∆θ / ∆E = 1 / (m x c)

This means the gradient of the graph = 1 / (m x c)

so, c = SHCCu = 1 / (m x gradient)

From the graph the gradient = (38 - 30) / (3800 - 500) = 8 / 3300 = 0.002424

therefore specific heat capacity of copper = SHCCu = 1 / (1.10 x 0.002424) = 1 / 0.002666 = 376 J/kgoC

Sources of error

However well insulated, the system will always be losing a small amount of its thermal energy store as it is being heated up. The system should be well insulated e.g. cotton wool or bubble wrap sheeting.

You always need to repeat experiments to be more sure of your data, but you should always be aware of sources of error and how to minimise them.

The heat energy has to conduct throughout the block and be evenly distributed, I doubt if that's the case, so the measured temperature reading might be different than the average temperature of the whole block.

The better the heat conduction of the solid, the faster the heat spreads, so better the results, so an aluminium or copper block should be ok.

The results would not be as good with a poorer conductor like concrete?

Its difficult to eliminate heat losses so the temperature rise might be a bit less than that expected for perfect insulation, but you should always use insulation around ALL of the surface of the block for this specific heat capacity experiment.

Experiment extension

You can repeat for any suitable material in solid block form.

You could also put other materials in a polystyrene container eg sand, soil  etc.

You can swap the block and insulation for an insulating polystyrene cup filled with a know mass of liquid.

It would need a lid with two holes in it for the heating element and accurate thermometer.

The procedures and calculations would be the same to determine the specific heat capacity of a liquid.

3b. Measuring the specific heat capacity of a liquid like water You can use a similar set-up to that described above for measuring the SHC of a solid block.

Instead of the block you can use a polystyrene cup (good insulation) with a lid.

Measure a mass of liquid into the polystyrene cup = mass of cup + liquid - mass of empty cup (measured on a mass balance).

You can use water for convenience.

Place the cup in an insulated box or beaker.

The double thermal insulation is essential to minimise the loss of heat energy to the surroundings.

Do extra diagram with a joulemeter?

The procedure is identical to that described for a solid.

In SHC experiments you can include in the power supply circuit a joulemeter to measure the energy transferred, which makes the calculation a lot easier.

By using a joulemeter you don't need the voltmeter and a ammeter, plus the extra calculation.

energy transferred = mass of water x specific heat capacity of water x temperature rise

energy transferred = E (J) = m x c x ∆θ = mass of water (kg) x SHCH2O (J/kgoC) x ∆T

rearranging gives: SHCH2O = ∆E / (mass of water x ∆T)

Let the temperature rise a good 10 degrees and repeat the experiment at least twice to get an average - for the most accurate result.

If you have no joulemeter, then, as in the diagram, take measurements from a voltmeter and ammeter.

The use the equation: total energy transferred (J) = P x t = I x V x t = current (A) x p.d. (V) x time in seconds

3c. Measuring the heat capacity of a solid using an indirect method (A)

(Procedures (iv)(A) and (iv)(B)  relate to calculation Q5 in the set of specific heat capacity questions above)

This method relies on the transfer between a brass weight's thermal energy store and the thermal energy store of water. Illustration of procedure A

Method (procedure A)

A brass weight is accurately weighed and placed in beaker of ice-cold water (tap water plus lumps of ice)

This is left for some time and occasionally stirred, until the brass weight is at the same temperature as the water.

250 g of water is weighed into a beaker and heated until its about 80oC (no need to boil, increases hazard).

After a gentle stir, the temperature of the warm water is taken (T2), immediately before transfer.

The temperature (T1) of the iced-water is also taken just before transferring the brass weight from the iced-water to the warm water using tongs or thin wire and loop.

After transferring the cold brass weight to the warm water, allow time for the thermal heat transfer into the brass weight to complete and after a gentle stir read the final temperature (T3)

Results and calculation (A)

Typical results:

Mass of brass weight 200 g (0.200 kg), mass of water 250 g (0.250 kg), SHC water = 4180 J/kgoC

Initial temperature of iced water T1 = 0.5oC

Initial temperature of the warm water (T2) = 79.5oC

Final temperature of water/brass in 2nd beaker (T3) = 74.0oC (cooler, as brass weight absorbs heat)

E = Thermal energy transfer to brass weight = Thermal energy transfer from water (all in J)

E = m x c x ∆T = 0.200 x SHCbrass x (T3 - T1) = 0.250 x 4180 x (T2 - T3) = 5747.5

E = 0.200 x SHCbrass x (74.0 - 0.5) = 0.250 x 4180 x (79.5 - 74.0) = 5747.5

0.20 x SHCbrass x 73.5 = 5747.5

14.7 x SHCbrass = 5747.5

SHCbrass = 5747.5 / 14.7 = 391 J/kgoC (3 sf)

Improvements and sources of error (A)

(i) Could not be quite sure if the brass weight has completely cooled down to ~0oC (T1).

(ii) The hot water in the beaker is losing heat all the time giving a greater cooling effect than just from the brass weight - might be better to use a lower start temperature in the 2nd beaker e.g. 40-50oC.

3c. continued: Alternative procedure, but a similar methodology (B) Illustration of procedure B

Method (procedure B)

You could do a similar experiment to (A) by putting the brass weight into hot water for some to heat up to ~80-90oC.

You could heat the water directly in a beaker to ~80-90oC.

Allow time for the brass weight to heat up.

After a careful gentle stir, measure the initial temperature of the brass weight (T1, ~90oC) in the hot water.

Measure the initial temperature of the cold water in the 2nd beaker (T2, ~20oC)

Pick up the brass weight with tongs or a thin wire and loop and transfer it to a beaker of water at room temperature.

Allow time for the thermal energy store transfers to take place.

After a careful gentle stir, measure the final temperature of the brass weight/water (T3) in the 'warmed up' water.

You can then do a similar calculation as above e.g.

Results and calculation (B)

Typical results:

Mass of brass weight 200 g (0.200 kg), mass of water 250 g (0.250 kg), SHC water = 4180 J/kgoC

Initial temperature of heated water/brass weight (T1) = 85.0oC

Initial temperature of the cold water (T2) = 20.5oC

Final temperature of water/brass in 2nd beaker(T3) = 24.9oC

E = Thermal energy transfer from hot brass weight = Thermal energy transfer to water (all in J)

E = m x c x ∆T = 0.200 x SHCbrass x (T1 - T3) = 0.250 x 4180 x (T3 - T2) = ?

E = 0.200 x SHCbrass x (85.0 - 24.9) = 0.250 x 4180 x (24.9 - 20.5) = 4598

0.20 x SHCbrass x 60.1 = 4598

12.02 x SHCbrass = 4598

SHCbrass = 4598 / 12.02 = 383 J/kgoC (3 sf)

Improvements and sources of error (B)

(i) Could not be quite sure if the brass weight has completely warmed up to ~80-90oC (T1)

(ii) Not absolutely sure the thermal energy transfer from the hot brass weight to the water is complete - if left too long the 2nd beaker of water will start to cool give a greater temperature change than should be measured.

(iii) The beaker absorbs some of the extra thermal energy transferred to the water's thermal energy store.

4. Applications of heat capacity data - examples of thermal energy storage systems

The greater the heat capacity of a material, the more heat energy it can hold for a given mass of material.

This means that high heat capacity materials can store lots of energy when heated and can then release a lot if cooled down. In other words, materials with a high specific heat capacity are good for storing heat energy - a good material for a thermal energy store.

Materials used in heaters/heating systems, usually have a high specific heat capacity eg water (SHC H2O = 4180 J/kgoC, very high) is used in central heating systems and is easily pumped around house to distribute lots of heat where needed, an excellent 'mobile' thermal energy store.

Water is also used as a coolant in car engines because it can absorb a lot of thermal energy for a given temperature increase. The thermal energy store of the engine block is reduced and the thermal energy store of the water increased. The thermal energy in the water is then transferred to the surrounding air to increase its thermal energy store via the radiator grill.

The good old fashioned hot water bottle is a nice convenient thermal energy store to heat up the bed.

Concrete (SHC 750-960 J/kgoC, quite high) is used in night storage heaters (using cheap night-time electricity).

The greater the mass of concrete, the greater its rise in temperature rise (safely!) the greater its capacity to store thermal energy, to be later released into the house in daytime..

Oil-filled heaters are used for a small scale heat storage (SHC oil = 900 J/kgoC, not as good as water) but will convect in the oil radiator and steadily release heat.

An archaeological note!

Prehistoric man learned thousands of years ago that hot stone retained a lot of thermal energy.

The heat capacity of natural stone is usually around 840 J/kgoC.

Large stones were heated in a fire and dropped into stone line cooking troughs like the one shown below.

The heat from the thermal energy store of the stone increases the thermal energy store of the cooler water, so boiling the water and cooking food like meat placed in the water filled trough.

It may seem crude, but brass cooking pots were something of a luxury item for many prehistoric people! This stone cooking trough is by the Bronze age stone circle (shown below) at Drombeg, Co. Cork, Ireland.

Several of them were constructed on this site and fed and connected by a diverted spring stream.

They can be found all over Ireland and the UK, and presumably on continental Europe.

Native American Indians also used the same technique by dropping hot stones into a wooden bowl of food and water. • Check out your practical work you did or teacher demonstrations you observed in Unit P1.1, all of this is part of good revision for your module examination context questions and helps with 'how science works'.

• Passing white light through a prism and detecting the infrared radiation with a thermometer.

• Demonstration using balls in a tray to show the behaviour of particles in substances in different states i.e. gas, liquid and solid.

• Measuring the cooling effect produced by evaporation by putting wet cotton wool over the bulb of a thermometer or temperature probe.

• Plan and carry out an investigation into factors that affect the rate of cooling of a can of water, eg shape, volume, and colour of can using Leslie’s cube to demonstrate the effect on radiation of altering the nature of the surface.

• Investigating thermal conduction using rods of different materials.

Heat Transfer and explaining physical changes and physical properties using a particle model

The density of materials and the particle model of matter gcse physics revision notes

gcse physics revision notes

Particle theory models, internal energy, heat transfer in state changes and latent heat and particle motion in gases (written more from a 'physics' point of view)

OCR GCSE 9-1 Gateway Science Physics: P1.2d Be able to define the term specific heat capacity and distinguish between it and the term specific latent heat. Investigation of the specific heat capacity of different metals or water using electrical heaters and a joulemeter. P1.2e Be able to apply the relationship between change in internal energy of a material and its mass, specific heat capacity and temperature change to calculate the energy change involved. IGCSE revision notes specific heat capacity KS4  physics Science notes on specific heat capacity GCSE  physics guide notes on specific heat capacity for schools colleges academies science course tutors images pictures diagrams for specific heat capacity science revision notes on specific heat capacity for revising  physics modules  physics topics notes to help on understanding of specific heat capacity university courses in technical science careers in physics jobs in the industry technical laboratory assistant apprenticeships technical internships in engineering physics USA US grade 8 grade 9 grade10 AQA  physics science GCSE notes on specific heat capacity Edexcel  physics science notes on specific heat capacity for OCR 21st century  physics science OCR GCSE Gateway  physics science notes WJEC gcse science CCEA/CEA gcse science experiment to determine specific heat capacity               Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial. 