# Specific Heat Capacity and Latent Heat

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Specific Heat Capacity and Latent Heat
AS level Physics Notes on: “Calculations involving change of energy DQ = mcDq where m is the mass of the substance, c is its specific heat capacity and Dq is the change in temperature DQ = ml where l is specific latent heat and m is the mass of substance changing state. “

Heat Travel Revision Heat travels from hot to cold
The bigger the temperature difference the faster the rate of transfer

Internal Energy The internal energy (sometimes called the random thermal energy) of a substance is the sum of the kinetic energy and the potential energy of its particles.

Internal Energy Potential energy is due to the interaction of neighbouring particles. This is therefore very significant in solids and liquids but less so in gases. In fact ideal gases have NO potential energy – just kinetic!

Internal Energy Kinetic energy is due to the movement of the particles in the substance. The faster they move the higher the KE and the higher the temperature

Specific Heat Capacity
When you give heat energy to a substance it gets hotter if the heat is used to increase the KE of the particles. How much hotter depends on two things: Its mass and What it is made of

Specific Heat Capacity
It makes sense that the more of it you have the more energy you would have to give it to raise the temperature a certain amount of a substance and that different materials would need different amounts of heat to get hotter as their structure and the number of particles in a certain mass is different for each type. Values are therefore quoted for the heat needed to raise the temperature of 1kg of a substance by 1K

c = specific heat capacity (J kg-1 K-1)
The Equation DQ = mcDq m = mass (kg) c = specific heat capacity (J kg-1 K-1) Q = heat energy (J) q = temperature (K) The D means ‘difference in’ because if you put heat in you get an increase – if heat comes out of the system the temperature goes down. You can therefore use signs to indicate gain or loss.

Definition The Specific Heat Capacity (c) of a substance is the quantity of energy required to raise the temperature of 1kg of the substance by 1K.

SHC - c The biggest changes in temperature of a given mass of a substance will occur in those that have low specific heat capacities - because it doesn't take much energy for them to get hotter!

This is about ten times that of a metal.
Water Water has a very large SHC 4200 J kg-1 K-1 This is about ten times that of a metal.

Water It makes an excellent heat store.
To store heat you want the thing it is stored in not to get too hot – otherwise it will lose heat very quickly to the surroundings – remember heat loss to the surroundings depends on the diference in temperature between the object and its surroundings (DT)

Water Water has a large SHC Therefore is makes an excellent coolant.
A good coolant will take a lot of heat away from whatever you are trying to cool without getting too hot itself. Once the coolant is the same temperature as the thing you want cooling it will stop taking heat from it!

Latent Heat Means ‘hidden heat’
Is absorbed or evolved when a substance changes state. Doesn’t result in a change of temperature. Some substances require more energy per kilogram to change the degree of freedom of their particles than others. Relates to internal potential energy NOT the average kinetic energy of particles

Heating Ice If ice is heated steadily and its temperature recorded at regular intervals an interesting graph is obtained.

Latent Heat – hidden heat that does not register as a temperature rise
Thus heat is used to raise the temperature of something if it is not changing state. If it IS changing state then that energy is used for that purpose and the substance does not get any hotter. It gives the particles in the substance more ‘freedom’ rather than increasing their kinetic energy. It is increasing their potential energy – the internal energy of the substance.

Work The heat energy is used to do work against intermolecular forces.
In substances that expand when they change state (not always the case!) the energy is used to do work against external forces

Equation It makes sense that the more of the substance you have to melt/boil the more energy you will have to supply. Therefore we have quoted values for how much energy is required to change the state of one kilogram of various substances As the values relate to 1kg the values are termed as the specific latent heat – symbol l (for latent!) The unit of l is therefore J/kg DQ = ml

l = specific latent heat (J/kg)
Equation DQ = ml l = specific latent heat (J/kg) m = mass (kg) Q = Heat energy (J) The D indicates that heat can be input to change ‘up’ a state or given out when a substance changes ‘down’ a state

Definitions Specific latent heat is the energy required to change the state of 1 kg of a substance. Specific latent heat of fusion is the energy required to change the state of 1 kg of a substance form a solid to a liquid. Specific latent heat of vaporisation is the energy required to change the state of 1 kg of a substance form a liquid to a gas.

Calculations The layout of calculations is important
Think about what is happening. What gains energy? What loses energy? List the energy gains and losses then equate them. Use symbols for the unknowns and the answer will ‘all come out in the wash’!

Example A kettle of power 2kW holds 2 litres of water at room temperature – 20oC. If the kettle is left on for 5 minutes will it boil dry? Cwater = 4200 J kg-1 K-1 l water = 2260 kJ kg-1

Where does the energy given to the water come from?
A kettle of power 2kW holds 2 litres of water at room temperature – 20oC. If the kettle is left on for 5 minutes will it boil dry? Cwater = 4200 J kg-1 K lwater = 2260 kJ kg-1 Where does the energy given to the water come from? The electricity supply via the kettle’s electrical element How much energy is given to the water? 2000 J every second for 5 minutes Calculate this! 2000 x 5 x 60 = 6 x 105 J

DQ = mcDq Energy input = 6 x 105 J
A kettle of power 2kW holds 2 litres of water at room temperature – 20oC. If the kettle is left on for 5 minutes will it boil dry? Cwater = 4200 J kg-1 K lwater = 2260 kJ kg-1 Energy input = 6 x 105 J How much energy is required to bring the water to boiling point? What equation do we need? DQ = mcDq Okay, so how much energy is needed? (Remember that 1 litre of water has a mass of 1kg and the SHC of water is 4200 J kg-1 K-1) Q = 2 X 4200 X 80 = 6.72 x 105 J

Energy required to boil the water = 6.72 x 105 J
A kettle of power 2kW holds 2 litres of water at room temperature – 20oC. If the kettle is left on for 5 minutes will it boil dry? Cwater = 4200 J kg-1 K lwater = 2260 kJ kg-1 Energy input = 6 x 105 J Energy required to boil the water = 6.72 x 105 J What does this mean? The water wouldn’t even boil – let alone boil dry!

What would happen if the power of the kettle used was 3kW instead?
A kettle of power 3kW holds 2 litres of water at room temperature – 20oC. If the kettle is left on for 5 minutes will it boil dry? Cwater = 4200 J kg-1 K lwater = 2260 kJ kg-1 What would happen if the power of the kettle used was 3kW instead? Try it yourself!

Energy required to boil the water = mcDq = 2 x 4200 x 80
A kettle of power 3kW holds 2 litres of water at room temperature – 20oC. If the kettle is left on for 5 minutes will it boil dry? Cwater = 4200 J kg-1 K lwater = 2260 kJ kg-1 Energy supplied = 3000 x 5 x 60 = 9 x 105 J Energy required to boil the water = mcDq = 2 x 4200 x 80 = 6.72 x 105 J Energy left over to change water into steam = (9.00 – 6.72) x 105 J = 2.28 x 105 J

A kettle of power 3kW holds 2 litres of water at room temperature – 20oC. If the kettle is left on for 5 minutes will it boil dry? Cwater = 4200 J kg-1 K lwater = 2260 kJ kg-1 Energy supplied = x 5 x 60 = 9 x 105 J Energy required to boil the water = ml = 2 x = J Energy left over to change water into steam = J A surplus of – = J Therefore it would not boil dry, even then – but it would boil!!

Summary You should be able to understand what is happening to the particles in a substance when heat is added to it or taken away.

Summary If the substance is remote from its melting or boiling point it will change temperature (getting hotter or colder) as the particles vibrate faster or slower on absorbing the heat energy - the kinetic energy component changes significantly the potential energy component is virtually the same..

Summary The amount of heat energy required to make a temperature change of 1K will depend on the mass of the substance The more you have the more energy you will need, natch! It will also depend on what it is. Some structures react to heat input more dramatically than others (this is indicated in the SHC 'c' of the substance) - low 'c' substances have particles that are easier to 'vibrate'!

Summary If the substance is at its boiling/melting point the energy given/taken away will not be used to get hotter or colder. It will be used to change state. If absorbed at the change of state point (given to the substance) it will not make the particles vibrate any faster, but will make them freer from each other.

Summary If being taken away, again it will not change the vibration of the particles but rather will make them more structured - less free - and change their state to do that. So when latent heat is involved we are looking at the change of potential energy not kinetic energy.

Summary You should be able to do calculations involving heat being given and/or taken away. Remember to add up all of the components that have been 'given energy' and all of those that have 'given energy away' and equate them in an exchange of heat problem.

Summary You should be able to define specific heat capacity and latent heat and recall the equations.

AQA Exam Question 1.00 kg water is brought to the boil inside an electric kettle. Throughout the heating process the heating element supplies energy at a constant rate. Assume that the energy supplied to the kettle itself is negligible. Calculate the rate at which energy is supplied to convert water to steam if 0.50g of steam is produced each second after the water has reached boiling point. b) Estimate the rate at which the temperature increases just before the water reaches boiling point. c) State and explain the main reason why the rate of increase in temperature is not constant during the heating process.

0.5g water turns to steam each second. l = 2260 kJ kg-1 = 2260 J g-1
Calculate the rate at which energy is supplied to convert water to steam if 0.50g of steam is produced each second after the water has reached boiling point. 0.5g water turns to steam each second. l = 2260 kJ kg-1 = 2260 J g-1 Therefore energy required to do this is x 0.5/1000 = J This is supplied each second therefore the rate is 1.13 kW

To raise 1 kg water by 1K you need 4200 J
b) Estimate the rate at which the temperature increases just before the water reaches boiling point. To raise 1 kg water by 1K you need 4200 J 1130J is supplied in a second… Therefore the number of degrees Kelvin that the temperature increases by, in that last second is: 1130/4200 = 0.27K And the rate of increase is 0.27K/s

c) State and explain the main reason why the rate of increase in temperature is not constant during the heating process. Heat loss to the surroundings depends on the temperature difference between the hot body and the temperature of the surroundings. The bigger the difference the greater the rate at which heat loss will occur.

AEB Question from 1984 – Typical synoptic question.
An X-ray tube is operating at a potential of 125 kV and 10 mA. (a) If only 1% of the electrical power is converted to X-rays, at what rate is the target being heated per second? (b) If the target has a mass of 0.3kg and is made of a material with a specific heat capacity of 150J kg-1 K-1 at what rate does its temperature rise? (c) What assumption have you made in calculating the value for part b?

Energy from electrical supply per second = power P = IV
(a) If only 1% of the electrical power is converted to X-rays, at what rate is the target being heated per second? Energy from electrical supply per second = power P = IV = 10 X 10-3 x = 1250 W 99% used in heating = 1240W (3sf)

(b) If the target has a mass of 0
(b) If the target has a mass of 0.3kg and is made of a material with a specific heat capacity of 150J kg-1 K-1 at what rate does its temperature rise? E = mcDq Each second 1240 J energy is supplied to the anode target. 1240 = 0.3 x 150 x Dq Therefore the temperature rises at a rate of: Dq = 1240/(0.3 x 150) = 27.6K per second

(c) What assumption have you made in calculating the value for part b?
The assumption is that there are no thermal losses to the surroundings.