SITEMAP   School-college Physics Notes: Thermal energy 4.7 Particle models

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Thermal energy & particle theory: 4.7 Everyday examples of using latent heat - ice cooling drinks, refrigerator, specific latent heat calculations

Doc Brown's Physics exam study revision notes

4.7a Some everyday examples of latent heat - internal energy transfers

• Whenever materials at different temperatures are placed in contact with each other, there will be an internal energy transfer of thermal energy from the hotter material to the colder material.

• (1) Using ice to cool a drink

• When you add ice to a drink to cool it, an internal energy change takes place involving the latent heat of melting.

• The ice is at a lower temperature than the liquid drink.

• The higher energy liquid particles transfer kinetic energy to the ice, increasing its internal energy.

• Sufficient thermal energy - the latent heat of melting, is absorbed by the ice to melt it.

• The energy is needed to weaken the intermolecular forces between the water molecules in the ice sufficiently to cause melting - when the particles have sufficient energy to break free from the inter-particle attractive forces.

• The ice warms up and your drink cools down - an internal thermal energy transfer!

• (2) Refrigerator - freezer

• In a refrigerator system, an electric pump is used to compress a gas to liquefy it - this pump-compressor is forcing condensation to take place and releases the latent heat of condensation.

• The liquid is then allowed to evaporate, absorbing its latent heat of vapourisation.

• This thermal internal energy is taken from the contents of the fridge-freezer.

• The internal energy of the freezer contents is reduced, lowering the temperature of the food.

• You can feel the warm air at the back of your fridge, this is from the release of the latent heat of condensation.

4.7b Defining specific latent heat

• The specific latent heat of a substance is the quantity of energy need to change 1 kg of the material from one state to another without change in temperature.

• (a) In heating a material to effect a state change e.g. melting or boiling, the specific latent heat must be added.

• (b) In cooling a material to effect a state change e.g. condensing or freezing, the specific latent heat must be removed (released) from the system.

• Specific latent heat values differ from substance to substance because of different values of inter-particle forces (intermolecular bonding) and also the state change itself for a specific substance (solid <=> liquid OR liquid<=> gas).

• Generally speaking latent heats of boiling/condensing are numerical much greater than latent heats of melting/freezing.

• The latent heat for the state changes solid <=> liquid is called the specific latent heat of fusion (for melting or freezing).

• The latent heat for the state changes liquid <=> gas is called the specific latent heat of vaporisation (for condensing, evaporation or boiling)

• Specific heat capacity is dealt with on separate pages

4.7c Examples of worked-out specific latent heat calculations

• To work out the energy needed or released to change the state of mass m of a substance the following formula applies

• thermal energy transfer = mass x specific latent heat

• E (J) = m (kg) x L (J/kg)

• E = mL

• (sometimes quoted as Q = mL, where Q = amount of thermal energy transferred)

• You may also need the

• energy transferred = mass x specific heat capacity x temperature change

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

• SHC often denoted by c, ∆T often denoted as ∆θ

• so the equation is simply E = mc∆T  or  E = mc∆θ

• Its a bit of a nuisance that the same notation isn't used uniformly, but get used to it!

• You may also need the electricity power equations P (W) = I (A) x V (V)  or  P (W) = E (J) / t (s)

• Some examples of latent heat calculations

• Q1 The latent heat of fusion of water is 334 000 J/kg (334 kJ/kg).

•  Q2 The specific latent heat of vaporisation of water is 2 265 000 J/kg (2265 kJ/kg).

• Q3 For aluminium, the latent heat of fusion is 397 000 J/kg and the latent heat of vaporisation is 11 400 000 J/kg.

• Q4 What mass of ice can be melted by 1 million J of heat energy?

• Q5 In an experiment 5 g of solid gold required 322 J of heat energy to melt it at 1063oC.

• Examples Q6 and Q7 are a bit more tricky to work out, so see if you follow the arguments!

• Q6 (a) How much energy is needed to convert 500 g of ice at 0oC to steam at 100oC?

You need three other pieces of information to complete the calculation and three intermediate calculations to get to the final answer.

The SHC of water is 4180 J/kgoC, the latent heat of fusion of water (ice) is 334 000 J/kg and the latent heat of vaporisation of water is 2 265 000 J/kg and 550 g = 0.50 k.

(i) The energy to melt the ice:

(ii) The energy to raise the temperature of the water from 0oC to 100oC:

(iii) The energy to boil the water at 100oC:

(iv) Finally, add all of (i) to (iii) together.

• -

• (b) If you monitored the temperature rise with time as the ice was being heated up, sketch the temperature - time graph you might expect and explain its features

• Worked out ANSWERS to the latent heat questions

• -

• Q7 A block of ice at -10oC was melted and further heated up to 20oC.

• The SHC of ice is 2100 J/kgoC, the SHC of water is 4180 J/kgoC and the latent heat of fusion of water (ice) is 334 000 J/kg.

• (a) If 200 000 J of thermal energy were supplied to the ice, what was the original mass of ice?

• This involves multiple stages of calculation and some clear logical thought!

• Let m be the mass of ice.

• (i) Energy needed to warm the ice from -10oC to 0oC.

• (ii) Energy needed to melt the ice

• (iii) Energy needed to warm the water from 0oC to 20oC.

• (iv) The total energy needed

• (v) Now the total energy needed = total energy supplied

• Worked out ANSWERS to the latent heat questions

• -

• (b) If you monitored the temperature rise with time, sketch the temperature - time graph you might expect and explain its features.

• Worked out ANSWERS to the latent heat questions

• -

• Q8 A 500 Watt heating element is used to heat up 1.50 kg of a solid until it just reaches its melting point, but before it actually starts melting.

Keywords, phrases and learning objectives for thermal energy, uses of latent heat and specific latent heat calculation questions

Be able to explain how a refrigerator works in terms of latent heat.

Know and explain the latent heat changes when water is cooled to make ice.

Be able to do specific latent heat calculations given appropriate data e.g. applying to state changes.

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Worked out ANSWERS to the velocity-graph questions
• Some examples of latent heat calculations

• Q1 The latent heat of fusion of water is 334 000 J/kg (334 kJ/kg).

• How much energy is needed to melt 5.5 kg of ice?

• E = mL

• E = 5.5 x 334 000 = 1 837 000 J  = 1837 kJ = 1840 kJ or  1.84 x 106 J (3 sf)

• -

•  Q2 The specific latent heat of vaporisation of water is 2 265 000 J/kg (2265 kJ/kg).

• How much energy is needed to boil 250 g of water at 100oC?

• 250g = 250/1000 = 0.25 kg

• E = mL = 0.25 x 2 265 000 = 566 250 J = 566 kJ or 5.66 x 105 J (3 sf)

• -

• Q3 For aluminium, the latent heat of fusion is 397 000 J/kg and the latent heat of vaporisation is 11 400 000 J/kg.

• How much energy is needed to completely vapourise 1.5 kg of aluminium

• For melting: E = mL = 1.5 x 397 000 = 595 500 J, 595.5 kJ

• For vaporisation:: E = mL = 1.5 x 11 400 000 = 17 100 000 J, 17 100 kJ

• Total energy needed = 595.5 + 17 100 = 17 700 kJ or 1.77 x 107 J (3 sf)

• -

• Q4 What mass of ice can be melted by 1 million J of heat energy?

• Latent heat of fusion of water is 334 000 J/kg

• E = mL, rearranging gives m = E / L

• m = 1 000 000 / 334 000 = 3.0 kg

• -

• Q5 In an experiment 5 g of solid gold required 322 J of heat energy to melt it at 1063oC.

• Calculate the latent heat of fusion of gold.

• 5 g = 5 / 1000 = 0.005 kg

• E = mL, rearranging gives L = E / m

• L = 322 / 0.005 = 64 400 J/kg

• -

• Examples Q6 and Q7 are a bit more tricky to work out, so see if you follow the arguments!

• Q6 (a) How much energy is needed to convert 500 g of ice at 0oC to steam at 100oC?

You need three other pieces of information to complete the calculation and three intermediate calculations to get to the final answer.

The SHC of water is 4180 J/kgoC, the latent heat of fusion of water (ice) is 334 000 J/kg and the latent heat of vaporisation of water is 2 265 000 J/kg and 550 g = 0.50 k.

(i) The energy to melt the ice:

• E = mL

• E = 0.5 x 334 000

• E = 167 000 J

(ii) The energy to raise the temperature of the water from 0oC to 100oC:

• E = mc∆T

• E = 0.5 x 4180 x 100 = 209 000 J

(iii) The energy to boil the water at 100oC:

• E = mL

• E = 0.5 x 2 265 000 = 1 132 500 J

(iv) Finally, add all of (i) to (iii) together.

• Total energy needed = 167 000  +  209 000  +  1 132 500 = 1 508 000 J = 1.51 x 106 J  (3 sf)

• -

• (b) If you monitored the temperature rise with time as the ice was being heated up, sketch the temperature - time graph you might expect and explain its features.

•

• Initially the ice is melting at 0oC and the ice/water mixture stays at a constant temperature as the latent heat of fusion is absorbed.

• This is an increase in internal energy that weakens the inter-particle forces sufficiently to cause melting.

• Then the liquid water steadily rises in temperature until the boiling point is reached at 100oC.

• The kinetic energy of the molecules is steadily increasing, increasing the temperature of them.

• The temperature of the water then remains constant at 100oC as the water boils away it is absorbing the latent heat of vapourisation.

• This is a further increase in internal energy that further weakens the inter-particle forces sufficiently to cause boiling.

• Q7 A block of ice at -10oC was melted and further heated up to 20oC.

• The SHC of ice is 2100 J/kgoC, the SHC of water is 4180 J/kgoC and the latent heat of fusion of water (ice) is 334 000 J/kg.

• (a) If 200 000 J of thermal energy were supplied to the ice, what was the original mass of ice?

• This involves multiple stages of calculation and some clear logical thought!

• Let m be the mass of ice.

• (i) Energy needed to warm the ice from -10oC to 0oC.

• E = mc∆T

• E = m x 2100 x 10 = 21 000 m J

• (ii) Energy needed to melt the ice

• E = mL

• E = m x 334 000 = 334 000 m J

• (iii) Energy needed to warm the water from 0oC to 20oC.

• E = mc∆T

• E = m 4180 x 20 = 83 600 m J

• (iv) The total energy needed

• Etot = 21 000  +  334 000  +  83 600 = 438 600 m J

• (v) Now the total energy needed = total energy supplied

• So: 438 600 m = 200 000

• m = 200 000 / 438 600 = 0.456 kg of ice

• -

• (b) If you monitored the temperature rise with time, sketch the temperature - time graph you might expect and explain its features.

• Initially the solid ice steadily heats up from -10oC until its melting point is reached at 0oC.

• The kinetic energy of vibration of the solid particles is increasing.

• The temperature stays constant as the ice melts in absorbing the latent heat of fusion.

• This is an increase in internal energy that weakens the inter-particle forces sufficiently to cause melting.

• When all the ice is melted the water steadily increases in temperature to 20oC as the molecules gain kinetic energy of movement from place to place.

• Q8 A 500 Watt heating element is used to heat up 1.50 kg of a solid until it just reaches its melting point, but before it actually starts melting.

• If it takes 10.0 minutes more of heating to melt all the solid, what is the latent heat of fusion of the solid?

• P (W) = E / t (J/s)

• energy supplied = E = P x t = 500 x 10 x 60 = 300 000 J

• latent heat fusion = 300 000 / 1.5 = 200 000 J/kg

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