Uploaded on

 

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
No Downloads

Views

Total Views
795
On Slideshare
0
From Embeds
0
Number of Embeds
2

Actions

Shares
Downloads
37
Comments
0
Likes
1

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. 2 Latent heat1
  • 2. Specific Latent Heat• In Thermodynamics ! We saw that energy is neededto break inter-atomic attractions when a substancemelts or boils.• This energy is called latent heat.• The temperature is constant during this change ofstate.• The following equation is used to calculate energyneeded for a change of state.Heat transferred = mass x specific latent heat capacityΔQ (J) = m (kg) x L (J kg-1)Specific latent heat, L is the energy needed to changethe state of 1 kg of the substance (without change intemperature). 2
  • 3. Specific latent heat• Latent heat of fusion refers to the change from asolid to a liquid (melting)• Latent heat of vaporisation refers to the changefrom a liquid to a gas (boiling).• Example 1 – the specific latent heat of fusion of iceis 330 000 J kg-1. How much energy is needed to melt0.65 kg of ice?3ΔQ = mL= 0.65 kg x 330 000 J kg-1= 210 000 J
  • 4. Measuring specific latent heat• Look at the diagram in example 2.• The water is boiling and stays at a constant 100 oC• Any energy delivered from the heater will turn thewater into steam.• Knowing the power of the heater,• the energy delivered (J) = power (W) x time (s)• Example 2: the power of the emersion heater is60 W. In 5 minutes, the top pan balance reading fallsfrom 282 g to 274 g. calculate the specific latentheat of vaporisation of water.4
  • 5. Measuring specific latent heat• Energy, ΔQ = power x time• = 60 W x (5 x 60) s = 18 000 J• Mass of water evaporated = 282 g – 274 g• = 8 g = 8 x 10-3kg• ΔQ = mL hence L = ΔQ / m• = 18 000 J / 8 x 10-3kg = 230000 J kg-1.• A similar method can be used for finding the latentheat of fusion.5Funnel is filled with ice at 0 oC, andthe mass of water produced in 5minutes is noted.In a warm room, some ice will meltwith energy received from theroom.
  • 6. Measuring specific latent heat• To allow for this, experiment is repeated twice forsame length of time.• Once with heater off and once with it on.• First experiment is the control – indicates how muchice melts by absorbing energy from the room.• Ice melted by heater is the difference between themasses in the two experiments.Specific heat capacity• For a substance being heated, the rise in temperaturedepends on:1. Mass of substance being heated2. How much energy has been put in3. What the substance is6
  • 7. Specific heat capacity• More energy is needed to the same temperature riseto 1 kg of water than 1 kg of copper.7ΔU = m x c x ΔTΔU is the change in internal energy (J)m is the mass of substance (kg)c is the specific heat capacity (J kg-1K-1)ΔT is the temperature change (K)If the energy is supplied as heat, equation can be writtenas: ΔQ = m x c x ΔTSpecific heat capacity depends on the substanceWater has a value of 4200 J kg-1K-1, while copper hasA value of 380 J kg-1K-1.
  • 8. Specific heat capacity• Defined as the energy needed to raise thetemperature of 1 kg of substance by 1 K.• Example 3: 0.5 kg of water is heated from 10 oC to100 oC. How much does its internal energy rise?• Rise in temperature = 100 oC – 10 oC = 90 oC• This is also a rise of 90 K (oC and K are the same size)• ΔU = m x c x ΔT• = 0.5 kg x 4200 J kg-1K-1x 90 K• = 190 000 J.Measuring specific heat capacity• An experiment can be done to find the specific heatcapacity using an emersion heater.8
  • 9. Worked example 4In an experiment shown in the diagram below, a 60 Wimmersion heater was used. The beaker contained 1kg of water at 21 oC. After 5 minutes, the heater wasswitched off. The temperature of the water went upto 25 oC. What is the specific heat capacity of thewater?9to power supplythermometer1 kg of waterimmersion heater
  • 10. Solution10Change in internal energy, ΔU = power x time= 60 W x (5 x 60) s= 18 000 JRise in temperature = 25 oC – 21 oC = 4 oC = 4 KΔU = m.c.ΔT 18 000 J = 1 kg x c x 4 Khence c = 18 000 J4 K x 1 kg= 4500 JKg-1K-1The result is too large. Can you see why?The 18 000 J from the electrical heater does not all goto increase the internal energy of the water.Some heat will escape to increase internal energy of theroom, there is an increase in the rest of the apparatus.
  • 11. Energy losses in measuring specific heat capacitiesThe increase in internal energy of water is less than18 000 J, and can be calculated from:ΔU = m.c.ΔT = 1.0 kg x 4200 J kg-1x 4 K= 17 000 JThe extra 1000 J supplied by the heater heats upthe apparatus and the surroundings.Comparing specific heat capacitiesSpecific heat capacities can also be determined formetal blocks of different materials.To reduce heat losses, metal blocks are lagged withinsulating materialTable on slide 11 gives the specific heat capacities of2 metals, measured this way11
  • 12. 12Substance Aluminium Ironc / J kg-1K-1880 4701 kg of Al needs almost twice as much energy as 1 kg ofFe for each 1 K rise – why so?If number of atoms of Al and Fe are the same, energyneeded will be similarAl atom is about half the mass of Fe atom, so contains 2 xas many atoms per kg as FeFor comparison, the number of atoms used is theAvogadro constant (6 x 1023atoms), which makes 1 moleThe molar heat capacity is the energy needed to raisethe temperature of one mole of a metal by 1 K.Molar heat capacities are similar – same number of atoms
  • 13. Questions1. A horseshoe of mass 0.8 kg is heated from 20 oC to 900 oC.How much does its internal energy rise? (The specific heatcapacity of steel is 470 J kg-1K-1).2. A kilogram of water falls 807 m (the height of the Angel Fallsin Venezuela) and gains 7900 J of kinetic energy. Assumingthat this is all converted to internal energy in the water,calculate the rise in temperature of the water. The specificheat capacity of the water is 4200 J kg-1K-1.3. A kettle contains 1.5 kg of water at 18 oC.(a) How much heat energy is needed to raise the temperature ofthe water to 100 oC?(b) Assuming no energy is lost to the room, calculate how long thiswill take if the power rating of the kettle is 2000 W.(c) What else must you assume in part (b)?The specific heat capacity of water is 4200 J kg-1K-1.13
  • 14. 144. Once the kettle in question 3 has reached 100 oC, the water willboil. Assume the lid has been left off, so the kettle does notswitch itself off.(a) How much energy is needed to boil away 0.5 kg of water?(the specific latent heat of vaporisation of water, L = 2.3 x 106J kg-1).(b) Assuming no energy is lost, calculate how this will take if thepower rating of the kettle is 2000 W).5. The specific latent heat of fusion of ice is 330 000 J kg-1.(a) Calculate the energy needed to melt 50 g of ice at its meltingpoint.(b) A galss contains 0.4 kg of lemonade at 20 oC, and 50 g of ice at0 oC are put in to cool it. Show that the temperature of thelemonade drops to about 10 oC.(You assume that the specific heat capacity of lemonade is4200 J kg-1K-1and that the energy needed to melt the ice allcomes from the lemonade).