Book 1 Section 3.1 Latent heat Ice storage cooling system Cooling curve Latent heat Latent heat and particle motion Check-point 1 Specific latent heat Check-point 2 Check-point 3 1 2 3 4 Book 1 Section 3.1 Latent heat
Ice storage cooling system The London underground train is very hot in summer! Melting ice is used to cool water Water cools the air that blows out under the passenger seats. Why melting ice is used but not cold water? Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Change of state Matter exists in 3 states: solid, liquid, gas E.g. water fusion vaporization at melting point at boiling point ice water steam solidification condensation at freezing point at boiling point Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Expt 3a Cooling curve of octadecan-1-ol Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Experiment 3a Cooling curve of octadecan-1-ol Melt some octadecan-1-ol using a water bath. Remove the water bath and record the temp T of octadecan-1-ol every minute until it falls to 45 C. Plot the cooling curve of temperature against time. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Experiment 3a Cooling curve of octadecan-1-ol The expt can also be carried out using a data-logger. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Experiment 3a Cooling curve of octadecan-1-ol Video Video 3.1 Expt 3a - Cooling curve of octadecan-1-ol Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat 1 Cooling curve Cooling curve of octadecan-1-ol: AB steadily — liquid cooling (temperature falling ) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat 1 Cooling curve Cooling curve of octadecan-1-ol: BC is flat — liquid solidifying (temperature unchanged) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat 1 Cooling curve Cooling curve of octadecan-1-ol: CD steadily — solid cooling to room temperature (temperature falling) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat 1 Cooling curve freezing point freezing point: read from the flat part BC Simulation 3.1 Heating and cooling curve of water Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat According to the cooling curve, when a substance is solidifying… …its temp remains unchanged even though it is losing energy to the surroundings. During change of state: Energy given out/absorbed is called latent heat. means ‘hidden’ Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Ice-water mixture stays at 0 C until all the ice is melted. energy is absorbed from air to change the ice to water temperature unchanged This energy is called latent heat of fusion of ice. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Energy supplied continuously to keep water boiling… energy is absorbed to change the water to steam temperature unchanged This energy is called latent heat of vaporization of water. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat steam condensation vaporization releases latent heat of vaporization absorbs latent heat of vaporization water solidification fusion absorbs latent heat of fusion releases latent heat of fusion ice Heating curve Example 1 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 1 Heating curve A liquid substance is heated. Its T-t graph is shown on the right. (a) Why is the curve flat at stage BC ? The substance is boiling at stage BC. Therefore, its temp remains constant. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 1 Heating curve (b) Comment on the following statements: (i) ‘The substance does not absorb energy at stage BC.’ Incorrect. It absorbs latent heat of vaporization. (ii) ‘The melting point of the substance is 82 C.’ Incorrect. That is the boiling point of the substance. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat 3 Latent heat and particle motion liquid solid Regular arrangement of particles breaks up particles can only vibrate particles can move around Energy has to be supplied to oppose the attractive forces among the particles. PE Book 1 Section 3.1 Latent heat
3 Latent heat and particle motion Transfer of energy does not change KE. Temperature remains unchanged Latent heat = change in PE during change of state Simulation 3.2 Particle motion and change of state Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 1 – Q1 1 kg of ice at 0 C is converted into steam at 100 C. Describe the energy change of the water molecules and the temperature change of the body at each step below. (use the words increase, decrease, and unchanged) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 1 – Q1 1 kg ice 1 kg water melting unchanged Kinetic energy ____________ Potential energy ____________ Internal energy ____________ Temperature ____________ increase increase unchanged Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 1 – Q1 1 kg water 1 kg water heating up increase Kinetic energy ____________ Potential energy ____________ Internal energy ____________ Temperature ____________ unchanged increase increase Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 1 – Q1 steam 1 kg water boiling unchanged Kinetic energy ____________ Potential energy ____________ Internal energy ____________ Temperature ____________ increase increase unchanged Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 1 – Q2 The figure shows the cooling curve of a substance. Which of the following must be correct? A The melting point of the substance is T. B The boiling point of the substance is T. C The substance absorbs latent heat at stage PQ. D The substance releases latent heat at stage QR. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat 4 Specific latent heat The specific latent heat of a substance is the energy transferred by heating to change the state of 1 kg of the substance without a change in temperature. Latent heat for 1 kg of a substance Symbol: l (Unit: J kg-1) Q m l = Q = ml Book 1 Section 3.1 Latent heat
a Specific latent heat of fusion of ice Specific latent heat of fusion of ice (lf ) = energy needed to change 1 kg of ice to water (without temperature change) = 3.34 105 J kg-1 3.3 Measuring specific latent heat of fusion of ice Simulation Video 3.2 It’s freezing! Measuring the specific latent heat of fusion of ice Expt 3b Book 1 Section 3.1 Latent heat
experimental apparatus Experiment 3b Measuring the specific latent heat of fusion of ice Set up the apparatus. Switch on the heater for some time. Find the mass of ice melted by the heater. Record the energy supplied to the heater. Find lf . control apparatus experimental apparatus crushed ice (roughly equal amounts) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Experiment 3b Measuring the specific latent heat of fusion of ice Ice also absorbs energy from the surroundings. Control is necessary. 3.3 Expt 3b - Measuring the specific latent heat of fusion of ice Video Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Experiment 3b Measuring the specific latent heat of fusion of ice Precautions: Use crushed ice. Ensure melting ice (0 C) is used. Pack the ice in the two funnels before switching on the heater. After switching off the heater, Place a small piece of wire gauze or steel wool at the neck of the funnels. remove the beaker until the drip rates are steady and about the same. Book 1 Section 3.1 Latent heat
a Specific latent heat of fusion of ice Finding specific latent heat of fusion of ice Example 2 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 2 Finding specific latent heat of fusion of ice Results of expt 3b: Mass of water (kg) Experimental beaker 0.050 Control beaker 0.014 Joulemeter reading (J) Initial 15 000 Final 29 200 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 2 Finding specific latent heat of fusion of ice (a) Specific latent heat of fusion of ice = ? Mass of ice melted by heater = 0.050 – 0.014 = 0.036 kg Energy supplied = 29 200 – 15 000 = 14 200 J Q m = 14 200 0.036 lf = = 3.94 105 J kg-1 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 2 Finding specific latent heat of fusion of ice (b) Experimental value of lf = 3.94 105 J kg-1 Standard value of lf = 3.34 105 J kg-1 % error = ? Suggest two possible sources of error. Difference between experimental value and standard value = 3.94 105 – 3.34 105 = 0.6 105 J kg-1 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 2 Finding specific latent heat of fusion of ice 0.6 105 3.34 105 % error = 100 % = 18 % Possible sources of error : Difficulty to keep the water dripping down the two funnels at the same rate Energy lost to the surroundings Book 1 Section 3.1 Latent heat
a Specific latent heat of fusion of ice Heating ice Example 3 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 3 Heating ice How much energy is required to change 0.5 kg of ice at 0 C to water at 80 C ? Total energy required = latent heat (ice at 0 C → water at 0 C) + energy to raise temp (water: 0 C → 80 C) = mlf + mcT = 0.5 3.34 105 + 0.5 4200 80 = 3.35 105 J Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 2 – Q1-3 Jimmy melts three materials X, Y and Z of equal mass with identical heaters and under same conditions. The T-t graphs of X, Y and Z are as follows. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 2 – Q1 Assume: energy loss = 0 Which material(s) has/have the highest melting point? ( X / Y / Z ) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 2 – Q2 Assume: energy loss = 0 Which material(s) has/have the largest value of specific latent heat of fusion? ( X / Y / Z ) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 2 – Q3 Assume: energy loss = 0 Which material(s) release(s) the largest amount of energy (per kg) when it/they freeze ? ( X / Y / Z ) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 2 – Q4 To freeze 0.7 kg of water at 10 C to ice, what is the minimum energy to be removed from it? A 29 kJ B 234 kJ C 263 kJ D 2338 kJ Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 2 – Q5 Energy is supplied to melt some ice cubes at 0 C. Mass of ice m = 0.3 kg Energy supplied Q = 107 000 J Specific latent heat of fusion of ice = ? Q m = 107 000 0.3 lf = = 3.57 105 J kg-1 Book 1 Section 3.1 Latent heat
b Specific latent heat of vaporization of water Specific latent heat of vaporization of water (lv ) = energy needed to change 1 kg of water to steam (without temperature change) = 2.26 106 J kg-1 Book 1 Section 3.1 Latent heat
b Specific latent heat of vaporization of water 3.4 Measuring specific latent heat of vaporization of water Simulation Measuring the specific latent heat of vaporization of water Expt 3c Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Experiment 3c Measuring the specific latent heat of vaporization of water Set up the apparatus. Take the reading from the electronic balance after the water boils. Start counting the rotations of the disc on the kW h meter. Stop counting after several rotations. Quickly take the reading of the electronic balance. Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Experiment 3c Measuring the specific latent heat of vaporization of water Calculate the energy supplied to the water and the mass of water boiled away. Find lv . Precaution: Switch on the heater unless the heating part of it is totally immersed in water. 3.4 Expt 3c - Measuring the specific latent heat of vaporization of water Video Book 1 Section 3.1 Latent heat
b Specific latent heat of vaporization of water Finding specific latent heat of vaporization of water Example 4 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 4 Finding specific latent heat of vaporization of water Results of an experiment: Initial mass of water = 0.82 kg Final mass of water = 0.72 kg Kilowatt-hour meter calibration = 600 turns/kW h Number of rotations counted = 41 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 4 Finding specific latent heat of vaporization of water Find lv . (1 kW h = 3.6 × 106 J) Mass of water vaporized = 0.82 – 0.72 = 0.10 kg Kilowatt-hour meter calibration = 600 turns/kW h Energy supplied per revolution of the disc = = 6000 J 3.6 × 106 600 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 4 Finding specific latent heat of vaporization of water Energy supplied to boil the water = 6000 × 41 = 246 000 J Q m lv = = 246 000 0.10 = 2.46 106 J kg-1 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 4 Finding specific latent heat of vaporization of water (b) Experimental value of lv = 2.46 106 J kg-1 Standard value of lv = 2.26 106 J kg-1 % error = ? Suggest two possible sources of error. Difference between experimental value and standard value = 2.46 106 – 2.26 106 = 0.2 106 J kg-1 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 4 Finding specific latent heat of vaporization of water 0.2 106 2.26 106 % error = 100 % = 9 % Possible sources of error : Steam condensing on the heater drips back into the cup. Energy lost to the surroundings Book 1 Section 3.1 Latent heat
b Specific latent heat of vaporization of water Heating water Example 5 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 5 Heating water How much energy is required to change 0.5 kg of water at 0 C to stream at 100 C ? Total energy required = energy to raise temp (water: 0 C → 100 C) + latent heat (water at 100 C → steam at 100 C) = mcT + mlv = 0.5 4200 100 + 0.5 2.26 106 = 1.34 106 J Book 1 Section 3.1 Latent heat
b Specific latent heat of vaporization of water Making coffee with steam Example 6 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 6 Making coffee with steam An expresso coffee machine injects 0.025 kg of steam at 100 C into a cup of cold coffee of mass 0.15 kg at 20 C. Final temp of the expresso coffee = ? (Specific heat capacity of the coffee = 5800 J kg–1 C ) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 6 Making coffee with steam (c of coffee = 5800 J kg–1 C ) Assume: energy lost to surroundings = 0 Let T be the final temperature of the coffee. energy lost by steam energy gained by coffee = 0.025 2.26 106 + 0.025 4200 (100 – T ) = 0.15 5800 (T – 20) T = 86.6 C The temperature T of the coffee is 86.6 C. Book 1 Section 3.1 Latent heat
b Specific latent heat of vaporization of water Simulation 3.5 Ice storage cooling system Steam and hot water Example 7 Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 7 Steam and hot water (a) Find the energy released when water of mass 15 g at 100 C is cooled to 50 C. Energy released = mcT = 0.015 4200 (100 – 50) = 3150 J Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Example 7 Steam and hot water (b) Find the energy released when steam of mass 15 g at 100 C is cooled to water at 50 C. Steam at 100 C first changes to water at the same temp and releases latent heat. Total energy released = mlv + mcT = 0.015 2.26 106 + 3150 = 37050 J Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 3 – Q1 Consider a cup of water (mass m) being heated from 0 C to 100 C. Specific heat capacity of water = 4200 J kg-1 C-1 Since Q = mcT and Q = mlv lv= cT = 4200 × 100 = 420 000 J kg1 Is the student correct? ( Yes / No ) Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 3 – Q2 Find the energy needed for changing 1 kg of ice at 0 C to steam at 100 C. Energy needed = _____________ + ____________ + _______________ = _____________ 3.34 105 J 4200 100 J 2.26 106 J 3.014 106 J Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat Check-point 3 – Q3 A 1.5-kW heater is immersed in 2 kg of water at 100 C. Minimum time it takes to vaporize all the water = ? Q P 2 2.26 106 1500 = Min time = = 3010 s Book 1 Section 3.1 Latent heat
Book 1 Section 3.1 Latent heat The End Book 1 Section 3.1 Latent heat