JK-TK/Newdata huggingface dataset:
(i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol. 14 A draught shield tries to minimize the loss by protecting wind from wobbling the flame. (ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc 5.Calculate the heating value of the fuel. Heating value = molar heat of combustion => 96.5517 kJmole-1 = 2.0989 kJg-1 Molar mass of fuel 46 g 6.Explain other factors used to determine the choice of fuel for domestic and industrial use. (i) availability and affordability-some fuels are more available cheaply in rural than in urban areas at a lower cost. (ii)cost of storage and transmission-a fuel should be easy to transport and store safely. e.g LPG is very convenient to store and use. Charcoal and wood are bulky. (iii)environmental effects –Most fuels after burning produce carbon(IV) oxide gas as a byproduct. Carbon(IV) oxide gas is green house gas that causes global warming. Some other fuel produce acidic gases like sulphur(IV) oxide ,and nitrogen(IV) oxide. These gases cause acid rain. Internal combustion engines exhaust produce lead vapour from leaded petrol and diesel. Lead is carcinogenic. (iv)ignition point-The temperature at which a fuel must be heated before it burns in air is the ignition point. Fuels like petrol have very low ignition point, making it highly flammable. Charcoal and wood have very high ignition point. 7.Explain the methods used to reduce pollution from common fuels. (i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0).
(i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,).
Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution.
Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g.
1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.
Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)
16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal.
A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)
16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal. To determine the molar standard enthalpy/heat of displacement(∆Hᶿd) of copper Procedure Place 20cm3 of 0.2M copper(II)sulphate(VI)solution into a 50cm3 plastic beaker/calorimeter. Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 .
Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 . Repeat the experiment to complete table 1 below Table 1 Experiment I II Final temperature of solution(T2) 30.0oC 31.0oC Final temperature of solution(T1) 25.0oC 24.0oC Change in temperature(∆T) 5.0 6.0 Questions 1.(a) Calculate: (i)average ∆T Average∆T = change in temperature in experiment I and II =>5.0 + 6.0 = 5.5oC 2 (ii)the number of moles of solution used Moles used = molarity x volume of solution = 0.2 x 20 = 0.004 moles 1000 1000 (iii)the enthalpy change ∆H for the reaction Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 5.5 = 462 Joules = 0.462 kJ 1000 (iv)State two assumptions made in the above calculations. Density of solution = density of water = 1gcm-3 Specific heat capacity of solution=Specific heat capacity of solution=4.2 kJ-1kg-1K This is because the solution is assumed to be infinite dilute. 2. Calculate the enthalpy change for one mole of displacement of Cu2+ (aq) ions. Molar heat of displacement ∆Hd = Heat produced ∆H
17 Number of moles of fuel => 0.462 kJ = 115.5 kJmole-1 0.004 3.Write an ionic equation for the reaction taking place. Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq) 4.State the observation made during the reaction. Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless. Brown solid deposits are formed at the bottom of reaction vessel/ beaker. 5.Illustrate the above reaction using an energy level diagram.
Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless. Brown solid deposits are formed at the bottom of reaction vessel/ beaker. 5.Illustrate the above reaction using an energy level diagram. Zn(s) + Cu2+(aq) Energy ∆H = -115.5 kJmole-1 (kJ) Cu(s) + Zn2+(aq) Reaction progress/path/coordinates 6. Iron is less reactive than Zinc. Explain the effect of using iron instead of Zinc on the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution. No effect. Cu2+ (aq) are displaced from their solution.The element used to displace it does not matter.The reaction however faster if a more reactive metal is used. 7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes
18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings.
7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes
18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings. Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>2.204 kJ = 0.0206moles 126 moles Molarity of the solution = moles x 1000 Volume of solution used = 0.0206moles x 1000 = 0.5167 M 40 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature.
The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings. Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>2.204 kJ = 0.0206moles 126 moles Molarity of the solution = moles x 1000 Volume of solution used = 0.0206moles x 1000 = 0.5167 M 40 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature. Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>25.08kJmole-1x 0.5 moles = 1.254 kJ x 1000 =1254J Mass of solution (m) = Heat produced ∆H specific heat capacity (c)x ∆T => 1254J = 99.5238 g 4.2 x 3 Volume = mass x density = 99.5238 g x 1 = 99.5238cm3 Note: The solution assumes to be too dilute /infinite dilute such that the density and specific heat capacity is assumed to be that of water. Graphical determination of the molar enthalpy of displacement of copper Procedure:
19 Place 20cm3 of 0.2M copper(II)sulphate (VI) solution into a calorimeter/50cm3 of plastic beaker wrapped in cotton wool/tissue paper. Record its temperature at time T= 0. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds . Place all the (1.5g) Zinc powder provided. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes. Determine the highest temperature change to the nearest 0.5oC.
Place all the (1.5g) Zinc powder provided. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes. Determine the highest temperature change to the nearest 0.5oC. Sample results Time oC 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0 Temperature 25.0 25.0 25.0 25.0 25.0 xxx 36.0 35.5 35.0 34.5 Sketch graph of temperature against time 36.5 Extrapolation Temperature point ∆T oC 130 Time(seconds) Questions 1. Show and determine the change in temperature ∆T From a well constructed graph ∆T= T2 –T1 at 150 second by extrapolation ∆T = 36.5 – 25.0 = 11.5oC 2.Calculate the number of moles of copper(II) sulphate(VI)used given the molar heat of displacement of Cu2+ (aq)ions is 125kJmole-1 Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 11.5 = 966 Joules = 0.966 kJ 1000 Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>.966 kJ = 0.007728moles 125 moles 7.728 x 10-3moles 2. What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 20 4.The actual concentration of copper(II)sulphate(VI) solution was 0.4M.Explain the differences between the two. Practical value is lower than theoretical.
What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 20 4.The actual concentration of copper(II)sulphate(VI) solution was 0.4M.Explain the differences between the two. Practical value is lower than theoretical. Heat/energy loss to the surrounding and that absorbed by the reaction vessel decreases ∆T hence lowering the practical number of moles and molarity against the theoretical value 5.a) In an experiment to determine the molar heat of reaction when magnesium displaces copper ,0.15g of magnesium powder were added to 25.0cm3 of 2.0M copper (II) chloride solution. The temperature of copper (II) chloride solution was 25oC.While that of the mixture was 43oC. i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH- ions to form one mole of H2O/water.
i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH- ions to form one mole of H2O/water. Neutralization is thus a reaction of an acid /H+ (H3O+)ions with a base/alkali/ OH- ions to form salt and water only. 21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH- ions) and exist more as molecules. Neutralization is an exothermic(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH- ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations.
21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH- ions) and exist more as molecules. Neutralization is an exothermic(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH- ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations. (ii) (i)for strong acid/base/alkali neutralization, no energy is used to dissociate /ionize since molecule is wholly/fully dissociated/ionized into free H+ H3O+ and OH- ions.The overall energy evolved is comparatively higher/more than weak acid-base/ alkali neutralizations. For strong acid-base/alkali neutralization, the enthalpy of neutralization is constant at about 57.3kJmole-1 irrespective of the acid-base used. This is because ionically: OH-(aq)+ H+(aq) -> H2O(l) for any wholly dissociated acid/base/alkali Practically ∆Hᶿn can be determined as in the examples below: To determine the molar enthalpy of neutralization ∆Hn of Hydrochloric acid Procedure Place 50cm3 of 2M hydrochloric acid into a calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper. Record its temperature T1.Using a clean measuring cylinder, measure another 50cm3 of 2M sodium hydroxide. Rinse the bulb of the thermometer in distilled water. Determine the temperature of the sodium hydroxide T2.Average T2 andT1 to get the initial temperature of the mixture T3. Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture.