CHEM J-8 June 2014

Transcription

1 CHEM J-8 June 2014 A 1.0 kg sample of copper metal is heated to C. The copper sample is immersed in a volume of water initially at 25.0 C. What volume of water is required so that the final temperature of the copper is 40.0 C? Show all working. Data: Specific heat capacity of Cu(s) is 0.39 J K 1 g 1. Specific heat capacity of H 2 O(l) is J K 1 g 1. The density of water is 1.0 g ml 1. 3 The copper cools from o C to 40.0 o C so ΔT = K. As the specific heat capacity of Cu(s) is 0.39 J K -1 g -1, the heat lost by 1.0 kg during this is given by: q = mc ΔT = ( g) (0.39 J K -1 g -1 ) (-60.0 K) = J This heat is gained by the water which is warmed from 25.0 o C to 40.0 o C a change of 15.0 K. As the specific heat capacity of water is J K -1 g -1 and a temperature change of 15.0 K is required: q = mc ΔT = m (4.184 J K -1 g -1 ) (15.0 K) = J m = 370 g The density of water is 1.0 g ml -1 so the volume required is: volume = mass / density = 370 g / 1.0 g ml -1 = 370 ml Answer: 370 ml THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

2 CHEM N-9 November 2014 Lead shot was traditionally made by dropping molten lead into a tank of water. A piece of lead, initially at 327 C is dropped into ml of water raising its temperature from 25 to 35 C. What was the weight of the lead? Data: Specific heat capacity of Pb is J K 1 g 1 Specific heat capacity of H 2 O(l) is J K 1 g 1 The density of water is 1.0 g ml ml of water corresponds to g. When this quantity warms from 25 o C to 35 o C, the heat change is: q water = mcδt = (200.0 g) (4.184 J g -1 K -1 ) ((35-25) K) = 8400 J The mass of lead is x g, the heat change when it cools from 35 o C to 327 o C is: q lead = mcδt = (x) (0.126 J g -1 K -1 ) ((35-327) K) = -36.7x J The heat gained by the water is lost by the lead so these two quantities must be equal in magnitude: q water = - q lead 8400 J = 36.7x J x = 230 g Answer: 230 g THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

3 CHEM J-8 June L of water is heated to 95 o C and then solid copper, initially at 25 o C, is immersed in it. What mass of copper was added if the final temperature of the water was 84 o C? Show all working. Data: Specific heat capacity of Cu(s) is 0.39 J g 1 K 1. Specific heat capacity of H 2 O(l) is J g 1 K 1. The density of water is 1.0 g ml L of water corresponds to g. When this quantity cools from 95 o C to 84 o C, the heat change is: q water = mcδt = ( g) (4.184 J g -1 K -1 ) ((84-95) K) = J If the mass of copper used is x g, the heat change when it warms from 25 o C to 84 o C is: q copper = mcδt = (x) (0.39 J g -1 K -1 ) ((84-25) K) = +23x J The heat lost by the water is gained by the copper so these two quantities must be equal in magnitude: 23x = so x = 2.0 kg Answer: 2.0 kg

4 CHEM J-9 June 2012 A 2.5 kg block of aluminium is heated to 80.0 C and then placed into a thermally insulated water bath consisting of 10.0 L of water at 25.0 C. Calculate the final temperature of the water once equilibrium has been reached. Show all working. Data: Specific heat capacity of Al(s) is J g 1 K 1. Specific heat capacity of H 2 O(l) is J g 1 K 1. The density of water is 1 g ml 1. 3 The aluminium will cool down and the water will heat up when the two are mixed. The final temperature, T f, will be the same for both. For the aluminium, q aluminium = m c ΔT = (2500 g) (0.900 J g -1 K -1 ) ΔT aluminium For the water, q water = m c ΔT = ( g) (4.18 J g -1 K -1 ) ΔT water As the heat lost by the aluminium is gained by the water, q water = -q aluminium : or ( g) (4.18 J g -1 K -1 ) ΔT water = - (2500 g) (0.900 J g -1 K -1 ) ΔT aluminium ΔT water = ΔT aluminium Using ΔT water = (T f 25.0) o C and ΔT aluminium = (T f 80.0) o C gives: T f = 27.8 o C Answer: 27.8 o C

5 CHEM N-10 November 2012 How many 2.0 L casks of wine and/or juices can be cooled on a hot Sydney day from 30 o C to a drinkable 10 o C with one 10.0 kg bag of ice taken from a freezer at 30 o C? Assume the specific heat of the wine and/or juice is the same as that of water, that the cardboard or plastic containers have negligible heat capacity, and that the density of the wine and juices is 1.0 g ml 1. Δ fus H(H 2 O) = 6.0 kj mol 1 ; C p (H 2 O(s)) = 2.2 J K 1 g 1 ; C p (H 2 O(l)) = 4.2 J K 1 g 1 6 The bag of ice will undergo three process in being raised from -30 o C to +10 o C: (a) The ice will be heated from -30 o C to 0 o C for which ΔT = +30 K: q (a) = mc p ΔT = ( g) (2.2 J K -1 g -1 ) (30 K) = kj (b) The ice will melt. As the molar mass of H 2 O is ( ) g mol -1 or g mol -1. The number of moles of ice that will be melted is: number of moles = mass / molar mass = ( g) / ( g mol -1 ) = 555 mol As Δ fus H(H 2 O) = 6.0 kj mol 1, the amount of energy required to melt this amount is: q (b) = (555 mol) (6.0 kj mol -1 ) = kj (c) The melted ice will be heated from 0 o C to +10 o C for which ΔT = +10 K: q (c) = mc p ΔT = ( g) (4.2 J K -1 g -1 ) (10 K) = kj Overall, the heat absorbed by the ice is therefore: q ice = q (a) + q (b) + q (c) = ( kj) + ( kj) + ( kj) = kj This heat is taken from the wine casks so: q wine = kj The mass of each cask is 2.0 kg and the required temperature change is from +30 o C to +10 o C for which ΔT = -20 K. If x casks are cooled: Hence: q wine = mc p ΔT = ( g) x (4.2 J K -1 g -1 ) (-20 K) = kj x = 26.3 i.e. 26 casks can be cooled. Answer: 26 casks ANSWER CONTINUES ON THE NEXT PAGE

6 CHEM N-10 November 2012 What other assumption have you made in your calculation? It has been assumed that all the heat that melts the ice and then warms the water comes solely from the casks being cooled. This clearly will not be the case - the system is exposed to the atmosphere so that people can access the drinks.

7 CHEM J-15 June 2010 Consider two blocks of steel: block A is 1.00 kg and block B is 600. g. Both blocks start from the same temperature and are heated so that 600. J flows into each of the blocks in the form of heat. What is the final difference in temperature, T A -T B, between block A and block B. The specific heat of steel is J g 1 K 1. Show all working. 2 The specific heat capacity C can be used to calculate the increase in temperature, T from a heat change, q as: q = m C T where m is the mass of the substance. For block A, m = g so that the temperature increase is: T = q / mc = (600. J) / ( g) (0.460 J g -1 K -1 ) = 1.30 K For block A, m = g so that the temperature increase is: T = q / mc = (600. J) / (600. g) (0.460 J g -1 K -1 ) = 2.17 K The final temperature difference, T A -T B, is therefore ( ) K = K Note that block B is lighter so ends up as the hotter of the two. Answer: 0.87 K

8 CHEM N-7 November 2010 Water solutions of NaOH (100.0 ml, 2.0 M) and HCl (100.0 ml, 2.0 M), both at 24.6 C, were mixed together in a coffee cup calorimeter. The temperature of the solution rose to 38.0 C during the reaction process. Write a balanced ionic equation to describe the reaction in the calorimeter. 5 H + (aq) + OH (aq) H 2 O(l) Is the process an endothermic or exothermic reaction? exothermic Assuming a perfect calorimeter, determine the standard enthalpy change for the neutralisation reaction. Assume the density of water is 1.00 g ml 1. The heat capacity of water is 4.18 J g 1 K 1. The temperature rise is ( ) C = 13.4 C. The temperature difference corresponds to 13.4 K. After mixing, the total volume of solution is ( ) ml = ml. With a density of 1.00 g ml -1, this corresponds to 200. g. The heat change is given by q = m C ΔT where C is the specific heat capacity. Hence: q = (200. g) (4.18 J g -1 K -1 ) (13.4 K) = J ml of 2.0 M NaOH (or HCl) contains: number of moles = concentration volume = (2.0 mol L -1 ) ( L) = 0.20 mol As J are released by 0.20 mol, the is the molar enthalpy change is therefore: ΔH = -(11200 J) / (0.20 mol) = J mol -1 = 56 kj mol -1 Answer: -56 kj mol -1

9 CHEM J-12 June 2009 A radiator generates 150 J to heat up air inside a sealed container with volume of 2.00 L and initially at 25 C and atmospheric pressure. What will be the pressure inside the container after heating? Assume that air is composed of 80 % nitrogen and 20 % oxygen by volume. Molar heat capacities: N J K 1 mol 1 and O J K 1 mol 1 5 Using the ideal gas law, the number of moles in 2.00 L of air at 25 C ( = K) at atmospheric pressure is: n = PV RT = 1.00 atm (2.00 L) L atm K 1 mol 1 = mol (298.0 K) The temperature change, ΔT, is related to the heat change, q, through, q q = n C p ΔT or ΔT = nc p The ideal gas law shows that the number of moles is directly proportional to the volume. The molar heat capacity of air is therefore the average of individual molar heat capacities: Hence, C p (air) = 0.80C p (N 2 ) C p (O 2 ) = [( ) + ( )] J K -1 mol -1 = J K -1 mol -1 ΔT = q nc p = 150 J mol (29.19 J K 1 mol 1 ) = 62.8 K As the initial temperature is 298 K and the air is being heated, the final temperature is ( ) K = K. Finally, at this temperature, the ideal gas gives the pressure as: P = nrt V = mol L atm K 1 mol 1 (360.8 K) (2.00 L) = 1.21 atm Pressure: 1.21 atm If this heated air is injected into a balloon, it will rise. Use the ideal gas equation to explain why this happens. From the ideal gas law, n V = P RT. As the density is proportional to n, this shows V that the density is inversely proportional to the temperature. The heated air has a lower density than the air in the atmosphere.

10 CHEM J-6 June 2006 At room temperature and pressure (RTP), 1 mole of an ideal gas occupies L. Calculate the molar volume of the same ideal gas in the stratosphere, where the pressure is atm and the temperature is 200 K. 2 Using the ideal gas equation, PV = nrt, the volume of 1 mole of gas at P = atm and T = 200 K is: V = nrt 1 ( ) (200) P (0.020) 2 = = 820L = L Answer: 820 L= L Two blocks of metal, as shown in the table below, are placed in intimate contact in an insulated environment. 4 Metal Iron Copper Mass (g) Initial T ( C) c (J g 1 K 1 ) In which direction will the heat flow? Write from Fe to Cu or from Cu to Fe. As the copper is at a higher initial temperature, heat will flow from Cu to Fe What is the final temperature of the system? Heat will flow until the temperatures are equal. The heat lost by the copper is equal to the heat gained by the iron. The heat is related to the temperature change by the equation, q = m c T. The heat lost by the copper in going from C to the final temperature T f is: q = m Cu c Cu T = (100.0 T f ). The heat gained by the iron in going from 0.0 C to the final temperature T f is: q = m Fe c Fe T = (T f 0.0). Setting the two heat changes to be equal gives: (100.0 T f ) = (T f 0.0) or T f = 13.5 T f Hence, T f = 36.4 C Answer: 36.4 C

11 CHEM J-7 June 2005 Methane, CH 4, represents an increasingly important fuel. Write the balanced chemical equation for the combustion of methane. 4 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) Calculate the mass of CO 2 that would be produced by the combustion of 1.00 kg of methane. The molar mass of CH 4 is (12.01 (C) (H)) g mol -1 = g mol -1. The number of moles of CH 4 in 1.00 kg is therefore: number of moles = g gmol 1 = 62.3 mol In the reaction, 1 mol of CH 4 leads to 1 mol of CO 2. Hence 62.3 mol of CO 2 will be emitted from this number of moles of CH 4. The molar mass of CO 2 is (12.01 (C) (O)) g mol -1 = g mol -1. The mass of 62.3 mol is therefore: mass of CO 2 = (62.3 mol) (44.01 g mol -1 ) = 2740 g = 2.74 kg Answer: 2.74 kg Calculate the volume of CO 2 produced at 0 C and 1 atm. Assuming that CO 2 behaves like an ideal gas, the volume is given by PV = nrt. Using n = 62.3 mol, T = 0 C = 273 K and P = 1 atm: 1 1 nrt (62.3mol) ( Latm K mol ) (273K) V = P (1atm) = 1400 L = L (As the pressure has been used in atmospheres, R = L atm K -1 mol -1 has been used leading to the volume in litres). Answer: L In an inefficient combustion reaction some of the methane gas may escape into the atmosphere, thereby decreasing the amount of CO 2 produced. Would such a leakage lead to a greater or lesser enhancement of the Greenhouse Effect? Why? Methane is a much more efficient greenhouse gas than carbon dioxide so the Greenhouse Effect will be enhanced by release of CH 4 instead of CO 2.

12 CHEM N-7 November 2004 Two students determined the specific heat capacity of steel as follows. Student 1 poured 200 ml of recently boiled water into a styrofoam cup calorimeter and measured the temperature of the water to be 90.0 C. A steel teaspoon (25 g) at room temperature (20.0 C) was placed into the cup, where it was completely submerged. After equilibrium was presumably reached, the temperature was measured to be 88.7 C. Student 2 took the same cup and filled it with 200 ml of water at room temperature (20.0 C) and placed the same teaspoon from the freezer ( 40.0 C) into the cup. After equilibrium was presumably reached, the temperature recorded was 19.2 C. Determine the specific heat capacity of steel using the experimental data of each student. The specific heat capacity of water is J g 1 K 1. Assume the density of water is 1.00 g ml 1 at all temperatures. 5 The specific heat capacity is related to the heat and temperature change by: q = c m ΔT In student 1 s experiment, the heat lost by the water is gained by the spoon. The temperature of the water changes by ( ) C = 1.3 C. The temperature of the spoon changed by ( ) C = 68.7 C. Equating the heat changes gives: c water m water ΔT water = c steel m steel ΔT steel (4.184 J g -1 K -1 ) (200 g) (1.3 K) = c steel (25.0 g) (68.7 K) c steel = 0.63 J g -1 K -1 In student 2 s experiment, the heat lost by the water is gained by the spoon. The temperature of the water changes by ( ) C = 0.8 C. The temperature of the spoon changed by ( ) C = 59.2 C. Equating the heat changes gives: c water m water ΔT water = c steel m steel ΔT steel (4.184 J g -1 mol -1 ) (200 g) (0.8 K) = c steel (25.0 g) (59.2 C) c steel = 0.45 J g -1 K -1 Student 1: 0.63 J g -1 K -1 Student 2: 0.45 J g -1 K -1 Which student's value is more accurate? Give reasons for your answer. The largest source of error is likely to be lose of heat (student 1) or gain of heat (student 2) from the surroundings. As student 2 s experiment is carried out with water close to room temperature, this effect is likely to be much smaller than for student 1 where the water is at 90 C.