Unit 3 Energy Changes and Rates of Reaction

Transcription

1 Unit 3 Energy Changes and Rates of Reaction Solutions to Practice Problems in Chapter 5 Energy Changes Calculating the Absorption of Heat (Student textbook page 281) 1. How much heat must be added to a 36.2 g sample of iron to increase its temperature by 250.0ºC? You have to calculate the amount of heat that must be added to a sample of iron to increase its temperature by 250.0ºC. You know the mass of iron: m = 36.2 g You know the change in the temperature: T = C You have the specific heat capacity of iron from Table 5.1 on page 280 of the student text: c Fe = J/g C Substitute the given information into the formula Q = mc T. Solve for Q. Q mct J (36.2 g ) (250.0C g C ) J 4.06 kj The iron must gain 4060 J or 4.06 kj of heat energy. Check that the information has been substituted correctly. The answer has the correct number of significant digits. Unit 3 Part B MHR 1

2 2. How much heat must be added to g of steam at 126.0ºC to increase its temperature to 189.5ºC? You must determine the amount of heat that must be added to a sample of steam to increase its temperature from 126.0ºC to 189.5ºC. You know the mass of steam: m = g You know the initial temperature: T initial = C You know the final temperature: T final = C You have the specific heat capacity of steam from Table 5.1 on page 280 of the student textbook: c = 2.02 J/g C H2O(g) Determine the temperature change of the steam. Use the formula Q = mc T to calculate the quantity of heat. T Tfinal Tinitial 189.5C126.0C 63.5C Q mct J (128.6 g ) 2.02 (63.5C g C ) J 16.5 kj The steam must gain J or 16.5 kj of heat energy. Check that the information has been substituted correctly. The answer has the correct number of significant digits. 2 MHR Chemistry 12 Solutions Manual

3 3. A g sample of ethanol at 25.0 C is heated until it reaches 50.0 C. How much thermal energy does the ethanol gain? You must calculate the thermal energy gained by a sample of ethanol. You know the mass of ethanol: m = g You know the initial temperature: T initial = 25.0 C You know the final temperature: T final = 50.0 C You have the specific heat capacity of ethanol from Table 5.1 on page 280 of the student textbook: c ethanol = 2.44 J/g C Determine the temperature change of the ethanol. Use the formula Q = mc T to calculate the quantity of heat gained. T Tfinal Tinitial 50.0C 25.0C 25.0C Q mct 2 J (110 g ) 2.44 (25.0C g C ) J 6.10 kj The ethanol gains J or 6.10 kj of heat energy. Check that the information has been substituted correctly. The answer has the correct number of significant digits. Unit 3 Part B MHR 3

4 4. Beaker A contains 50 g of liquid at room temperature. The beaker is heated until the liquid increases in temperature by 10 C. Beaker B contains 100 g of the same liquid at room temperature. The beaker is also heated until the liquid increases in temperature by 10 C. In which beaker does the liquid absorb more heat? Explain your answer. You must compare the heat absorbed by two samples of the same liquid. You know the mass of sample A is 50 g and that the mass of sample B is 100 g. Both liquids increase in temperature by 10 C. Use the formula Q = mc T to compare the quantities of heat gained. The liquid in beaker B absorbs twice as much heat because it has twice the mass. Q A mact (50 g) ct Q B mbct (100 g) ct 2 50 gct 2 Q A Therefore, Q B = 2Q A. The answer is reasonable since Q and m are directly related in the equation Q = mc T. 4 MHR Chemistry 12 Solutions Manual

5 5. How much heat is released when the temperature of 789 g of liquid ammonia decreases from 82.7 C to 25.0 C? You must determine the quantity of heat released when a sample of liquid ammonia is cooled. You know the mass of ammonia: m = 789 g You know the initial temperature: T initial = 82.7 C You know the final temperature: T final = 25.0 C You have the specific heat capacity of ammonia from Table 5.1 on page 280 of the student textbook: c = 4.70 J/g C NH 3 ( ) Determine the change in temperature of the liquid ammonia. Use the formula Q = mc T to calculate the quantity of heat gained. T Tfinal Tinitial 25.0C82.70C 57.7C Q mct J (789 g ) 4.70 ( 57.7 C g C ) J kj The quantity of heat released is J or kj. The answer seems reasonable and has the correct number of significant digits. The negative sign indicates that heat energy was given off. Unit 3 Part B MHR 5

6 6. A solid substance has a mass of g. It is cooled by C and loses kj of heat. What is the specific heat capacity of the substance? Identify the substance using the values in Table 5.1 (student textbook page 280). You must determine the specific heat capacity of a sample of solid. You know the mass of the solid: m = g You know the change in temperature of the solid: T = C You know the quantity of heat lost: Q = kj Substitute the given information into the formula Q = mc T. Then solve for c. Use the specific heat capacity information in Table 5.1 to identify the solid. T = C Q mct kj ( g) c( 25.00C) 4937 J ( g) c( 25.00C) c J/g C 0.79 J/g C is the specific heat capacity of granite. The specific heat capacity is in the range listed for solids in Table 5.1 and seems reasonable. The answer shows close agreement with the value listed in Table MHR Chemistry 12 Solutions Manual

7 7. The specific heat capacity of a compound used in fireworks is J/g C. If it takes J to heat this material from 20.0 C to C, what mass of compound was used? You must determine the mass of a sample that has absorbed a given amount of heat energy. You know the specific heat capacity of the compound: c = J/g C You know the initial temperature: T initial = 20.0 C You know the final temperature: T final = C You know the quantity of heat absorbed: Q = J Determine the temperature change of T Tfinal Tinitial the compound C 20.0C 905C Use the formula Q = mc T to calculate Q mct the mass. 3 J J m0.800 (905.0C g C ) J m J C g CC m11.1 g The mass of the compound is 11.1 g. The mass seems reasonable for a firecracker and the answer has the correct number of significant digits. Unit 3 Part B MHR 7

8 8. One litre of water at 1.00 C is warmed by the addition of 4.00 kj of heat energy. What is the final temperature of the water? (1.00 L of water has a mass of 1.00 kg) You must find the final temperature of a sample of water that has gained heat energy. You know the mass of water: m = 1.00 kg, or g You know the initial temperature: T initial = 1.00 C You know the amount of heat absorbed: Q = 4.00 kj, or 4000 J You have the specific heat capacity of water from Table 5.1 on page 280 of the student textbook: c H2O( ) c 2 H O( ) = 4.19 J/g C Use the formula Q = mc T to determine the change in temperature, T. Q = mc T 3 J 4000 J ( g ) 4.19 T g C T C Calculate the final temperature, T final C Tfinal Tinitial Tfinal 1.00C T 1.95C final The final temperature of the water is 1.95 C. The answer is reasonable for this amount of heat energy added to 1.00 kg of water. The answer has the correct number of significant digits. 8 MHR Chemistry 12 Solutions Manual

9 9. On a warm day, how much solar energy does a kg piece of concrete absorb as heat energy if its temperature increases from C to C? You need to calculate the quantity of heat absorbed by a piece of concrete. You know the mass of concrete: m = kg You know the initial temperature: T initial = C You know the final temperature: T final = C You have the specific heat capacity of concrete from Table 5.1 on page 280 of the student textbook: c concrete = 0.88 J/g C Determine the change of temperature of the concrete. Use the formula Q = mc T to calculate the quantity of heat absorbed. T Tfinal Tinitial 14.50C13.60C 0.90C Q mct J (3982 g ) 0.88 (0.90C g C ) J The concrete absorbed 3154 J or 3.2 kj of energy. The answer is reasonable and shows the correct number of significant digits. Unit 3 Part B MHR 9

10 10. You have samples of the air and hydrogen gas at room temperature, both having a mass of g. a. Compare the change in temperature of these two samples if each gains J of thermal energy. b. Suggest a reason for the difference in the temperature changes. a. You must calculate and compare the temperature change of samples of hydrogen and of air that have the same mass and gain the same amount of thermal energy. b. You must suggest a reason for the difference in temperature. You know the mass of hydrogen: m H 2 = g You know the mass of the air: m air = g You know the amount of heat absorbed by each sample: Q = 500 J You have the specific heat capacities of hydrogen and air from Table 5.1 on page 280 of the student textbook: c = J/g C and c air = 1.01 J/g C H 2 a. Use the formula Q = mc T to calculate the change in temperature, T, for each sample. Q maircairt air J 500 J (10.00 g ) 1.01 T g C T 49.5C air air Q mhchth 500 J (10.00 g ) T 3.496C H J T g C H ÄT for air is 49.5ºC and ÄT for hydrogen is 3.49ºC; the change in temperature of the hydrogen gas is about 14 times less than the change in temperature of the air. 10 MHR Chemistry 12 Solutions Manual

11 b. The molecules that make up air, primarily oxygen, O 2 (g), and nitrogen, N 2 (g), have a greater mass than hydrogen molecules, H 2 (g). There will be more molecules in 10.0 g of H 2 (g) than in 10.0 g of air. The absorbed energy will be distributed among a larger number of H 2 (g) molecules and each will gain less energy than in the air sample. Since temperature is a measure of the average kinetic energy of the particles in a sample, the H 2 (g) sample will experience a smaller temperature change as calculated. (Going further: the average molar mass of molecules in air is about 29 g/mol. The molar mass of hydrogen is 2 g/mol, about 14 times less.) The answer is reasonable since the average molar mass of air is about 14 times the molar mass of hydrogen gas. Unit 3 Part B MHR 11