Please work out parts a-d in clear handwriting - A 1-80 kg sample of i.docx
Please work out parts a-d in clear handwriting - A 1-80 kg sample of i.docx
Please work out parts a-d in clear handwriting . A 1.80 kg sample of iron initially at 94.0°C is dropped into a 405 mL beaker of water initially at 14.5°C. If the iron and water come to thermal equilibrium, assuming no loss of heat, what is the final temperature? The specific heat capacities of iron and water are 0.450 and 4.18 J/g3Ç, respectively. The density of water is 1.00 g/m a) 28.9-Q c) 14.63 d) 40236 e) 84.23S A 25.5-g aluminum block is warmed to 95.4°C and plunged into an insulated beaker containing 45.2 g water initially at 22.0°C. The aluminum and water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum? The specific heat capacities of aluminum and water are 0.903 and 4.18 J/g%, respectively a) 26.5°C bl,30.0°C c) 28.7\"C d) 24.1°C e) 22.1°C . The air within a cylinder equipped with a piston absorbs 565 J of heat and expands from an initial volume of 0.10 L to a final volume of 1.90 L against an external pressure of 0.904 atm. What is the change in internal energy of, the air within the cylinder? HINT: 1L atm-101.3 J a) 383 J d) 400. J e) 730. J . The combustion of benzoic acid AEcARASi3228 kJ/mol C,H&02) can be used to determine the heat capacity of a bomb calorimeter. A 3.05 g tablet of C,HsO2 undergoes combustion in a bomb calorimeter; the temperature of the bomb calorimeter increases from 22.1°C to 36.6°C. Calculate the heat capacity of the bomb calorimeter. a) 5.56 kJ/3c c) 73.1 kJ/3c d) 4.98 kJ/3c e) 5.72 kl/°c. Solution 1. Heat change = mass * specific heat * change in temperature Heat lost by metal + heat gained by water = 0 1.80 * 1000 * 0.450 * ( t2 - 94.0 ) + 405 * 4.18 * ( t2 - 14.5 ) = 0 810 t2 - 76140 + 1692.9 t2 - 24547 = 0 t2 = 100687 / 2502.9 t2 = final temperature =40.2 0 C (d) 2. 25.5 * 0.903 ( t2 - 95.4 ) + 45.2 * 4.18 * ( t2 - 22.0 ) = 0 23.0 t2 - 2196.7 + 188.94 t2 - 4156.6 = 0 t2 = 6353.3 / 211.94 t2 = Final temperature = 30.0 0 C (b) 3. Accroding to first law of thermodynamics, deltaE = q - P * ( V2 - V1) = 565 - [0.904 * ( 1.90 - 0.10 ) * 101.3] = 400. J (d) 4. q = heat capacity of calorimeter * change in temperature 3228 * ( 3.05 / 122.1 ) = Ccal * ( 36.6 - 22.1) Ccal = Heat capacity of calorimeter = 5.56 kJ/ 0 C (a) .
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Please work out parts a-d in clear handwriting - A 1-80 kg sample of i.docx
- 1. Please work out parts a-d in clear handwriting . A 1.80 kg sample of iron initially at 94.0°C is dropped into a 405 mL beaker of water initially at 14.5°C. If the iron and water come to thermal equilibrium, assuming no loss of heat, what is the final temperature? The specific heat capacities of iron and water are 0.450 and 4.18 J/g3Ç, respectively. The density of water is 1.00 g/m a) 28.9-Q c) 14.63 d) 40236 e) 84.23S A 25.5-g aluminum block is warmed to 95.4°C and plunged into an insulated beaker containing 45.2 g water initially at 22.0°C. The aluminum and water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum? The specific heat capacities of aluminum and water are 0.903 and 4.18 J/g%, respectively a) 26.5°C bl,30.0°C c) 28.7"C d) 24.1°C e) 22.1°C . The air within a cylinder equipped with a piston absorbs 565 J of heat and expands from an initial volume of 0.10 L to a final volume of 1.90 L against an external pressure of 0.904 atm. What is the change in internal energy of, the air within the cylinder? HINT: 1L atm-101.3 J a) 383 J d) 400. J e) 730. J . The combustion of benzoic acid AEcARASi3228 kJ/mol C,H&02) can be used to determine the heat capacity of a bomb calorimeter. A 3.05 g tablet of C,HsO2 undergoes combustion in a bomb calorimeter; the temperature of the bomb calorimeter increases from 22.1°C to 36.6°C. Calculate the heat capacity of the bomb calorimeter. a) 5.56 kJ/3c c) 73.1 kJ/3c d) 4.98 kJ/3c e) 5.72 kl/°c. Solution 1. Heat change = mass * specific heat * change in temperature Heat lost by metal + heat gained by water = 0 1.80 * 1000 * 0.450 * ( t2 - 94.0 ) + 405 * 4.18 * ( t2 - 14.5 ) = 0 810 t2 - 76140 + 1692.9 t2 - 24547 = 0 t2 = 100687 / 2502.9 t2 = final temperature =40.2 0 C (d)
- 2. 2. 25.5 * 0.903 ( t2 - 95.4 ) + 45.2 * 4.18 * ( t2 - 22.0 ) = 0 23.0 t2 - 2196.7 + 188.94 t2 - 4156.6 = 0 t2 = 6353.3 / 211.94 t2 = Final temperature = 30.0 0 C (b) 3. Accroding to first law of thermodynamics, deltaE = q - P * ( V2 - V1) = 565 - [0.904 * ( 1.90 - 0.10 ) * 101.3] = 400. J (d) 4. q = heat capacity of calorimeter * change in temperature 3228 * ( 3.05 / 122.1 ) = Ccal * ( 36.6 - 22.1) Ccal = Heat capacity of calorimeter = 5.56 kJ/ 0 C (a)