Question

Research Paper Criterion 1. The first draft paper should contain the following a DO NOT include a title page with the first draf (minimum 800 words), Insert name, class, ar right corner. The header will be removed to b. in text footnotes (citations) corresponding to With numerical citations, references should b repeated, do not give it number; rathe ams. In-text Chem 111 Lab Start to Name: Sufian Allah od Calelations Due Date: 11/22 Station Reaction tenergy. heat of wald ange, SL ad Hessly -5 ue Bibliography CHM 111 PS 2019 Name Sultan Allah correspond Due date 11/20 Unit Thermodynamics Problem Set Heat and Physical Changes Use the specific heat table to answer questions 1-7. Show work on all problems on separate paper. ards the po Co (Jg "C) ally arrang Substance Co (lig "C) Substance Co (JigoC) Substance 0448 carbon 0.716 ethanol 2.438 iron 1.86 aluminum odicals ar 0.900 2 200 methane steam appar han 2014 water 0.388 4.184 web sit 1. Answer hy vọED : 1. Find the specific heat of a substance having a mass of 560.0 g and is placed in water to cool from 84.6 °C to 34 5 "C? The heat released e Quant is 1.20 X 10'J. 2. While taking a bath your body temperature rose from 37.0°C to 2. Answer 39.8 °C. The Cap is of human flesh is 3 941 Jig. "C and the energy text an absorbed was 1.25 X 10°. What is the mass of your water shriveled body? n your phy 3. Find the specific heat of a metal that's heated from 50.0 C to 3. Answer 3 will 95.0 °C. The heat required was 1.10 X 10'J of energy. The mass of 94 the metal sample is 170.09. alues 4. The carbon cooling rods in Chernobyl's nuclear reactors had a mass 4. Answer of 1,050.0 g how much heat could each rod absorb if the temperature pe changed from 1000.0 C to 1500.0 "C? st 5. What is the final temperature of 250.0 g of ethanol ingested into a 5. Answer ind human if it has an initial temperature of 25.0 "C? It absorbed 4500.0 J of heat. Enthalpy of Closed Systems 6. How much carbon, if heated to 180.0 °C, then cooled to a final temperature of 40.5 "C. The mass of the water in the calorimeter 6. Answer is 210.0 g and has an initial temperature of 24.0 °C? 7. 94.4 g of hot iron with an initial temperature of 152.6 °C is placed into 7. Answer 98.1 g of a flammable liquid at 24.6 °C. The final temperature of both is 45.2 "C. What is the liquid's specific heat? Enthalpy of Chemical Processes Reactions producing heat are exothermic (heat is leaving). The information needed to solve enthalpy problems is a balanced equation, enthalpy change, and/or heat of formation. Heat can be treated like a reactant or product in reactions if the conversion factor of kJ/mole is present. 8. Butane has a heat of combustion of 2880 kJ/mole, how many kJ are 8. Answer produced by 2.75 moles of butane? Plan: mols → kJ 9. How much heat is produced from 3.50 moles of O2 in the burning of 9. Answer acetylene: 2C2H2 + 502 > 4CO2 + 2H2O + 1299 kJ Plan: mols → mol/mol → kJ 10. How many grams of butane are required to produce 6550 kJ of heat? 10. Answer The heat of combustion for butane is 2880 kJ/mole. Plan: kJ → mols → grams 11. Propane produces 2220 kJ/mole of heat in the following reaction: 11. Answer How many grams of oxygen is needed to produce 1.00 X 10 kJ of heat? CsHs + 502 → 4H2O + 3CO2 Plan: kJ → mols → mol/mol → mols grams.

Answer 1

where, Answer :- ol formula used. - q = MCAT. 9- The amount of heat m. mass of sample c. Specific heat. AT: change in temperabere Data given. q? 1:20 x 10 y 1200 J. m = 560.0 g. o can be calculated a AT= 84.6°C - 34.5°C - 50.10c Put these values and we have c = eq/mot. - 120J - 56000g x 50.1°C 0:04277 c. 4.277x10 2 5/9 9. MCAT_ where, q1:10 X 10'Je llwJ. me 170.09 AT 95.0°c - 50.0°C = 45.00c Ica Scanned with CamScanner

MAT. 170.0g x 45.0°c 0.1438 J/goc c= where meat m . 1050.0 g. for carbon rods = 0.716 9°c AT = 15eo.ooc-loo.ooc 5w0.0°C. q = 1050. og x 0.716 I x 500.0°C q 37596 s. q 3:759 x 105 05 qmcal_ qe 4500.0 J. m = 250.0 g. for ethanol = 2.438 J/g c mc 450.03 250.0 g x 2.438 8 / g c. - cs Scanned with CamScanner 7.38 °C

AT = final temp. - Initial temp. where, AT = 7:38°C - Initial temp. = 25.0°C so, final temp. = AT + Initial temp. 7:38°C + 25-0°c - 32:38 °C Thus, final temp. is 32.38°C - q MCAT ma q САТ, where, - 3.941 J/goc a 1:25 x 103 J.: 1257 ) ATV 39.8°C - 37.0°c = 2.8°C so, m= 12505 3.941 3 x 2.8 . m..2 113.278 g convert g to kg 1 kg loro g. - 113.978 0i113 Scanned with lovo CS CamScanner kg.