Copy of MHI POW Unit 3 Q2 KG - Google Docs

# Copy of MHI POW Unit 3 Q2 KG - Google Docs - SCH4U Unit 3...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SCH4U Unit 3 Malaysia’s National Drink Note . Adapted from Teh Tarik, by Maxilla, (2007­2015), Daily Burn, Inc. a. What is the specific heat capacity of teh tarik ? Assume that the cup of teh tarik was heated from room temperature to 80.0°C. Given data: Initial Temperature: 25.0 °C Final Temperature: 80.0 °C 35 × 10 1 KJ/cup ÷ 227 g = ΔT= 55.0 °C Mass= 1 cup = 227.0 grams (4.184 KJ ) 1 cup Change in Enthalpy per gram = 1 Kcal × 83 Kcal 1 cup × 227 grams ΔQ= (4.184 KJ ) 1 Kcal × ΔQ= m ∙ c ∙ ΔT 2 83 kcal = 3.5 × 10 KJ We want c * ΔQ m∙ΔT = c 3.5 × 10 2 KJ × 1000 J = 3.5 × 10 5 J 1 KJ 3.5 × 10 5 J ÷ (227.0 g ∙ 55.0 °C) = c c = 28 J /g ∙ c ∴ T he specific h eat capacity of t eh tarik is 28 J /g ∙ c Created by: Ms. Kimberley Gagnon / k [email protected] = 1.5 KJ/g SCH4U Unit 3 b. A student used 0.23 moles of an unknown fuel to heat the cup of teh tarik (mass of mug is145.0 g). The transfer of heat is only 65.3 % efficient. i. What is the unknown fuel? Assume we have a Glass cup ErrorGlass specific heat capacity = 0.84 J/g ∙ c Mass of glass= 145.0 g Q of system= Q of cup + Q t eh tarik = [(145.0g) ∙ (0.84 J/g ∙ c) ∙ (55.0 °C)] + [ 3.5 × 10 2 KJ] =(6.7 KJ)+(3.5 × 10 2 KJ) =3.6 × 10 2 KJ energy output ef f iciency = energy input× 100% 65.3% × E nergy input = Energy output | 0.655 × E nergy input = Energy output 100% Energy output 2 E nergy input = 0.655 | Energy input = 3.6 × 10 KJ ÷ 0.655 Energy input = 5.5 × 10 2 KJ 2 ΔH =− Q | ΔH = ­5.5 × 10 KJ ΔH =n ΔHcomb | ΔHn = ΔHx ΔHcomb = ­5.5 × 10 2 KJ/0.23mol =­2.4 × 10 3 KJ/mol * round to ­2219.2 KJ/mol = Propane ∴ the unknown fuel is Propone ii. How many grams of the fuel did the student have to use to heat the cup of teh tarik ? M ass of P ropane U sed moles × M olar mass of C3H8 M ass of P ropane 0.23 mol 44.1 g = 1.00 mol Used mass of propane= 1.0 × 10 1 g ∴ the student had to used 1.0 × 10 1 g of propane c. How many kilograms of carbon dioxide emissions will be produced from heating this cup of t eh tarik? 5.5 × 10 2 KJ × 66.61 Kg = 3.7 × 10 4 Kg KJ ∴ 3.7 × 10 4 Kg of carbon dioxide emissions will be produced. Created by: Ms. Kimberley Gagnon / k [email protected] SCH4U Unit 3 Reference A. Clancy, C., Amdemichael, T., Aurora, A., & Anderson, M. (2011). Chemistry 12. Toronto, Ont.: McGraw­Hill Ryerson. page 290­340 B. Cup to Gram Conversions ­ Allrecipes Dish. (2014). Retrieved August 11, 2016, from ­to­gram­conversions/ C. Kcal to kilojoules conversion. (n.d.). Retrieved August 11, 2016, from ­to­kj.htm Hint 1: Refer to page 325 and to Table 5.4; round the Heat of Combustion to the same number of significant figures as your answer so you can determine which fuel was used. Hint 2: Refer to Table 5.6 Created by: Ms. Kimberley Gagnon / k [email protected] ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern