I am having trouble answering part 6. Some of the questions might require soluti

Question

I am having trouble answering part 6. Some of the questions might require solutions from part 5, which I have already done. Please answer the part 6 sub-questions. Thank you! 5. Assume that you are heating water in a calorimeter with perfect insulation, from the combustion of ethanol is transferred to the water. Assume that the temperatui of water rose by 10.0°C. Answer the following questions, and show all work elearly a. Calculate the heat required to raise the water temperature. The specific heat cap insulation, that is, all the heat the temperature of 200 mL specific heat capacity of water an Il work clearly esc heat capacity of water Q: mx CAT100 4.184 (10)8363 8.3683 b. Calculate the moles of ethanol required to release that amount of heat. The heat of combustion of ethanol is-1366 kJ/mol. 6.13x163,hel 366 c. Calculate the mass of ethanol required to release that amount of heat. The molar mass of ethanol is 46.07 g/mol. d. Calculate the volume of ethanol required. The density of ethanol is 0.78 g/mL nass 0.182 0.98 6. Next, assume your calorimeter is not perfect, and that burning 0.51g of ethanol resulted in a 10.0°C rise in the temperature of the 200 mL of water in your calorimeter. Answer the following questions, and show all work clearly Calculate the energy, (kJ), released by buning 0.51g of ethanol. a. of water to raise the water temperature by b. Calculate the energy, (kJ), captured by the 200 mL of water to raise the water t 10,0°C of the heat released was captured by the water? c. Calorimeter Efficiency: What percentage

Explanation / Answer

(a) Moles of ethanol present = 0.51gm/46 gmmol^-1 = 0.011 moles

enthalpy of combustion of ethanol = -1366kJ/mol

So, heat released when 0.51 gm ethanol is burned = 1366kJ/mol * 0.011 moles = 15.026 kJ = 15026 J

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heat absorbed by water = m *Cp * dT

= 200gm * 4.18 J/gm/oC * 10^oC = 8360 J

density of water = 1gm/mL . So, 200mL water = 200gm water

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% of heat released = 8360 J*100/ 15026 J = 55.64 %

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energy lost = 15026 J-8360 = 6666J

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rest amount of heat was taken up by the surrounding air to increase the air temperature.

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