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4 years ago

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Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

a renewable resource and ethanol-blended fuels, like gasohol, appear to burn more efficiently in combustion engines. The heat of combustion of ethanol is 326.7 kcal/mol. The heat of combustion of octane, C8H18, is 1.308×103 kcal/mol. How much energy is released during the complete combustion of 463 grams of octane ? kcal Assuming the same efficiency, would 463 grams of ethanol provide more, less, or the same amount of energy as 463 grams of octane? Submit Answer

1 answer:

weeeeeb [17]4 years ago

3 0

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = <u>5.30 * 10³ kcal</u>

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = <u>3.28 * 10³ kcal</u>

<u />

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

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Answer:

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Explanation:

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$T_{f_{a}}$ : is the final temperature of the water = 25.5 °C

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$C_{b}$ : is the specific heat of metal =?

$T_{i_{b}}$ : is the initial temperature of the metal = 89.0 °C

$T_{f_{b}}$ : is the final temperature of the water = 25.5 °C

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

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$C_{b} = -\frac{72.0 g*4.184 J/(g^{\circ} C)(25.5 ^{\circ} C - 19.2 ^{\circ} C)}{141 g(25.5 ^{\circ} C - 89.0 ^{\circ} C)} = 0.212 J/(g^{\circ} C)$

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!

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Answer:

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