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When 1.0 g of gasoline burns, it releases 11 k cdot cal of heat. The density of gasoline is 0.74...

Question:

When {eq}1.0 \ g {/eq} of gasoline burns, it releases {eq}11 k \cdot cal {/eq} of heat. The density of gasoline is {eq}0.74 \ g / mL. {/eq}

a) How many mega joules are released when {eq}3.0 {/eq} of gasoline burns?

b) If a television requires {eq}150 \ kJ / h {/eq} to run, how many hours can the television run on the energy provided by {eq}3.0 {/eq} gal of gasoline?

Fuel value

The amount of heat that a fuel can release when is burned per gram of fuel is called the fuel value (FV). This parameter is similar to the enthalpy change of reaction but the later is measured on a molar basis {eq}\Delta H_{rx} {/eq}.

When fuel is burned for a period of time it can produce power (P), power is the amount of heat or energy applied or released by a system per unit of time. The SI unit of power is the watt, which is one joule applied per second.

Answer and Explanation:

We need to convert the amount of heat released by 1g of gasoline to MJ, so we need to remember that 1MJ=239kcal:

{eq}FV_{gas}= 11kcal/g \times \frac {1MJ}{239kcal}=0.046MJ/g {/eq}

Now, if we need to know the amount of heat released by 3g of gasoline we multiply the FV by the number of grams (mass, m):

{eq}-q= FV \times m = 0.046MJ/g \times 3g =-0.1380MJ {/eq}

a) Burning 3g of gasoline releases 0.138MJ of heat.

Now, we need to adjust that fuel value to heat produced per mL because we are asked to calculate the amount of heat produced by a defined volume of gas. We can convert the g of gasoline to mL of gasoline with the density of 0.74g/mL:

{eq}FV_{vol}= 0.046MJ/g \times \frac {0.74g}{1mL}=0.034MJ/mL {/eq}

Then, if 1gal=3785mL, we can calculate the amount of heat released by 3gal of gas:

{eq}-q= 3gal \times (\frac {3785mL}{1gal}) \times (\frac {0.034MJ}{1mL})= 386.07MJ {/eq}

Then, to convert that amount of heat to kJ so we can relate it with the power consumed by the TV, we need to remember that 1MJ=1000kJ. Then we can divide that amount of heat by the power consumed to obtain the number of hours that the TV can run.

{eq}t_{h}= \frac {386.07MJ (\frac {1000kJ}{1MJ})}{150 \frac {kJ}{h}}=2573h {/eq}

We would be able to run the TV for 2753h, considering that ALL of the energy is used exclusively on running the TV, with no waste.


Learn more about this topic:

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Energy Conservation and Energy Efficiency: Examples and Differences

from Geography 101: Human & Cultural Geography

Chapter 13 / Lesson 9
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