The complete combustion of 1.00mL of octane increases the temperature of 250.00g of water by 22.7 degrees Celsius. The density of octane is 0.70g/mL. What is the efficiency of the octane in heating the water?
I attempted this question and I have -400 something percent efficiency...
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To find the efficiency of octane in heating the water, we need to calculate the energy released by the combustion of octane and compare it to the energy absorbed by the water.
Step 1: Calculate the mass of octane burned.
Given the density of octane is 0.70g/mL and the volume of octane used is 1.00mL, we can find the mass:
mass = density x volume
mass = 0.70g/mL x 1.00mL
mass = 0.70g
Step 2: Calculate the energy released by the combustion of octane.
The energy released can be calculated using the heat capacity of water and the temperature change. The specific heat capacity of water is 4.18 J/g°C.
energy released = mass x specific heat capacity x temperature change
energy released = 250.00g x 4.18 J/g°C x 22.7°C
energy released = 236,607 J
Step 3: Calculate the energy absorbed by the water.
The energy absorbed by the water can be calculated using the equation:
energy absorbed = mass x specific heat capacity x temperature change
energy absorbed = 250.00g x 4.18 J/g°C x 22.7°C
energy absorbed = 236,607 J
Step 4: Calculate the efficiency.
efficiency = (energy absorbed / energy released) x 100%
efficiency = (236,607 J / 236,607 J) x 100%
efficiency = 100%
Therefore, the efficiency of the octane in heating the water is 100%.
To find the efficiency of octane in heating the water, we need to calculate the amount of heat absorbed by the water and compare it to the energy released by the combustion of octane.
First, let's calculate the heat absorbed by the water using the specific heat capacity of water, which is 4.18 J/g°C:
Heat absorbed by water = mass of water × specific heat capacity of water × change in temperature
Given:
Mass of water = 250.00 g
Change in temperature = 22.7°C
Heat absorbed by water = 250.00 g × 4.18 J/g°C × 22.7°C
Heat absorbed by water = 236,185 J
Next, let's calculate the energy released by the combustion of octane. Since octane has a density of 0.70 g/mL, the mass of octane combusted can be found by multiplying its density by its volume.
Volume of octane combusted = 1.00 mL
Mass of octane combusted = volume of octane combusted × density of octane
Mass of octane combusted = 1.00 mL × 0.70 g/mL
Mass of octane combusted = 0.70 g
The energy released by the combustion of octane can be calculated using its heat of combustion, which is typically given as -5471 kJ/mol.
Energy released by octane combustion = moles of octane combusted × heat of combustion
First, let's calculate the number of moles of octane combusted using its molar mass, which is 114.32 g/mol.
Moles of octane combusted = mass of octane combusted / molar mass of octane
Moles of octane combusted = 0.70 g / 114.32 g/mol
Moles of octane combusted ≈ 0.00611 mol
Energy released by octane combustion = 0.00611 mol × -5471 kJ/mol × 1000 J/1 kJ
Energy released by octane combustion ≈ -33,376 J
Finally, the efficiency can be calculated by dividing the heat absorbed by the water by the energy released by the octane combustion, and multiplying by 100 to express it as a percentage:
Efficiency = (Heat absorbed by water / Energy released by octane combustion) × 100
Efficiency = (236,185 J / -33,376 J) × 100
Efficiency ≈ -708.09%
Note: The negative sign indicates that the energy was released instead of being absorbed, which commonly happens in combustion reactions. However, it's important to double-check the values used in the calculations and ensure the correct signs are assigned to effectively represent the efficiency value.