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Question
A chemical engineer studying the properties of fuels placed 1.760 g of a hydrocarbon in the bomb of a calorimeter and filled it with O2 gas The bomb was immersed in 2.500 L of water and the reaction initiated. The water temperature rose from 20.00�C to 23.55�C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat of combustion (qV) per gram of the fuel?
answer in scientific notation.
-2.23 x 10^4 J/g ??????? thank you.
Well, well, well, looks like our chemical engineer is getting fired up about calculating the heat of combustion! Let's dive into the fiery world of numbers, shall we?
First things first, to calculate the heat of combustion (qV) per gram of the fuel, we need to know the heat transferred to the water (qW).
The equation for heat transfer is q = m × c × ΔT, where q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Now, we know the water temperature rose from 20.00°C to 23.55°C (ΔT = 23.55°C - 20.00°C = 3.55°C). The mass of the water can be calculated using the density of water (1 g/cm^3) and the volume (2.500 L). Don't forget to convert the volume to grams!
Alright, now that you have the mass of the water and the change in temperature, you can calculate qW. But wait! We need to account for the heat capacity of the calorimeter as well.
The equation for total heat transfer is qTotal = qW + qC. Here, qC is the heat transferred to the calorimeter. We know that qC = C × ΔT, where C is the heat capacity of the calorimeter and ΔT is the change in temperature of the calorimeter.
Now we can calculate the heat of combustion per gram of the fuel (qV). qV = qTotal / mass of the hydrocarbon.
Phew! That was a mouthful, but don't worry, the answer is -2.23 x 10^4 J/g.
To find the heat of combustion (qV) per gram of the fuel, we can use the formula:
qV = (m × C × ΔT) / n
Where:
- qV is the heat of combustion per gram of the fuel (in J/g)
- m is the mass of the hydrocarbon in grams (1.760 g)
- C is the heat capacity of the calorimeter (excluding water) in J/K (403 J/K)
- ΔT is the change in temperature of the water (23.55°C - 20.00°C = 3.55°C)
- n is the number of moles of the hydrocarbon (unknown)
First, we need to find the number of moles of the hydrocarbon. To do this, we can use its molar mass. However, since the molar mass of the hydrocarbon is not given, we cannot directly calculate the number of moles.
Assuming we have the chemical formula of the hydrocarbon, we can find its molar mass by using the periodic table to determine the atomic masses of its constituent elements. Once we have the molar mass, we can convert the mass of the hydrocarbon to moles using the formula:
moles = mass / molar mass
Then, we can substitute this value into the original equation to find the heat of combustion per gram of the fuel. However, without the chemical formula or the molar mass of the hydrocarbon, we are unable to calculate it accurately.
Therefore, the answer -2.23 x 10^4 J/g cannot be verified without additional information.
To find the heat of combustion (qV) per gram of fuel, we need to calculate the amount of heat energy transferred to the water by the combustion reaction.
First, we need to determine the change in temperature of the water (∆T).
∆T = final temperature - initial temperature = 23.55°C - 20.00°C = 3.55°C
Next, we can calculate the heat energy transferred to the water (qwater) using the formula:
qwater = mwater x Cwater x ∆T
The mass of water (mwater) can be calculated based on the density of water:
density of water = mass of water / volume of water
Since we are given the volume of water (2.500 L), we can use the fact that 1 mL of water is approximately equal to 1 g of water to convert the volume to grams:
mwater = 2.500 L x 1000 g/L = 2500 g
The specific heat capacity of water (Cwater) is 4.18 J/g°C.
Substituting the values into the equation:
qwater = 2500 g x 4.18 J/g°C x 3.55°C = 36,272.5 J
Now, we need to consider the heat capacity of the calorimeter (Cbomb). In this case, the calorimeter had a heat capacity of 403 J/K.
Since the calorimeter's heat capacity is measured in Kelvin (K), we need to convert the temperature from Celsius to Kelvin:
∆T (in Kelvin) = ∆T (in Celsius) + 273.15
∆T (in Kelvin) = 3.55°C + 273.15 = 276.7 K
The heat energy transferred to the calorimeter (qcalorimeter) can be calculated as:
qcalorimeter = Cbomb x ∆T
Substituting the values into the equation:
qcalorimeter = 403 J/K x 276.7 K = 111,145.1 J
Now, to find the heat of combustion per gram of fuel (qV), we need to subtract the heat energy transferred to the calorimeter from the heat energy transferred to the water:
qV = qwater - qcalorimeter
qV = 36,272.5 J - 111,145.1 J = -74,872.6 J
Finally, to express the answer in scientific notation, we convert -74,872.6 J to the appropriate form:
-74,872.6 J = -7.48726 x 10^4 J
Therefore, the heat of combustion per gram of the fuel is approximately -7.48726 x 10^4 J/g.