The amount of energy released when one gram of oil is burned is[tex]\boxed{{\text{6}}{\text{.12 kJ}}}[/tex].
Further explanation
Heat capacity is defined as the amount of heat required to change the temperature of a pure substance by one degree. Its S.I. unit is J/K.
Molar heat capacity is defined as the amount of heat required to raise the temperature of one mole of substance by one degree. Specific heat capacity is defined as the amount of heat needed to increase the temperature of 1 gram of a pure substance by one degree.
The expression to calculate amount of heat released or absorbed as follows:
[tex]{\text{q}}={\text{mC}}\Delta{\text{T}}[/tex] …… (1)
Here, q is the amount of heat.
m is the mass of substance.
C is specific heat.
[tex]\Delta{\text{T}}[/tex]is change temperature.
Calorimeter is the device that measures the change in amount of heat absorbed or released during chemical reaction followed by change in temperature. Bomb calorimeter is the device which measure the amount of heat absorbed or released during chemical reaction at constant volume.
The formula to calculate the amount of heat change in calorimeter is as follows:
[tex]{{\text{q}}_{{\text{calorimeter}}}}={{\text{C}}_{{\text{calorimeter}}}}\times\Delta{{\text{T}}_{{\text{calorimeter}}}}[/tex] …… (2)
The expression to calculate amount of heat absorbed by water is as follows:
[tex]{\text{q}}=\left({{{\text{m}}_{{\text{water}}}}}\right)\left({{{\text{C}}_{{\text{water}}}}}\right)\left({\Delta{\text{T}}}\right)[/tex] …… (1)
Given, [tex]{{\text{C}}_{{\text{water}}}}[/tex] is 4.184 [tex]{\text{J/g}}\cdot^\circ{\text{C}}[/tex].
[tex]{{\text{m}}_{{\text{water}}}}[/tex]is 1 kg.
Change in temperature [tex]\left({\Delta{\text{T}}}\right)[/tex]is [tex]5\;^\circ{\text{C}}[/tex].
Substitute the value of [tex]{{\text{C}}_{{\text{water}}}}[/tex], [tex]{{\text{m}}_{{\text{water}}}}[/tex] and [tex]\Delta{\text{T}}[/tex]in equation (1).
[tex]\begin{aligned}{{\text{q}}_{{\text{water}}}}&=\left({{\text{1}}\;{\text{kg}}}\right)\left({\frac{{1000\;{\text{g}}}}{{{\text{1}}\;{\text{kg}}}}}\right)\left({\frac{{{\text{4}}{\text{.184 J}}}}{{{\text{g}}\cdot ^\circ{\text{C}}}}}\right)\left({5\;^\circ{\text{C}}}\right)\\&{\text{=20920 J}}\\\end{aligned}[/tex]
The amount of heat absorbed by water is 20920 J.
The value of [tex]{{\text{C}}_{{\text{calorimeter}}}}[/tex]is[tex]0.10\;{\text{kJ/}}^\circ{\text{C}}[/tex].
The value of [tex]\Delta{{\text{T}}_{{\text{calorimeter}}}}[/tex]is [tex]{5^\circ}{\text{C}}[/tex].
Substitute these values in equation (2).
[tex]\begin{aligned}{{\text{q}}_{{\text{calorimeter}}}}&=\left({\frac{{0.10\;{\text{kJ}}}}{{^\circ{\text{C}}}}}\right)\times 5\;^\circ{\text{C}}\\&={\text{0}}{\text{.50 kJ}}\\\end{aligned}[/tex]
Amount of heat absorbed by calorimeter is[tex]{\mathbf{0}}{\mathbf{.50 kJ}}[/tex].
The formula to calculate the total amount of heat released to burn 3.5 g of sample is as follows:
[tex]- {{\text{q}}_{{\text{rxn}}}}={{\text{q}}_{{\text{water}}}}+{{\text{q}}_{{\text{calorimeter}}}}[/tex] …… (3)
Substitute 20920 J for [tex]{{\text{q}}_{{\text{water}}}}[/tex] and 0.50 J for [tex]{{\text{q}}_{{\text{calorimeter}}}}[/tex] in equation (3)
[tex]\begin{aligned}-{{\text{q}}_{{\text{rxn}}}}&=20920{\text{ J}}+{\text{0}}{\text{.50 kJ}}\left({\frac{{1000\;{\text{J}}}}{{{\text{1}}\;{\text{kJ}}}}}\right)\\&={\text{21420 J}}\\\end{aligned}[/tex]
The total amount of energy released to burn 3.5 g of oil is 21420 J.
The amount of energy released to burn 1 g of oil is calculated as follows:
[tex]\begin{aligned}{\text{Energy released per gram}}&=\frac{{21420{\text{ J}}}}{{3.5{\text{ g}}}}\\&={\text{6120 J}}\\\end{aligned}[/tex]
The conversion factor to convert energy from J to kJ is as follows:
[tex]1{\text{ kJ}}={\text{1000}}\;{\text{J}}[/tex]
The amount of energy released after one gram of oil gets burned is calculated as follows:
[tex]\begin{aligned}{\text{Energy released per gram}}\left({{\text{kJ}}}\right)&=\left({{\text{1 kJ}}}\right)\left({\frac{{{\text{6120 J}}}}{{{\text{1000J}}}}}\right)\\&={\text{6}}{\text{.12 kJ}}\\\end{aligned}[/tex]
Hence, [tex]{\mathbf{6}}{\mathbf{.12 kJ}}[/tex] of energy is released when one gram of oil is burned.
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Answer details:
Grade: Senior school.
Subject: Chemistry.
Chapter: Thermodynamics
Keywords: Heat capacity, molar heat capacity, specific heat capacity, released, absorbed, calorimeter, bomb calorimeter, water, mass, temperature, degree, 6.12 kj, 6120 j, 0.50 and 2120 J.
The heat released per gram of oil shale is 6120 J/g.
To determine the heat released per gram of oil shale, calculate the heat absorbed by both the water and the calorimeter, then divide the total heat by the mass of the oil shale.
The heat released per gram of oil shale is 6120 J/g.To determine how much heat is released per gram of oil shale when it is burned, follow these steps:
First, calculate the heat absorbed by the water.
The heat absorbed by the water (q water) can be calculated using the formula:qwater = mass * specific heat * ΔTHere, the mass of the water is 1000 g (since 1 kg = 1000 g), the specific heat of water is 4.184 J/g°C, and ΔT is the change in temperature (5.0°C).Thus, q water = 1000 g * 4.184 J/g°C * 5.0°C = 20920 J.
Next, calculate the heat absorbed by the bomb calorimeter.
Since the calorimeter has a heat capacity of 0.10 kJ/°C, convert this to J/°C (1 kJ = 1000 J): 0.10 kJ/°C = 100 J/°C.Then, calculate the heat (q calorimeter) absorbed by the calorimeter:
q calorimeter = heat capacity * ΔT = 100 J/°C * 5.0°C = 500 J.Add the heat absorbed by the water and the calorimeter to get the total heat released by the combustion of the oil shale:
q total = q water + q calorimeterq total = 20920 J + 500 J = 21420 J.Finally, determine the heat released per gram of oil shale by dividing the total heat by the mass of the oil shale:
Heat released per gram = q total / mass21420 J / 3.5 g = 6120 J/g.Therefore, the heat released per gram of oil shale is 6120 J/g.
Correct question is: A 3.5-g sample of colorado oil shale is burned in a bomb calorimeter, which causes the temperature of the calorimeter to increase by 5.0°c. the calorimeter contains 1.00 kg of water (heat capacity of H₂O = 4.184 j/g°c) and the heat capacity of the empty calorimeter is 0.10 kj/°c. how much heat is released per gram of oil shale when it is burned?
The process in which water changes from a liquid to a vapor is known as __________.
Magda has the two magnets shown below what will happen if magda tries to push the north poles of the two magnets toward one another
What is the percent yield of hbr if 85.00 g of hbr was formed from 30. g of h2?
Ok, lets see the definitions of percent yield, actual yield and theoretical yield.
Percent yield is the ratio of actual yield to theoretical yield.
Actual yield- amount of product produced in the EXPERIMENT.
Theoretical yield- max amount of product produced through CALCULATIONS
% yield= actual yield (from experiment)/theoretical yield (from calculation) *100
1st Step: Write the reaction
H2 + Br2 --> 2 HBr
2nd Step: Get the mass ratio of H2 and HBr to find theoretical yield
1 mole H2 gives 2 moles HBr ( molar mass of H2= 2 g/mol, HBr= 81 g/mol)
2 g H2 gives 2*81= 162 g HBr
30 g H2 gives 162*30/2 = 2430 g HBr ( The equation is 2 g H2/ 30 gH2= 162 g HBr/ x g HBr)
So theoretically 2430 g HBr are produced by calculation ( THEORETICAL YIELD)
By experiment 85 g HBr are produced as it is given at the question ( ACTUAL YIELD)
% yield= actual yield / theoretical yield *100 = 85/2430 *100= 3.5 %
The percent yield is 3.5 %.
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R = Gas constant = 8.314J/K/mo
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Explanation:
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edge 23
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b.the separation of a compound into its elements
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A chemical change involves the conversion of substances into different substances, as in the separation of a compound into its elements which requires breaking chemical bonds, unlike a physical change.
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Pls Help with this Chemistry question
Which of the following energy level changes would give off the most light energy
A.n=5 to n=1
B.n=4 to n=5
C.n=2 to n=5
D.n=5 to n=4
"The correct option is A. [tex]\( n=5 \) to \( n=1 \).[/tex]
In atomic physics, the energy of the emitted photon when an electron transitions from one energy level to another is given by the Rydberg formula:
[tex]\[ E = h \cdot f = R \cdot \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \][/tex] where \( h \) is Planck's constant, \( f \) is the frequency of the emitted photon, \( R \) is the Rydberg constant for hydrogen, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.
The energy of the photon is directly proportional to the frequency of the light emitted, and the frequency is inversely proportional to the wavelength of the light. Therefore, the greater the energy difference between the two levels, the higher the frequency of the emitted light and the more energy the light carries.
Let's calculate the energy released for each transition:
[tex]A. \( n=5 \) to \( n=1 \):[/tex]
[tex]\[ E_{5 \to 1} = R \cdot \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = R \cdot \left( 1 - \frac{1}{25} \right) = R \cdot \frac{24}{25} \][/tex]
B. [tex]\( n=4 \) to \( n=5 \):[/tex]
[tex]\[ E_{4 \to 5} = R \cdot \left( \frac{1}{5^2} - \frac{1}{4^2} \right) = R \cdot \left( \frac{1}{25} - \frac{1}{16} \right) = R \cdot \left( -\frac{9}{400} \right) \][/tex]
(Note: This transition actually absorbs energy, as the electron moves to a higher energy level.)
C.[tex]\( n=2 \) to \( n=5 \):[/tex]
[tex]\[ E_{2 \to 5} = R \cdot \left( \frac{1}{5^2} - \frac{1}{2^2} \right) = R \cdot \left( \frac{1}{25} - \frac{1}{4} \right) = R \cdot \left( -\frac{21}{100} \right) \][/tex]
(Again, this transition absorbs energy.)
D. [tex]\( n=5 \) to \( n=4 \):[/tex]
[tex]\[ E_{5 \to 4} = R \cdot \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R \cdot \left( \frac{1}{16} - \frac{1}{25} \right) = R \cdot \frac{9}{400} \][/tex]
Comparing the magnitudes of the energy changes, it is clear that the transition from \( n=5 \) to \( n=1 \) releases the most energy because it has the largest difference in energy levels. This corresponds to the largest absolute value of the energy change, which means it will give off the most light energy.
The energy released in the fission of a 235u nucleus is about 200 mev. how much rest mass (in kg) is converted to energy in this fission?
To convert 1.00 kg of 235U into energy, about 9.14 × 10^-4 kg of rest mass is converted.
Explanation:The energy released in the fission of a 235U nucleus is about 200 MeV. To find out how much rest mass is converted to energy, we need to calculate the number of 235U atoms in 1.00 kg. One mole of 235U has a mass of 235.04 grams, so there are 4.25 moles of 235U in 1.00 kg. Using Avogadro's number (6.02 × 10^23 U/mol), we can calculate that there are 2.56 × 10^24 atoms of 235U in 1.00 kg. The total energy released is the number of atoms times the given energy per U fission, which is:
(2.56 × 10^24 atoms)(200 MeV/atom) = 5.12 × 10^26 MeV
To convert this to kilograms, we can use the conversion factor 1 MeV = 1.783 × 10^-30 kg, so:
(5.12 × 10^26 MeV)(1.783 × 10^-30 kg/MeV) = 9.14 × 10^-4 kg
Therefore, about 9.14 × 10^-4 kg of rest mass is converted to energy in this fission.
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