Question

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?

Answer 1

The amount of energy released when one gram of oil is burned is$\boxed{{\text{6}}{\text{.12 kJ}}}$.

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:

${\text{q}}={\text{mC}}\Delta{\text{T}}$                                       …… (1)

Here, q is the amount of heat.

m is the mass of substance.

C is specific heat.

$\Delta{\text{T}}$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:

${{\text{q}}_{{\text{calorimeter}}}}={{\text{C}}_{{\text{calorimeter}}}}\times\Delta{{\text{T}}_{{\text{calorimeter}}}}$              …… (2)

The expression to calculate amount of heat absorbed by water is as follows:

${\text{q}}=\left({{{\text{m}}_{{\text{water}}}}}\right)\left({{{\text{C}}_{{\text{water}}}}}\right)\left({\Delta{\text{T}}}\right)$                                                …… (1)

Given, ${{\text{C}}_{{\text{water}}}}$ is 4.184 ${\text{J/g}}\cdot^\circ{\text{C}}$.

${{\text{m}}_{{\text{water}}}}$is 1 kg.

Change in temperature $\left({\Delta{\text{T}}}\right)$is $5\;^\circ{\text{C}}$.

Substitute the value of ${{\text{C}}_{{\text{water}}}}$, ${{\text{m}}_{{\text{water}}}}$ and $\Delta{\text{T}}$in equation (1).

$\begin{aligned}{{\text{q}}_{{\text{water}}}}&=\left({{\text{1}}\;{\text{kg}}}\right)\left({\frac{{1000\;{\text{g}}}}{{{\text{1}}\;{\text{kg}}}}}\right)\left({\frac{{{\text{4}}{\text{.184 J}}}}{{{\text{g}}\cdot ^\circ{\text{C}}}}}\right)\left({5\;^\circ{\text{C}}}\right)\\&{\text{=20920 J}}\\\end{aligned}$

The amount of heat absorbed by water is 20920 J.

The value of ${{\text{C}}_{{\text{calorimeter}}}}$is$0.10\;{\text{kJ/}}^\circ{\text{C}}$.

The value of $\Delta{{\text{T}}_{{\text{calorimeter}}}}$is ${5^\circ}{\text{C}}$.

Substitute these values in equation (2).

$\begin{aligned}{{\text{q}}_{{\text{calorimeter}}}}&=\left({\frac{{0.10\;{\text{kJ}}}}{{^\circ{\text{C}}}}}\right)\times 5\;^\circ{\text{C}}\\&={\text{0}}{\text{.50 kJ}}\\\end{aligned}$

Amount of heat absorbed by calorimeter is${\mathbf{0}}{\mathbf{.50 kJ}}$.

The formula to calculate the total amount of heat released to burn 3.5 g of sample is as follows:

$- {{\text{q}}_{{\text{rxn}}}}={{\text{q}}_{{\text{water}}}}+{{\text{q}}_{{\text{calorimeter}}}}$                   …… (3)

Substitute 20920 J for ${{\text{q}}_{{\text{water}}}}$ and 0.50 J for ${{\text{q}}_{{\text{calorimeter}}}}$ in equation (3)

$\begin{aligned}-{{\text{q}}_{{\text{rxn}}}}&=20920{\text{ J}}+{\text{0}}{\text{.50 kJ}}\left({\frac{{1000\;{\text{J}}}}{{{\text{1}}\;{\text{kJ}}}}}\right)\\&={\text{21420 J}}\\\end{aligned}$

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:

$\begin{aligned}{\text{Energy released per gram}}&=\frac{{21420{\text{ J}}}}{{3.5{\text{ g}}}}\\&={\text{6120 J}}\\\end{aligned}$

The conversion factor to convert energy from J to kJ is as follows:

$1{\text{ kJ}}={\text{1000}}\;{\text{J}}$

The amount of energy released after one gram of oil gets burned is calculated as follows:

$\begin{aligned}{\text{Energy released per gram}}\left({{\text{kJ}}}\right)&=\left({{\text{1 kJ}}}\right)\left({\frac{{{\text{6120 J}}}}{{{\text{1000J}}}}}\right)\\&={\text{6}}{\text{.12 kJ}}\\\end{aligned}$

Hence, ${\mathbf{6}}{\mathbf{.12 kJ}}$ of energy is released when one gram of oil is burned.

Learn more:

1. Which is most likely a covalent compound brainly.com/question/2083444.

2. Calculate the pH of 0.1 m compoundhttps://brainly.com/question/2114744.

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.

Answer 2

q(rxn) = - [q(water)+q(bomb)] q(rxn) = -{[ (1000g)(4.184)(5.0)] + [ (5.0)(0.10)]} q(rxn) = - (20920 + 0.5) Now we divide 3.5g q(rxn)= - (20920)/(3.5g) q(rxn) = 5977.14 And final answer, change is to Kilo joule unit -q(rxn) = 5.23 KJ/unit