Question

a sample of ethanol (c₂h₆o) has a mass of 0.2301 g. complete combustion of this sample causes the temperature of a bomb calorimeter to increase by 1.35 °c. the calorimeter has a mass of 2.000 kg and a specific heat of 2.45 j/g·°c. what is the heat of combustion for this sample?

□ kj

Explanation:

Step1: Convert calorimeter mass to grams

The calorimeter mass is \(2.000\space kg\). Since \(1\space kg = 1000\space g\), we convert it: \(2.000\space kg\times1000\space g/kg = 2000\space g\).

Step2: Calculate heat absorbed by calorimeter

The formula for heat absorbed \(q\) is \(q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat, and \(\Delta T\) is temperature change.
Substituting values: \(m = 2000\space g\), \(c = 2.45\space J/g^\circ C\), \(\Delta T = 1.35^\circ C\).
\(q = 2000\space g\times2.45\space J/g^\circ C\times1.35^\circ C\)
First, calculate \(2000\times2.45 = 4900\), then \(4900\times1.35 = 6615\space J\).

Step3: Convert heat to kJ and relate to sample

The heat released by the ethanol sample is equal to the heat absorbed by the calorimeter (but with opposite sign, but we need magnitude for heat of combustion). Convert \(6615\space J\) to \(kJ\): \(6615\space J\div1000 = 6.615\space kJ\).
This heat is from \(0.2301\space g\) of ethanol. But wait, actually, in bomb calorimetry, the heat released by the sample (\(q_{combustion}\)) is equal to \(-q_{calorimeter}\), but since we want the heat of combustion (heat per gram or per mole, here per gram? Wait, the question says "heat of combustion for this sample" – maybe per gram? Wait, no, let's check. Wait, the heat absorbed by calorimeter is \(q_{cal} = mc\Delta T\), and \(q_{combustion} = -q_{cal}\) (since combustion releases heat, calorimeter absorbs). But the mass of sample is \(0.2301\space g\). Wait, maybe the question is asking for the heat released per gram? Wait, no, let's recalculate. Wait, the calorimeter mass is \(2.000\space kg = 2000\space g\), specific heat \(2.45\space J/g^\circ C\), \(\Delta T = 1.35^\circ C\). So \(q_{cal} = 2000\times2.45\times1.35 = 6615\space J = 6.615\space kJ\). This is the heat absorbed by the calorimeter, so the heat released by the ethanol is \( - 6.615\space kJ\) (but magnitude is \(6.615\space kJ\) for the sample's combustion). Wait, but maybe the question is asking for the heat of combustion (heat per gram of ethanol)? Wait, no, the sample's mass is \(0.2301\space g\), so the heat released by the sample is \(6.615\space kJ\) (since calorimeter absorbs that heat from the sample's combustion). Wait, maybe I made a mistake. Wait, bomb calorimeter: the heat released by the reaction (combustion) is equal to the heat absorbed by the calorimeter system. So \(q_{rxn} = -q_{cal}\). \(q_{cal} = mc\Delta T\), so \(q_{rxn} = -mc\Delta T\). Then, the heat of combustion (per gram) would be \(q_{rxn}\) divided by the mass of the sample. Wait, let's do that.

First, \(q_{cal} = 2000\space g\times2.45\space J/g^\circ C\times1.35^\circ C = 6615\space J = 6.615\space kJ\). Then \(q_{rxn} = - 6.615\space kJ\) (negative because combustion is exothermic). The mass of ethanol is \(0.2301\space g\). So the heat of combustion (per gram) is \(q_{rxn}/m = -6.615\space kJ / 0.2301\space g\approx -28.75\space kJ/g\). Wait, but the question says "heat of combustion for this sample" – maybe total heat released by the sample? Wait, the sample's mass is \(0.2301\space g\), and the heat released is \(6.615\space kJ\) (since calorimeter absorbed that). Wait, maybe the question is asking for the total heat released by the sample, which is equal to the heat absorbed by the calorimeter. So \(6.615\space kJ\), approximately \(6.62\space kJ\) or more precise. Wait, let's recalculate \(2000\times2.45 = 4900\), \(4900\times1.35\): \(4900\times1 = 4900\), \(4900\times0.35 = 1715\), so total \(4900 + 1715 = 6615\space J = 6.615\space kJ\). So the heat of combustion (the heat released by the sample) is \(6.615\space kJ\), or if we consider per gram, but the question says "for this sample" – maybe the total heat released by the \(0.2301\space g\) sample, which is \(6.615\space kJ\), so approximately \(6.62\space kJ\) or more accurately \(6.615\space kJ\). Wait, but let's check the calculation again.

Wait, the calorimeter's mass is \(2.000\space kg = 2000\space g\), specific heat \(2.45\space J/g^\circ C\), \(\Delta T = 1.35^\circ C\). So \(q = mc\Delta T = 2000 \times 2.45 \times 1.35\). Let's compute that:

\(2000 \times 2.45 = 4900\)

\(4900 \times 1.35\):

\(4900 \times 1 = 4900\)

\(4900 \times 0.3 = 1470\)

\(4900 \times 0.05 = 245\)

So \(1470 + 245 = 1715\)

Then \(4900 + 1715 = 6615\space J = 6.615\space kJ\)

So the heat released by the sample is \(6.615\space kJ\) (since the calorimeter absorbed that heat from the sample's combustion). So the heat of combustion for this sample (the total heat released by the \(0.2301\space g\) sample) is \(6.615\space kJ\), which can be rounded to \(6.62\space kJ\) or kept as \(6.615\space kJ\).

Answer:

\(6.62\) (or \(6.615\))