Respuesta :
Answer : The temperature change observed is [tex]4.35^oC[/tex]
Explanation :
First we have to calculate the heat released by the combustion of benzoic acid.
[tex]q=n\times \Delta H[/tex]
where,
q = heat released by combustion of benzoic acid = ?
n = moles of benzoic acid = [tex]\frac{\text{Mass of benzoic acid}}{\text{Molar mass of benzoic acid}}=\frac{1.221g}{122.122g/mole}=0.0099mole[/tex]
[tex]\Delta H[/tex] = enthalpy of combustion = 3227 kJ/mole
Now put all the given values in the above formula, get:
[tex]q=(0.0099mole)\times (3227kJ/mole)=31.9473kJ=31947.3J[/tex]
Now we have to calculate the heat absorbed.
Heat released by combustion = Heat absorbed by the calorimeter + Heat absorbed by the water
[tex]q=[q_1+q_2][/tex]
[tex]q=[c_1\times \Delta T+m_2\times c_2\times \Delta T][/tex]
where,
q = heat released by the combustion = 31947.3 J
[tex]q_1[/tex] = heat absorbed by the calorimeter
[tex]q_2[/tex] = heat absorbed by the water
[tex]c_1[/tex] = specific heat of calorimeter = [tex]1365J/^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_2[/tex] = mass of water = 1.430 kg = 1430 g
[tex]\Delta T[/tex] = change in temperature = ?
Now put all the given values in the above formula, we get:
[tex]31947.3J=[(1365J/^oC\times \Delta T)+(1430g\times 4.18J/g^oC\times \Delta T)][/tex]
[tex]\Delta T=4.35^oC[/tex]
Therefore, the temperature change observed is [tex]4.35^oC[/tex]
The temperature change observed is [tex]4.35^oC[/tex].
Specific heat capacity:
It is the ratio of the amount of heat required to raise the temperature of a substance by one degree Celsius to the amount of heat required to raise the temperature of the equal amount of water by one-degree Celsius at the room temperature. It is given by:
[tex]Q=m*c*\triangle T[/tex]
→ Calculation for moles of benzoic acid:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\\\\text{Number of moles}=\frac{1.221g}{122g/mol}\\\\ \text{Number of moles}=0.0099\text{ mole}[/tex]
→ Calculation for heat absorbed:
Heat released by combustion = Heat absorbed by the calorimeter + Heat absorbed by the water
[tex]Q=[Q_1+Q_2]\\\\Q=[c_1*\triangle T+m_2*c_2*\triangle T][/tex]
where,
Q = heat released by the combustion = [tex]31947.3 J[/tex]
Q₁ = heat absorbed by the calorimeter
Q₂ = heat absorbed by the water
c₁ = specific heat of calorimeter = [tex]1365J/g^oC[/tex]
c₂ = specific heat of water = [tex]4.18J/g^oC[/tex]
m₂ = mass of water = 1.430 kg = 1430 g
[tex]\triangle T[/tex] = change in temperature = ?
On substituting the values in the above formula:
[tex]Q=[c_1*\triangle T+m_2*c_2*\triangle T]\\\\31947.3J=[(1365J/^oC *\traingle T)+(1430*4.18J/g^oC*\triangle T)]\\\\\triangle T=4.35^oC[/tex]
Therefore, the temperature change observed is [tex]4.35^oC[/tex].
Find more information about Specific heat here:
brainly.com/question/13439286
