Question

High-purity benzoic acid (C6H5COOH; ΔHrxn for combustion = −3227 kJ/mol) is used as a standard for calibrating bomb calorimeters. A 1.221-g sample burns in a calorimeter (heat capacity = 1365 J/°C) that contains exactly 1.430 kg of water. What temperature change is observed?

Answer 1

Answer : The temperature change observed is $4.35^oC$

Explanation :

First we have to calculate the heat released by the combustion of benzoic acid.

$q=n\times \Delta H$

where,

q = heat released by combustion of benzoic acid = ?

n = moles of benzoic acid = $\frac{\text{Mass of benzoic acid}}{\text{Molar mass of benzoic acid}}=\frac{1.221g}{122.122g/mole}=0.0099mole$

$\Delta H$ = enthalpy of combustion = 3227 kJ/mole

Now put all the given values in the above formula, get:

$q=(0.0099mole)\times (3227kJ/mole)=31.9473kJ=31947.3J$

Now we have to calculate the heat absorbed.

Heat released by combustion = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the combustion = 31947.3 J

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $1365J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water = 1.430 kg  = 1430 g

$\Delta T$ = change in temperature = ?

Now put all the given values in the above formula, we get:

$31947.3J=[(1365J/^oC\times \Delta T)+(1430g\times 4.18J/g^oC\times \Delta T)]$

$\Delta T=4.35^oC$

Therefore, the temperature change observed is $4.35^oC$