Question

Stearic acid $\left[C{H}_3\right(C{H}_2{)}_{16}C{O}_{2}H]$ is a fatty acid. $1.0g$ of stearic acid was burned in a bomb calorimetre. The bomb had a heat capacity of $652J{/}^{\circ }C$. If the temperature of $500g$ water $(c=4.18J/g{}^{\circ }C)$ rose from $25.0to39.3{}^{\circ }C$, how much heat was released when the stearic acid was burned? [Given $C_p\left(H_{2}O\right)=4.18J/g{}^{\circ }C]$

• A. $-39.21kJ$
• B. $-29.91kJ$
• C. $-108kJ$
• D. $-9.32kJ$

Answer 1

The correct option is A $-39.21kJ$

The heat released by reaction of burning of stearic acid in a calorimetre will be used to supplement the heat capacity of bomb calorimeter and to raise the temperature. So,
$q_{\text{calorimetre}}=q_{\text{bomb}}+q_{\text{water}}$
$q_{\text{calorimetre}}=[\text{Heat capacity of calorimeter}+(m_{\text{water}}\times c\left)\right]△T$
Putting the values we get,
$q_{\text{calorimetre}}=(652+500\times 4.18)\times 14.3$
$=39210Jor39.21kJ$
$\therefore q_{\text{reaction}}=-q_{\text{calorimetre}}=-39.21kJ$