complete combustion of a 0.600 - g sample of a compound in a bomb calorimeter releases 24.0 kj of heat. the…

Explanation:

Step1: Convert mass to grams and heat to joules

The mass of the calorimeter $m = 1.30\ kg=1.30\times10^{3}\ g$, and $q = 24.0\ kJ = 24.0\times10^{3}\ J$.

Step2: Rearrange the heat - capacity formula for $\Delta T$

Given $q = mC_{p}\Delta T$, we can solve for $\Delta T$ as $\Delta T=\frac{q}{mC_{p}}$. Substitute $q = 24.0\times 10^{3}\ J$, $m = 1.30\times10^{3}\ g$ and $C_{p}=3.41\ J/(g\cdot^{\circ}C)$ into the formula. $\Delta T=\frac{24.0\times 10^{3}\ J}{1.30\times 10^{3}\ g\times3.41\ J/(g\cdot^{\circ}C)}\approx5.4^{\circ}C$

Step3: Calculate the final temperature

We know that $\Delta T=T_{f}-T_{i}$, so $T_{f}=T_{i}+\Delta T$. Given $T_{i} = 25.5^{\circ}C$ and $\Delta T\approx5.4^{\circ}C$, then $T_{f}=25.5^{\circ}C + 5.4^{\circ}C=30.9^{\circ}C$

Answer:

$30.9^{\circ}C$