Question

a 295 - g aluminum engine part at an initial temperature of 13.00 °c absorbs 75.0 kj of heat. what is the final temperature of the part? (c of al = 0.900 j/(g·k))

Explanation:

Step1: Convert heat to joules

$75.0\ kJ = 75.0\times10^{3}\ J$

Step2: Use the heat - transfer formula $q = mc\Delta T$

We know $q = mc(T_{f}-T_{i})$, where $q$ is heat, $m$ is mass, $c$ is specific heat capacity, $T_{f}$ is final temperature and $T_{i}$ is initial temperature. We need to solve for $T_{f}$. First, we can re - arrange the formula to $T_{f}=\frac{q}{mc}+T_{i}$.
$m = 295\ g$, $c=0.900\ J/(g\cdot K)$ and $T_{i}=13.00^{\circ}C$. Since a change in Celsius is equal to a change in Kelvin for temperature differences, we can use the values directly.
$T_{f}=\frac{75.0\times 10^{3}\ J}{295\ g\times0.900\ J/(g\cdot K)}+ 13.00^{\circ}C$

Step3: Calculate the temperature change

$\frac{75.0\times 10^{3}\ J}{295\ g\times0.900\ J/(g\cdot K)}=\frac{75000\ J}{265.5\ J/K}\approx282.5\ K$ (or $^{\circ}C$ as the scale change is the same for differences)

Step4: Calculate the final temperature

$T_{f}=282.5^{\circ}C + 13.00^{\circ}C=295.5^{\circ}C$

Answer:

$295.5^{\circ}C$