Question

q3: what will be the change in temperature of a 25.0 g block of aluminum (specific heat of al = 0.892) when it absorbs 10.0 kj of heat?

Explanation:

Step1: Recall the heat - capacity formula

$q = mc\Delta T$, where $q$ is heat absorbed, $m$ is mass, $c$ is specific heat, and $\Delta T$ is change in temperature. We need to solve for $\Delta T$.

Step2: Rearrange the formula for $\Delta T$

$\Delta T=\frac{q}{mc}$

Step3: Convert units

The mass $m = 25.0\ g$, the specific heat $c=0.892\ J/(g\cdot^{\circ}C)$ and $q = 10.0\ kJ=10000\ J$.

Step4: Substitute values into the formula

$\Delta T=\frac{10000\ J}{25.0\ g\times0.892\ J/(g\cdot^{\circ}C)}$
$\Delta T=\frac{10000}{25.0\times0.892}\ ^{\circ}C$
$\Delta T=\frac{10000}{22.3}\ ^{\circ}C\approx448\ ^{\circ}C$

Answer:

$448\ ^{\circ}C$