a 5.25 g sample of metal gives off 10.4 j of energy as it cools from 49.5 °c to 40.5 °c. what is the…

Explanation:

Step1: Identify the formula

The formula for heat transfer is $q = mc\Delta T$, where $q$ is the heat energy, $m$ is the mass, $c$ is the specific - heat capacity, and $\Delta T$ is the change in temperature. We need to solve for $c$, so we can re - arrange the formula to $c=\frac{q}{m\Delta T}$.

Step2: Calculate the change in temperature

$\Delta T=T_{final}-T_{initial}=40.5^{\circ}C - 49.5^{\circ}C=-9.0^{\circ}C$.

Step3: Substitute the values into the formula

We know that $q = - 10.4\ J$, $m = 5.25\ g$, and $\Delta T=-9.0^{\circ}C$. Substituting these values into $c=\frac{q}{m\Delta T}$, we get $c=\frac{-10.4\ J}{5.25\ g\times(-9.0^{\circ}C)}$.

Step4: Calculate the specific heat

$c=\frac{-10.4\ J}{-47.25\ g\cdot^{\circ}C}\approx0.22\ \frac{J}{g\cdot^{\circ}C}$.

Answer:

$0.22\ \frac{J}{g\cdot^{\circ}C}$