Question

what is the mass of an iridium sample which absorbs 894 j of energy when its temperature increases by 12.3 °c? (c_ir = 1.30 \frac{j}{g \cdot °c}) m = ? g

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed or released 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 $m$, so rearrange the formula to $m=\frac{q}{c\Delta T}$.

Step2: Identify the given values

We know that $q = 894\ J$, $c = 1.30\ \frac{J}{g\cdot^{\circ}C}$, and $\Delta T = 12.3\ ^{\circ}C$.

Step3: Substitute the values into the formula

Substitute the values into $m=\frac{q}{c\Delta T}$: $m=\frac{894\ J}{1.30\ \frac{J}{g\cdot^{\circ}C}\times12.3\ ^{\circ}C}$. First, calculate the denominator: $1.30\times12.3 = 15.99$. Then, divide the numerator by the denominator: $m=\frac{894}{15.99}\approx55.91\ g$.

Answer:

$\approx 55.9\ g$ (or more precisely around 55.91 g)