Question

what is the specific heat of an object that weighs 86.7 g and absorbs 2566.2 j of heat when the temperature changes of 25.0 °c.

Explanation:

Step1: Recall the heat - capacity formula

The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat absorbed or released, $m$ is the mass of the object, $c$ 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: Identify the given values

We are given that $Q = 2566.2\ J$, $m = 66.7\ g$, and $\Delta T=25.0^{\circ}C$.

Step3: Substitute the values into the formula

First, substitute the values into $c=\frac{Q}{m\Delta T}$: $c=\frac{2566.2\ J}{66.7\ g\times25.0^{\circ}C}$.

Step4: Calculate the specific heat

$66.7\ g\times25.0^{\circ}C = 1667.5\ g\cdot^{\circ}C$. Then $c=\frac{2566.2\ J}{1667.5\ g\cdot^{\circ}C}=1.54\ J/g\cdot^{\circ}C$.

Answer:

$1.54\ J/g\cdot^{\circ}C$