# What is the specific heat capacity of a 17.5g sample of an unknown metal that releases 6700J of heat when it cools from 90 to 25 degrees C?

Jul 8, 2017

#### Answer:

The specific heat capacity of the metal is $\left(6 \textcolor{w h i t e}{.} \text{J")/("g"*""^@"C}\right)$.

#### Explanation:

You will use the calorimetry formula:

$q = c m \Delta T$,

where $q$ is energy in Joules (J), $c$ is the specific heat capacity, $m$ is mass, and $\Delta T$ is the change in temperature. $\Delta T = \left({T}_{\text{final"-T_"initial}}\right)$

Organize your data:

Known

$q \text{="-"6700 J}$ $\Leftarrow$ $q$ is negative because energy was released

$m = \text{17.5 g}$

DeltaT=(T_"final"-T_"initial")=(25^@"C"-90^@"C")=-65^@"C"

Unknown

$c$

Solution

Rearrange the formula above to isolate $c$. Insert your data into the formula and solve.

$c = \frac{q}{m \cdot \Delta T}$

$c = \left(- 6700 \text{J")/(17.5"g"xx-65^@"C")=(6color(white)(.)"J")/("g"*""^@"C}\right)$ rounded to one sig fig because of ${90}^{\circ} \text{C}$