# Question bce5b

May 3, 2017

The specific heat capacity of the material is $\text{0.6469 J·°C"^"-1""g"^"-1}$.

#### Explanation:

The formula for the amount heat $q$ absorbed by a substance is

color(blue)(bar(ul(|color(white)(a/a)q = mCΔTcolor(white)(a/a)|)))" "

where

$m$ is the mass of the substance
$C$ is the specific heat capacity of the material
ΔT is the temperature change

You can rearrange the formula to calculate the specific heat capacity:

C = q/(mΔT)

In this problem,

$q = \text{-86 285 J}$
$m = \text{863.3 g}$
ΔT = "(25.00 - 179.5) °C" = "-154.5 °C"#

$C = \text{-86 285 J"/"863.3 g × (-154.5 °C)" = "0.6469 J·°C"^"-1""g"^"-1}$

Calcium?