# What is the specific heat of a substance that absorbs 2.5*10^3 joules of heat when a sample of 1 * 10^4 g of the substance increases in temperature from 10°C to 80°C?

Nov 6, 2015

s=3.57xx10^(-3)J/(g*""^@C)

#### Explanation:

The amount of heat absorbed by the system could be calculated by:

$q = s \times m \times \Delta T$

where, $\textcolor{b l u e}{s}$ is the specific heat capacity
$\textcolor{b l u e}{m}$ is the mass of the object
$\textcolor{b l u e}{\Delta T} = {T}_{f} - {T}_{i}$ is the change on temperature. $\Delta T = 80 - 10 = {70}^{\circ} C$

The specific heat capacity is then: s=q/(mxxDeltaT)=(2.5xx10^3)/(1xx10^4xx70)=3.57xx10^(-3)J/(g*""^@C)

Here is a video that fully explains this concept:
Thermochemistry | Enthalpy and Calorimetry.