# How do you determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 293 to 313 K?

Jun 28, 2017

$\text{specific heat} = 0.137 \frac{J}{{g}^{o} C}$

#### Explanation:

Specific heat capacity of a substance is the amount of heat required to raise one gram of the substance by one degree Celsius. This means the first thing you need to do is convert your temperatures to Celsius.

""^@C = K - 273.15

${T}_{1} = 293 - 273.15 = {19.85}^{\circ} C$

${T}_{2} = 313 - 273.15 = {39.85}^{\circ} C$

$\Delta T = {T}_{2} - {T}_{1} = {20.00}^{\circ} C$

$\text{specific heat} = \frac{q}{m \times \Delta T}$,

where $q = \text{heat in Joules", " "m = "mass in grams}$

$\text{specific heat} = \frac{96}{35 \times 20.00} = 0.137 \frac{J}{{g}^{o} C}$