Question

In the laboratory a student finds that it takes 103 Joules to increase the temperature of 12.6 grams of solid diamond from 22.4 to 39.4 degrees Celsius. The specific heat of diamond calculated from her data is

Answer 1

Answer:

The correct solution is "$0.480 \ J/g^{\circ}C$".

Explanation:

Given:

q = 103 J

Mass,

m = 12.6 grams

Temperature,

$T_1=22.4$

$T_2=39.4$

$\Delta T=T_2-T_1$

      $=39.4-22.4$

      $=17^{\circ}C$

Now,

⇒  $C=\frac{q}{m\times \Delta T}$

⇒      $=\frac{103}{12.6\times 17}$

⇒      $=\frac{103}{214.2}$

⇒      $=0.480 \ J/g^{\circ}C$