Question

Sam heats an 8kg sample of sand, with a specific heat of 664 J/kg·C°, from 20° to 40°. What is the change in thermal energy?

Answer 1

Answer:

106.24 kJ.

Explanation:

Given that,

Mass of sample of sand, m = 8 kg

Specific heat of sand, c = 664 J/kg-°C

The temperature changes from 20° C to 40° C. We need to find the change in thermal energy. It is given by :

$Q=mc\Delta T\\\\Q=8\times 664(40-20)\\\\=106240\ J\\\\=106.24\ kJ$

So, the change in thermal energy is 106.24 kJ.