Recall specific heat of water is 4186 j/kg/C. Find the specific heat of sample.

Water. Sample

Mass 109 192

Internal temperature. 21. 67

Final temperature. 30.1. 30.1

Shown by explanation;

Explanation:

The heat of the sample = mass * specific heat capacity of the sample * temperature change (∆T)

Assumption; I assume the mass of the samples are : 109g and 192g

∆T = 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 * 4186 * 8.9 = 4060.84J

For 0.192g are;

∆T = 67-30.1-=36.9°c

0.192 * 4186*36.9=29656.97J