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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

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maxwellekoh

Answer:

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

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