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