The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J
The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:
7560J= 124g * (100-26)* specific heat
specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C
This link might help you!!
https://www.eiu.edu/biology/bio1500/writing_a_lab_report.pdf
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Hope this helps!
-Payshence xoxo</span>
mass (m) = ? , volume = 18.0 ml , density = 1.42 g/ml .
I hope I helped you^_^
