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OpenStudy (anonymous):
An unknown substance weighing 42.0 g was heated from 36.0°C to 65.0°C. In this process, the substance absorbed 5895 J of energy. What is the specific heat of the substance?
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OpenStudy (anonymous):
you can use q=mC(Tf-Ti) and rearrange it to solve for C. C=q/m(Tf-Ti).
OpenStudy (anonymous):
\[4.84 J/g C\]
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