Verified Answer :

EjerciciosFyQ

From the equation \(Q = m\cdot c_e\cdot \Delta T\) we can clear specific heat:

\(c_e = \frac{Q}{m\cdot \Delta T}\ \to\ c_e = \frac{313.5\ J}{10\ g\cdot (60 - 25)^\circ C} = \bf 0.896\frac{J}{g\cdot ^\circ C}\)

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