Question

Exactly 313.5 J will raise the temperature of 10.0 g of a metal from 25.0 C to 60.0 C. What is the specific heat of the metal?

Answer 1

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

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