Answer
$\frac{2.42J}{g.°C}$
Work Step by Step
For finding out the heat, q, we use the relation
q= c×m×∆T where c is the specific heat capacity, m is the mass and ∆T the change in temperature.
Rearranging, we get c= $\frac{q}{m.∆T}$
Given that q= 677 J, m= 25.0 g
and ∆T= (34.7-23.5)°C = 11.2°C
Then, specific heat c= $\frac{677J}{25.0g×11.2°C}=\frac{2.42J}{g•°C}$
