## Chemistry: Principles and Practice (3rd Edition)

$\frac{2.42J}{g.°C}$
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= $\frac{677J}{25.0g×11.2°C}=\frac{2.42J}{g•°C}$