Question

When 732.5 J of heat are added to 102G of a solid substance the temperature of the solid increases by 48.8 K what is the specific heat capacity of the solid

Answer 1

$Q = mcΔT \\ c = \frac{Q}{mΔT} \\ = \frac{732.5}{102 \times 48.8} \\ = 0.147\;J/gK$

Answer: 0.147 J/gK