Question

A 40 g gold coin is cooled from 50°C to 10°C (CAu is 0.13 J/g-°

Answer 1

When the Heat gain or lose = the mass * specific heat * ΔT

and when we have the mass of gold coin= 40 g 

and the specific Heat of gold=  0.13 J/g°

and ΔT = (Tf- Ti) = 10°C - 50°C = -40 °C

so by substitution:

∴Heat H = 40 g * 0.13 J/g° * -40
               = - 208 J