A40 g gold coin is cooled from 50°c to 10°c (cau is 0.13 j/g-° c. what is the h ? 325 j 208 j +208 j

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

It is called a neutralization reaction