A 40 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