Question

Addition of 50. J to a 10.0-g sample of a metal will cause the temperature of a metal to rise from 25ºC to 35ºC. The specific heat of the metal is closest to0.0005 J/(g·ºC)0.50 J/(g·ºC)2.5 J/(g·ºC)4.2 J/(g·ºC)

Answer 1

b) C = 0.50 J/(g°C)

Step-by-step explanation:

• Q = mCΔT

∴ Q = 50 J

∴ m = 10.0 g

∴ ΔT = 35 - 25 = 10 °C

specific heat (C) :

⇒ C = Q / mΔT

⇒ C = 50 J / (10.0 g)(10 °C)

⇒ C = 0.50 J/(g°C)