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)

Question

Gage Barker

Chemistry

20 June, 08:01

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)

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Brielle Berg · Answer 1

b) C = 0.50 J / (g°C)

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)