Question

A 0.200-kg block of a pure material is heated from 20.0°C to 65.0°C by the addition of 3.08 kJ of energy. Calculate its specific heat.

Answer 1

342.22 J/kg°C

Step-by-step explanation:

m = 0.2 kg

T₀ = 20°C

T = 65°C

Q = 3.08 kJ = 3080 J

Q = mc(T-T₀)

3080 = 0.2 × c × (65-20)

c = 3080 ÷ (0.2 × 45)

c = 342.22 J/kg°C