Answer to Question #56025 in General Chemistry for Brian

Question

A 49.50 g sample of a substance is initially at 20.7 °C. After absorbing 1633 J of heat, the temperature of the substance is 154.8 °C. What is the specific heat (c) of the substance?

Answer 1

c = Q / m * ∆T
∆T = T₂ – T₁
T₂ = t₂ + 273.15 = 154.8 °C+ 273.15 = 427.95 K
T₁ = t₁ + 273.15 = 20.7 °C + 273.15 = 293.85 K
∆T = 427.95 K - 293.85 K = 134.1 K
c = 1633 J / (0.0495 kg * 134.1 K) = 246 J / kg * K