Question #56025

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?

Expert's answer

c = Q / m * ∆T

∆T = T_{2} – T_{1}

T_{2} = t_{2} + 273.15 = 154.8 °C+ 273.15 = 427.95 K

T_{1} = t_{1} + 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

∆T = T

T

T

∆T = 427.95 K - 293.85 K = 134.1 K

c = 1633 J / (0.0495 kg * 134.1 K) = 246 J / kg * K

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