Billy and Susie heat a 78.9 g sample of Unobtanium at 18.0C. Using a lab burner, they burn propane and the unobtanium absorbs 13,240 J of heat untill the temperature reaches the melting point 109C. what is the specific heat of unobtanium

1.8 J/g·°C (2 sig.figs. per ΔT)

Explanation:

Given:

mass (m) = 78.9 grams

specific heat (c) = ?

Temp. Change (ΔT) = 109°C - 18°C = 91°C

Heat flow (q) = 13,240 Joules

q = m·c·ΔT => c = q/m·ΔT

∴c = 13,240J / 78.9g·91°C = 1.844 J/g·°C ≅ 1.8 J/g·°C (2 sig.figs. per ΔT)

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