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Heat Capacity and Calorimetry

AP Chemistry· difficulty 3/5

A 25.0 g unknown metal at 100.0 °C is placed into 50.0 g of water at 25.0 °C in a calorimeter. Final T = 27.5 °C. (c_water = 4.18 J/g·°C)

Mass metal: 25.0 g, Tᵢ=100.0 °C Mass water: 50.0 g, Tᵢ=25.0 °C T_f = 27.5 °C c_water = 4.18 J/g·°C Find c_metal

What is the specific heat of the metal?

  • A

    0.180 J/g·°C

  • B

    0.520 J/g·°C

  • C

    0.288 J/g·°C

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  • D

    0.450 J/g·°C

Explanation

q_water = 50.0·4.18·2.5 = 522.5 J. q_metal = -522.5 J = 25.0·c·(27.5−100.0) = 25.0·c·(−72.5). c = 522.5/(25.0·72.5) = 0.288 J/g·°C.

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