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Heat Capacity and Calorimetry

AP Chemistry· difficulty 3/5

-q_metal = q_water; c_water = 4.18 J/(g·°C).

Cu metal 100. g, 100°C c=0.385 J/g·°C → water 200. g 22.0°C final T = ?

Approximately what is the final temperature of the system?

  • A

    ~25.4°C

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  • B

    ~75°C

  • C

    ~22°C

  • D

    ~61°C

Explanation

100(0.385)(100-T) = 200(4.18)(T-22). Solving: 38.5(100-T)=836(T-22) → 3850-38.5T=836T-18392 → 22242 = 874.5T → T ≈ 25.4°C.

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