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1 October, 06:36

The specific heat of a certain type of metal is 0.128 J / (g ⋅ ∘ C). What is the final temperature if 305 J of heat is added to 93.4 g of this metal, initially at 20.0 ∘ C?

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  1. 1 October, 06:58
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    Final T° is 45.5°C

    Explanation:

    Formula for calorimetry is:

    Q = m. C. ΔT, where

    ΔT = Final T° - Initial T°

    C = Specific heat

    m = mass

    Let's replace with the data given

    305 J = 93.4 g. 0.128 J/g°C. (Final T° - 20°C)

    305 J / 93.4 g. 0.128 J/g°C = Final T° - 20°C

    25.5°C = Final T° - 20°C → Final T° = 25.5°C + 20°C = 45.5°C
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