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12 December, 02:39

When 300. cal of energy is lost from a 125 g object, the temperature decreases from 45.0°C to 40.0°C. What is the specific heat of this object in Cal/g 'C?

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  1. 12 December, 05:08
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    The answer to your question is C = 0.48 cal/g°C

    Explanation:

    Data

    Heat = Q = 300 cal

    mass = m = 125 g

    temperature 1 = T1 = 45°C

    temperature 2 = T2 = 40°C

    Specific heat = C = ? cal/g°C

    Formula

    H = mC (T2 - T1)

    -Solve for Specific heat (C)

    C = H / m (T2 - T1)

    -Substitution

    C = 300 / 125 (40 - 45)

    -Simplification

    C = - 300 / 125 (-5)

    C = - 300 / - 625

    -Result

    C = 0.48 cal/g°C
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