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12 October, 20:51

A sample of water drops from an initial temperature of 67.5 C to a final temperature of 25.2 C. If - 4425 J of heat were released from this sample of water, what is the mass of the sample?

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  1. 12 October, 21:29
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    mass = 25.03 g

    Explanation:

    Given dа ta:

    Initial temperature = 67.5°C

    Final temperature = 25.2°C

    Heat released = - 4425 j

    Mass of water = ?

    Solution:

    Specific heat capacity:

    It is the amount of heat required to raise the temperature of one gram of substance by one degree.

    Formula:

    Q = m. c. ΔT

    Q = amount of heat absorbed or released

    m = mass of given substance

    c = specific heat capacity of substance

    ΔT = change in temperature

    ΔT = 25.2°C - 67.5°C

    ΔT = - 42.3°C

    Specific heat of water = 4.18 j/g.°C

    Q = m. c. ΔT

    -4425 j = m * 4.18 j/g.°C * - 42.3°C

    -4425 j = m * - 176.81 j/g

    m = - 4425 j / - 176.81 j/g

    m = 25.03 g
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