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14 January, 01:39

A sample of water with a mass of 587.00 kg is heated with 87 kJ of energy to a temperature of 518.4 K. The specific heat of water is 1 J-1 kg K-1. What is the initial temperature of the water?

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  1. 14 January, 02:07
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    Answer : Initial temperature of the water is 518.3 K

    Explanation:

    The specific heat of water (Cp) can be expressed in 2 ways

    1) Cp = 1 J g⁻¹ K⁻¹ or

    2) Cp = 1 kJ kg⁻¹ K⁻¹

    The amount of heat absorbed by water is given by the following equation.

    Q = m*Cp*ΔT

    Here Q is the amount of energy absorbed in kilojoules

    m is the mass of water in kilograms

    Cp is specific heat of water

    ΔT is the difference is temperature of water.

    We have been given that

    Q = 87 kJ

    Cp = 1 kJ kg⁻¹ K⁻¹

    m = 587.00 kg

    Let us plug in these values in equation for Q

    Q = mxCpxΔT

    87 kJ = 587.00 kg x 1 kJ kg⁻¹ K⁻¹ x ΔT

    87/587.00 = ΔT

    ΔT = 0.15 K

    ΔT is calculated as Tf - Ti where Tf is the final temperature and Ti is the initial temperature.

    We know that Tf = 518.4 K

    And we calculated ΔT as 0.15 K.

    Let us plug these values in the equation

    ΔT = Tf - Ti

    0.15 K = 518.4 K - Ti

    Ti = 518.25 K

    Initial temperature of the water is 518.3 K
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