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4 September, 01:06

1 kg of water (specific heat = 4184 J / (kg K)) is heated from freezing (0°C) to boiling (100°C). What is the change in thermal energy?

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  1. 4 September, 04:13
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    Answer: 1560632 joules

    Explanation:

    The change in thermal energy (Q) required to heat ice depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

    Thus, Q = MCΦ

    Given that:

    Q = ?

    Mass of frozen water (ice) = 1kg

    C = 4184 J / (kg K)

    Φ = (Final temperature - Initial temperature)

    = 100°C - 0°C = 100°C

    Convert 100°C to Kelvin

    (100°C + 273) = 373K

    Then, Q = MCΦ

    Q = 1kg x 4184 J / (kg K) x 373K

    Q = 1560632 joules

    Thus, the change in thermal energy is 1560632 joules
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