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30 October, 22:23

A sample of water with a mass of 123.00 kg is heated from 25 C to 97 C. If the specific heat of water is 1 J-1 kg K-1 then how much energy is required?

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  1. 30 October, 22:25
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    8,860 J = 8.86 kJ

    Explanation:

    The heat energy required to heat a substance may be calculated from the formula:

    Q = m * C * ΔT

    Where:

    m = mass of substance = 123.00 kg of water C = specific heat = 1 J⁻¹ kg K⁻¹ ΔT increase of temperature = 97°C - 25°C = 72°C = 72 K.

    Then, substitute the numbers in the formula to get:

    Q = 123.00 kg * 1 J⁻¹ kg K⁻¹ * 72 K = 8,856 J ≈ 8,860 J or 8.86 kJ
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