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14 April, 23:45

Given that the specific heat of water is 4.18 kJ / (kg·°C), how much heat does it take to raise the temperature of 3.5 kg of water from 25°C to 55°C?

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  1. 15 April, 01:20
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    The heat required = 438900 J

    Explanation:

    mass = 3.5 Kg

    change in temperature=ΔT=55-25=30°C

    specific heat of iron = 4.18 kJ / (kg·°C) = 4180 J / (kg·°C)

    the formula for heat is given by

    Q = m C ΔT

    Q=3.5 (4180) (30)

    Q=438900 J
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