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1 March, 13:38

If the quantity of heat lost or gained is (ΔH) = mass * change in T * specific heat, what is the amount of heat required to raise 200.0 g of water from 22.00°C to 100.0°C? The specific heat of water is 4.180 Joules per gram-°C.

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  1. 1 March, 17:25
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    1) formula:

    ΔH = m * Cs * ΔT

    2) data

    m = 200.0 g

    ΔT = 100.0 °C - 22.00°C

    Cs = 4.180 J / (g * °C)

    3) Calculations:

    ΔH = 200 g * (100.0°C - 22.00°C) * 4.180 J / (g * °C) = 65,208 J

    Answer: 65,208 J
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