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11 December, 20:45

Calculate the heat needed to change 3.00 * 10^2 grams of water at 25 degrees celsius to steam at 100 degrees celsius

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  1. 11 December, 22:03
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    9.4*10⁴J

    Explanation:

    The specific heat capacity (SHC) of water, which is the amount of energy (J) that it takes to warm 1mL (1mL = 1g) of water by 1°C is 4.184 J/g*°C.

    Q = mcΔT relates energy to mass, SHC, and change in temperature.

    The mass of water is 300g, and will not change because no water will evaporate until it reaches 100°C.

    The change in temperature is 75°C:

    100°C - 25°C = 75°C

    (Desired temp. - initial temp. = change in temp.)

    Subbing these values into Q=mcΔT:

    Q = 300g * 4.184J/g*°C * 75°C

    Q = 94140J

    Scientific notation tells us that we have two significant figures, so:

    Q = 9.4 * 10⁴J
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