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550 g of water at 105°C is poured into an 855 g aluminum container with an initial temperature of 11°C. The specific heat of aluminum is 900 J / (kg K). How much heat flows from the water to the aluminum, assuming no heat is exchanged with the surroundings?

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  1. Today, 11:03
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    54.22 kJ

    Explanation:

    In this case, we need to calculate the heat. The expression to use is:

    Q = m * C * ΔT

    Now, with the specific heat of water (4186 J / kg K), we can calculate the temperature in which this occurs.

    So:

    Q = 0.550 * 4186 * (105 - T)

    Q = 2302.3 (105 - T)

    Q = 241,741.5 - 2302.3T

    Now with the Aluminium:

    Q = 0.855 * 900 * (T - 11)

    Q = 769.5T - 8464.5

    Now, with both equations, we solve for the final temperature:

    769.5T - 8464.5 = 241,741.5 - 2302.3T

    (2302.3 + 769.5) T = 241,741.5 + 8464.5

    3071.8T = 250,206

    T = 81.45 K

    This is the temperature which the change occurs. Now, let's determine the amount of heat from water to Al:

    Q = 241,741.5 - 2302.3 (81.45)

    Q = 54,219.17 J or simply 54.22 kJ.
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