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23 July, 20:30

An investigation has been completed, similar to the one on latent heat of fusion, where steam is bubbled through a container of water. Steam condenses and the lost energy heats the water and container. Given the following data, do the calculations requested.

Mass of the aluminum container 50 g

Mass of the container and water 250 g

Mass of the water 200 g

Initial temperature of the container and water 20°C

Temperature of the steam 100°C

Final temperature of the container, water, and condensed steam 50°C

Mass of the container, water, and condensed steam 261g

Mass of the steam 11g

Specific heat of aluminum 0.22 cal/g°C

Heat energy gained by the container

a. 770 cal

b. 550 cal

c. 330 cal

d. 220 cal

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  1. 23 July, 23:24
    0
    c. 330 cal

    Explanation:

    Given the information from the question. We need to calculate heat energy gained by the container

    The formula to be used is Q=M*C * (Change in T). Where Q = heat energy gained by the container, M = mass of aluminum, Change in T = Change in temperature and T = Specific heat of aluminum.

    Now we have Q=M*C * (Change in T) = 50g*0.22 * (50-20) = 330 cal

    The correct answer is option c. 330 cal.
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