Answer:
807 cal
Step-by-step explanation:
To determine the energy released as the mercury vapor cools from 356°C to 25°C, we need to determine the change in its internal energy. The internal energy of a substance is given by the equation ΔU = q + w, where ΔU is the change in internal energy, q is the heat absorbed or released by the substance, and w is the work done by the substance.
In this case, the heat absorbed by the mercury as it is heated from 25°C to 356°C is 110 cal, and the heat absorbed to vaporize the mercury is 697 cal, so the total heat absorbed is 110 + 697 = 807 cal. As the mercury vapor cools from 356°C to 25°C, it releases heat, so the heat released can be calculated as q = -807 cal.
Since no work is done in this process (the substance is not expanding or contracting against an external pressure), the work term can be ignored. Therefore, the change in internal energy of the mercury vapor as it cools from 356°C to 25°C is ΔU = q = -807 cal.
So, the mercury releases 807 cal of energy as it cools from 356°C to 25°C.
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