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A Styrofoam cup holding 125 g of hot water at 1.00×102∘C cools to room temperature, 20.0 ∘. What is the change in entropy of the room? (Neglect the specific heat of the cup and any change in temperature of the room.)

Sagot :

The change in entropy of a system will be 6.25 kJ/K. For further explanation you can read below.

The change in entropy of a system is given by the equation ΔS = Q/T, where Q is the heat transferred and T is the temperature at which the transfer occurs. In this case, the temperature of the hot water is initially 1.00×102 ∘C and it cools to room temperature, which is 20.0 ∘C. Note that entropy is a measure of the disorder or randomness of a system, and that this calculation assumes that the heat from the hot water is completely transferred to the room, increasing its disorder. It also assumes that the specific heat of the cup and any change in temperature of the room are negligible, which is generally a good approximation for a small cup of hot water in a large room.

We can therefore calculate the change in entropy of the room as follows:

ΔS = Q/T = (125 g * 1.00×102 ∘C - 125 g * 20.0 ∘C) / (20.0 ∘C)

      = 6.25 kJ/K

Learn more about entropy here, https://brainly.com/question/13146879

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