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Ber [7]
3 years ago
8

Plzz tell me the answer u need help im kinda confused

Chemistry
1 answer:
djverab [1.8K]3 years ago
8 0
Hi
I think it’s B
Hopefully this helps!!
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A gardener waters the plants in a garden using a garden hose during a drought. The gardener observes that most of the water give
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1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

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Freezing point =0 °C = (0 + 273 K)= 273 K

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Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

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Entropy change of universe = entropy change of the system+ entropy change of the surrounding

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Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
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