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Darina [25.2K]
3 years ago
15

It has been raining all day and the temperature is expected to freeze tonight. Mrs. Gussman treats her front steps with sodium c

hloride (NaCl). This results in a water and salt solution with a freezing point that is lower than pure water. If 850 grams of sodium chloride is added to 6.10 kilograms of water, what will be the freezing point for this solution? The freezing-point depression constant for water is Kf = –1.86°C/m. In solving this problem, consider that sodium breaks into Na+ and Cl– ions, doubling the molar concentration of the salt.
Chemistry
1 answer:
Yuri [45]3 years ago
3 0

Answer:

Di ko po mahanap

Explanation:

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
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Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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