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3 years ago

PLEASE HELP!!!! Two science questions!!!! 20 points and brainliest!!!

Chemistry

1 answer:

Angelina_Jolie [31]3 years ago

8 0

1. A cylinder is submerged in water as illustrated in the diagram. Which of the following statements is true?

The pressure at D is greater than at B. This is because D is trying to get back to the surface with an amount of pressure. B has less amount of pressure.

2. A cylinder is submerged in water as illustrated in the diagram. If the area of the top and the bottom of the cylinder is the same, which of the following statements is true?

buoyant force = (force D - force A).

Edit: If you ever notice. When you fill a sink full of water. Put a glass upside down and try forcefully to submerge it under the water it will resist. This is because of the pressure and air. The glass is less dense then water. (Also depends on the glass)

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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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