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s2008m [1.1K]
3 years ago
13

Why is it a good idea to start with room temperature water in the calorimeter?

Physics
2 answers:
densk [106]3 years ago
8 0

Explanation:

It is a good idea to start with room temperature water in the calorimeter because the room temperature water helps to determine the heating up/cooling down because of the environment as the experiment takes place. Because the calorimeter heat is the same as the heat of the water.

jeka57 [31]3 years ago
6 0

Answer:

it a good idea to have room temperature water in calorimeter because it has a high specific heat it is non-flammable, it can be measured easily, it is affordable and it can retain the heat for longer time.

Explanation:

For several reasons water at room temperature is used in the calorimeter.

  • High Specific Heat

It can store high amount of heat with low rise in temperature. It can store heat for longer time which allows the material in the calorimeter to absorb that heat.

  • High range of Freezing and Boiling point

It has a huge rage between its freezing point and its boiling point and thus we  can measure a large range of temperature without its phase change.

  • Abundance

Water is a pure compound available in abundance with ease.

  • Definition

According to the definition, one calorie is the amount of energy required by one gram of water to raise its temperature by one degree Celsius. Therefore water plays a vital role in calorie measurement.

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A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the
vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ ln[\frac{P_{1} }{P_{2} }]

   = <em>mRT</em>₁ln[\frac{P_{1} }{P_{2} }]      [∵<em>pV</em> = <em>mRT</em>]

   = (5) (0.287) (272.039) ln[\frac{2.0}{1.0}]

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Since the process is isothermal, Internal energy change is zero

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Q = Δ<em>U  </em>+ <em>W</em>

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   = 270.588 kJ

4 0
3 years ago
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