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romanna [79]
3 years ago
7

Barbara is conducting an experiment to observe heat flow. She places one piece of metal in a freezer at -18 degrees Celsius (°C)

and a second piece of metal in an oven at 200 °C. After 2 hours, she takes the metal pieces out of the freezer and oven and places them both in the same beaker of boiling water for two minutes. Which of the following best describes how some of the heat will initially move in this system?
Heat will flow from the frozen metal into the hot metal.


Heat will flow from the boiling water into the hot metal.


Heat will flow from the boiling water into the frozen metal.


Heat will flow from the frozen metal into the boiling water.
Chemistry
1 answer:
yulyashka [42]3 years ago
4 0

<u>Answer:</u> The correct answer is heat will flow from the boiling water into the frozen metal.

<u>Explanation:</u>

According to the law of conservation of energy, energy cannot be destroyed nor created but it can be transformed from 1 form to another form.

There are 3 processes of heat transfer:

  1. <u>Conduction:</u> This type of heat transfer occurs when there is direct contact between the two objects.
  2. <u>Convection:</u> This type of heat transfer occurs when there is a movement of fluid (liquid or gas) due to the movement of hot layers to the top and cold layers to the bottom which leads to convection currents.  
  3. <u>Radiation:</u> This type of heat transfer occurs when there is a direct transfer of energy through space.

The heat moves from a hot surface to a cold surface to maintain equilibrium.

We are given:

A metal is placed in a freezer (cold object) and another metal is placed in an oven (hot object) and then both the metals are placed in boiling water.

Initially, the heat will flow from the boiling water (hot object) into the frozen metal (cold object) to maintain equilibrium

Hence, the correct answer is heat will flow from the boiling water into the frozen metal.

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Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6
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Explanation:

The given data is as follows.

 Weight of solute = 75.8 g,   Molecular weight of solute (toulene) = 92.13 g/mol,    volume = 200 ml

  • Therefore, molarity of toulene is calculated as follows.

      Molarity = \frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}

                    = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}

                    = 4.11 M

Hence, molarity of toulene is 4.11 M.

  • As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

   Molality = \frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}

             = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}

             = 8.6 m

Hence, molality of given toulene solution is 8.6 m.

  • Now, calculate the number of moles of toulene as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{75.8 g}{92.13 g/mol}

                             = 0.8227 mol

Now, no. of moles of benzene will be as follows.

     No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{95.6 g}{78.11 g/mol}

                             = 1.2239 mol

Hence, the mole fraction of toulene is as follows.

         Mole fraction = \frac{\text{moles of toulene}}{\text{total moles}}

                             = \frac{0.8227 mol}{(0.8227 + 1.2239) mol}

                             = 0.402

Hence, mole fraction of toulene is 0.402.

  • As density of given solution is 0.857 g/cm^{3} so, we will calculate the mass of solution as follows.

         Density = \frac{mass}{volume}

     0.857 g/cm^{3} = \frac{mass}{200 ml}      (As 1 cm^{3} = 1 g)

                      mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

      Mass % = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

                   = \frac{75.8 g}{171.4 g} \times 100

                   = 44.22%

Therefore, mass percent of toulene is 44.22%.

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molar~mass=\frac{9.33~grams}{0.126~mol}

molar~mass=73.88\frac{grams}{mol}

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