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Bess [88]
3 years ago
7

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

J/g oF; water = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium
Chemistry
1 answer:
alisha [4.7K]3 years ago
7 0

Answer: The final temperature of both the weight and the water at thermal equilibrium is 50.26^{o}C.

Explanation:

The given data is as follows.

mass = 7.62 g,           T_{2} = 10.8^{o}C

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

       q = mC \times \Delta T    

          = 7.62 g \times 4.184 J/^{o}C \times (52.3 - T)

Now, heat gained by lead will be calculated as follows.

       q = mC \times \text{Temperature change of lead}  

           = 2.04 \times 0.128 \times (T - 11.0)

According to the given situation,

     Heat lost = Heat gained

7.62 g \times 4.184 J/^{o}C \times (52.3 - T) = 2.04 \times 0.128 \times (T - 11.0)

        T = 50.26^{o}C

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is 50.26^{o}C.

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Basile [38]

Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

              Q = kA \frac{\Delta T}{L}

  where,    A = area

                  L = length

                  k = thermal conductivity = 0.8 W/m

             \Delta T = change in temperature.

Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

                = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}

                = 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

                           = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}

                           = 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0
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Answer:

Option D. 4.02 kJ

Explanation:

A simple calorimetry problem

Q = m . C . ΔT

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C = Specific heat capacity

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Let's replace the data

Q = 125 g . 2.42 J/g∘C . (34.8°C -21.5 °C)

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We must convert the answer to kJ

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COF2 has a trigonal planar structure which are symmetric. The electronegativity of oxygen is slightly different regarding fluor. So as you can see in the image, the electronic density is specially displaced to the fluor atoms, but either to the oxygen atom.

COFH has a trigonal structure but differs from COF2 because there is an hydrogen who is donating it's electronic density, so in this zone the electronic density is less than over oxygen or fluor. That makes bond angles be different between them.

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