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frosja888 [35]
3 years ago
15

The melting point of copper is 1084°C. How does the energy of the particles in a certain amount of liquid copper compare to the

energy of the molecules in the same amount of liquid water? Why?
Chemistry
1 answer:
ipn [44]3 years ago
6 0
The melting point of the solid form of water, which is ice, is 0°C. When we convert both temperatures to kelvin by adding 273 to each we get the melting point of copper as 1357K and that of ice is 273K. Then, dividing the melting point of copper by the melting point of ice, both in absolute temperature scale. The answer would be 4.97. Thus, the energy of molecules of copper is approximately 5 times compared to that of water. 
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3 years ago
A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu
OLEGan [10]

Answer: The final temperature of nickel and water is  25.2^{o}C.

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, (t_{i}) = 23^{o}C,      

  Final temp, (t_{f}) = ?,

  Specific heat of water = 4.184 J/g^{o}C,      

Now, we will calculate the heat energy as follows.

           q = mS \Delta t

              = 55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, t_{i} = 100^{o}C,

   Final temperature, t_{f} = ?

 Specific heat of nickel = 0.444 J/g^{o}C

Hence, we will calculate the heat energy as follows.

          q = mS \Delta t

             = 15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)      

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

              q_{water}(gain) = -q_{alloy}(lost)

55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C) = -(15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C))

       t_{f} = \frac{25.9^{o}C}{1.029}

                 = 25.2^{o}C

Thus, we can conclude that the final temperature of nickel and water is  25.2^{o}C.

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Answer:

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