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. Assuming that all the heat lost by nickel is absorbed by the water, calculate the final temperature of the nickel and water. (C

Respuesta :

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

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, [tex](t_{i}) = 23^{o}C[/tex],      

  Final temp, [tex](t_{f})[/tex] = ?,

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

Now, we will calculate the heat energy as follows.

           q = [tex]mS \Delta t[/tex]

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

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, [tex]t_{i} = 100^{o}C[/tex],

   Final temperature, [tex]t_{f}[/tex] = ?

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

Hence, we will calculate the heat energy as follows.

          q = [tex]mS \Delta t[/tex]

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

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

              [tex]q_{water}(gain) = -q_{alloy}(lost)[/tex]

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

       [tex]t_{f} = \frac{25.9^{o}C}{1.029}[/tex]

                 = [tex]25.2^{o}C[/tex]

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

The final temperature of water and nickel has been 25.2 degrees Celsius.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat (c) has been expressed as:

[tex]q=mc\Delta T[/tex]

Where, the heat required by the object has been q.

The mass of the sample has been m

The change in temperature has been, [tex]\Delta T=\rm Final\;Temperature-Initial\;Temperature[/tex]

Computation for the final temperature of Nickel

The heat lost by nickel has been given as:

[tex]q_{lost}=150\;\times\;0.444\;\times\;(\rm Final\;temperature\;-\;100)\\[/tex]

The heat gained by the water sample has been given as:

[tex]q_{gain}=55\;\times\;4.184\;\times\;(\rm Final \;temperature-23)[/tex]

Since, heat gain has been equivalent to heat lost. Substituting the values as:

[tex]q_{gain}=q_{lost}\\55\;\times\;4.184\;\times\;(\rm Final\;temperature-23)=15\;\times\;0.444\;\times\;(Final\;temperature-100)[/tex]

Simplifying the equation as:

[tex]230.12\;(\rm Final\;temperature-23)=66.6\;(Final\;tempertaure-100)\\Final temperature=25.2\;^\circ C[/tex]

The final temperature of water and nickel has been 25.2 degrees Celsius.

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