Question

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

Answer 1

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  $25.2^{o}C$.