Question

If a piece of aluminum with a mass of 3.99 g and a temperature of 100.0 °C is dropped
in 10.0 cm3 of water at 21.0 °C, what will be the final temperature of the system?
(Hint: Remember the density of water is 1.00 g / mL)

Answer 1

Answer:

The final temperature of the system is 27.3°C.

Explanation:

Heat lost by aluminum = 3.99 × 0.91 × (100-T)

                                     = 3.631 (100-T)

Heat gained by water = 10 × 4.184 × (T-21)

                                    = 41.84 (T-21)

As,

                                Heat gained = Heat loss

                          or, 3.631(100-T) = 41.84(T-21)

                          or,363.1 -  3.631 T = 41.84 T - 878.64)

                          or, (41.84+ 3.631) T = 878.64 +363.1

                          or  T= $\frac{1241.74}{45.47}$

                         or, T = 27.3°C

Hence the final temperature is 27.3°C.

Answer 2

The final temperature of the system is 27.3°C this can be calculated by specifying the heat gained or lost.

The calculation for temperature:

Heat lost by aluminum = 3.99 × 0.91 × (100-T)

= 3.631 (100-T)

Heat gained by water = 10 × 4.184 × (T-21)

= 41.84 (T-21)

As,

Heat gained = Heat loss

3.631(100-T) = 41.84(T-21)

363.1 -  3.631 T = 41.84 T - 878.64)

(41.84+ 3.631) T = 878.64 +363.1

T = 27.3°C

Hence the final temperature is 27.3°C.

Find more information about heat gained here:

brainly.com/question/21704399