Question

A sample of ammonium nitrate, weighing 25.2 g, was added to a 150.0 mL container of H2O at 20.0 °C, and then is thoroughly dissolved. The final temperature of the mixture is 8.5 °C. Find how much heat (kJ) was lost/gained by the solution (surroundings).

Answer 1

Explanation:

It is known that the relation between heat energy, specific heat and change in temperature is as follows.

                            q = $m \times C \times \Delta T$

where,         q = heat released or absorbed

                    m = mass of substance

                    C = specific heat

             $\Delta T$ = change in temperature

It is given that mass is 25.2 g and it is added to 150 ml container so total mass will be (25.2 g + 150 g) = 175.2 g, $T_{1}$ is $20^{o}C$, $T_{2}$ is $8.5^{o}C$, and specific heat of water is 4.184 $J/g ^{o}C$.

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

                  q = $m \times C \times \Delta T$

                     = $175.2 g \times 4.184 J/g^{o}C \times (8.5 - 20)^{o}C$

                      = $175.2 g \times 4.184 J/g^{o}C \times -11.5^{o}C$

                     = -8430 J

As, 1 J = 0.001 kJ. Hence, -8430 J will also be equal to -8.43 kJ. The negative sign indicates that heat is being lost.

Thus, we can conclude that heat (kJ) was lost by the solution (surroundings) is 8.43 kJ.