Asample 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).

Explanation:

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

                            q =

where,         q = heat released or absorbed

                    m = mass of substance

                    C = specific heat

              = 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, is , is , and specific heat of water is 4.184 .

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

                  q =

                     =

                      =

                     = -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.

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