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

Step-by-step explanation:

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

q =
`m * C * \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 * C * \Delta T`

=
`175.2 g * 4.184 J/g^(o)C * (8.5 - 20)^(o)C`

=
`175.2 g * 4.184 J/g^(o)C * -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.