Question

Some water is placed in a coffee cup calorimeter. When 1.0 g of an ionic solid is added, the temperature of the solution increases from 21.58C to 24.28C as the solid dissolves. For the dissolving process, what are the signs?

Answer 1

Answer: The sign of $\Delta S_{sys}$ is positive, the sign of $\Delta S_{surr}$ is negative and the value of $\Delta S_{universe}$ is 0

Explanation:

We are given:

Mass of ionic solid = 1.0 g

Initial temperature of solution = 21.58°C

Final temperature of solution = 24.28°C

To calculate the entropy change of the reaction (system), we use the equation:

$\Delta S_{sys}=\frac{q}{\Delta T}$

where,

$\Delta S_{sys}$ = entropy change of the system

q = heat released or absorbed = $mc\Delta T$

$\Delta T$ = temperature change = $(24.28-21.58)^oC=2.7^oC$

Putting values in above equation:

$\Delta S_{sys}=\frac{1\times c\times 2.7^oC}{\Delta T}$

As, the sign of heat is coming out to be positive, so the sign of entropy change of the system will also be positive.

In the above process, the system is absorbing heat which is released by the surrounding but with the same magnitude. So, the sign of entropy change of the surrounding is negative.

We know that:

$\Delta S_{universe}=\Delta S_{surr}+\Delta S_{sys}$

The value of entropy change of system and surrounding is same but with opposite sign. So, the value of entropy change of universe will be 0.

Hence, the sign of $\Delta S_{sys}$ is positive, the sign of $\Delta S_{surr}$ is negative and the value of $\Delta S_{universe}$ is 0